Restalls Logic.textbook Exercise Solutions

Greg Restall’s Logic (McGill-Queen’s UP)
Exercise Solutions

Chapter 1: Propositions and Arguments
Basic
Question {1.1}
1. Proposition
2. Question
3. Proposition (though of course, it’s false)
4. Question
5. Exclamation
6. Wish
7. Command or request
8. Proposition (which states that I have a wish, so you might say it expresses a wish too. It differs from the
example in part 6 in that it can be true or false, and 6 cannot be.)
9. Proposition (though false, again)
10. Proposition (it can be true or false — though it might also express a wish or a command or something
else too.)
Question {1.2}
All of them are valid except affirming the consequent.
For affirming the consequent, consider the instance
•
•
•

If I’m doing logic, I’m happy.
I’m happy.
Therefore, I’m doing logic.

That’s not valid, as you can have the premises true, while the conclusion is false. For example, suppose I’m
also happy when having dinner with friends in a good cafe (and not doing logic). Then in this situation you

have the premises true but the conclusion false. As a result, this instance of the argument is invalid, and so,
the argument form is invalid.
However, this form has valid instances. Here’s one
•
•
•

If I’m doing logic I’m doing logic.
I’m doing logic.
Therefore, I’m doing logic.

This argument is stupid — the first premise is not needed and the conclusion is a restatement of the second
premise. However, it’s valid according our definition of validity. There’s no way that the premises could be
true and the conclusion false.
Question {1.3}
Form 1 is the form of all arguments with two premises both of which include “and”, and with any
conclusion. Arguments 1 and 2 have this form, but argument 3 does not, as the second premise of argument
3 is not of the form r and s.
All of the arguments have form 2, since they have two premises and one conclusion, and form 2 is the form
of any argument with this general shape.
Only argument 1 has form 3. Argument 2 does not have form 3 as there is no statement q shared by both
premises in this argument. Argument 3 does not have form 3 as the second premise of this argument is not
of the form q and r. Argument 1 has form 3 as we can take p to be “Greg lectures PHIL137”, q to be
“Caroline lectures PHIL137” and r to be “Caroline Lectures PHIL132” (see errata).

Advanced
Question {1.4}
I think that the first four express propositions, and that the fifth does only if you give it a context in which
the non-English words have some kind of meaning. But many disagree with me on this.
Some say that (1) has a “category mistake”. It makes no sense to think of the sum of 2 and the Pacific
Ocean.
Some think that (2) has a “reference failure” as the phrase “The present King of France” doesn’t refer to
anyone or anything, so he cannot be bald (and he cannot fail to be bald). See Chapter 12 for more on this
topic.
Some think that (3) cannot express a proposition, for if it’s true it is false, and if it’s false it’s true.
Some think that (4) cannot express a proposition, for there is nothing to decide whether it’s true or false.
Question {1.5}
Not every invalid argument form has valid instances. I think that the form:
•
•

p or not p
Therefore, q and not q

has no valid instances. Each instance of this argument will have a true premise and a false conclusion, and
therefore, be invalid.
Question {1.6}
This is a very interesting question. To answer it, you need to have a clear understanding of what counts as
the form of an argument. Some arguments are clearly valid, and have no valid form as far as we have seen
so far. The argument
•
•
•

All footballers are bipeds.
Socrates is a footballer.
So, Socrates is a biped.

is pretty obviously valid, but the only form it has — for the level of analysis we’ve looked at so far is:
•
•
•

p
q
So, r

since none of the premises are conditionals, biconditionals, conjunctions, disjunctions or negations.
However, it has some kind of form. (Exactly what, we’ll see in Chapter 8.)

Chapter 2: Connectives and Argument Forms
Solutions
Question {2.1}
1. Greg is in town.
2. Fred is very smart.
3. Minh is a bad student
4. Every car is fuel efficient.
5. That car is either red or a diesel. (This is a tricky one. The answer is not “That car is both red and a
diesel.” Do you understand why? )
Question {2.2}
1. It is -5 degrees. The clouds are grey.
2. He was tired. He wanted to keep going.
3. The waves were breaking. The surf was low.
4. There was a strike. The power was not cut.
5. Fred is not a mechanic. Jack is not a mechanic.
Question {2.3}
1. Eric is there. Yukiko is there.
2. The car is white. The car is yellow.
3. Brian is doing a Ph.D.
Brian is doing a Masters degree.
4. It rains. It doesn’t rain.
5. I’ll have coffee. I’ll have tea.
Question {2.4}
1. It is raining. I’ll walk home.
2. You look outside. (Perhaps better: You will look outside, to match the consequent.) You’ll see the nice
garden I planted.

3. I’m tired. I don’t do my logic well.
4. I do logic well. I’m awake. (Be careful with “only if"s.)
5. I’m awake. I do logic well.
6. You work hard. You will pass.
7. The world’s future will be assured. We get rid of nuclear weapons.
8. We get rid of nuclear weapons. The world’s future will be assured.
9. Oswald didn’t shoot Kennedy. Someone else did shoot Kennedy.
10. Oswald didn’t shoot Kennedy. Someone else did shoot Kennedy. (This is a tricky one. “Oswald hadn’t
shot Kennedy” is not a sentence, so it can’t be an antecedent, given our definition. Most people think that 9
and 10 have the same antecedent and consequent, but the conditional combines them slightly differently. In
the first case we wonder what happens if Oswald isn’t Kennedy’s assassin. In the second, we assume that
Oswald was Kennedy’s assassin and wonder how things would have been had he not shot Kennedy.)
Question {2.5}
1. Conditional of “Christine is happy” and “She is not thinking about her thesis”. The second proposition
here is a negation of “she is thinking about her thesis.”
2. Negation of “I know if she’s coming to the party”, and this is not a conditional. (If it were, what would
be the antecedent and consequent?)
3. Conditional with antecedent of “Theodore is enrolled in PHIL134, and he passes” and consequent “He
can go on to do advanced logic subjects.” The antecedent is a conjunction of “Theodore is enrolled in
PHIL134” and “Theodore passes PHIL134”.
4. Conditional with antecedent “Theodore isn’t enrolled in PHIL134, or he doesn’t pass” (the “if” in there
can be taken out) and consequent “Theodore cannot go on to do advanced logic subjects”. The antecedent
is a disjunction of “Theodore isn’t enrolled in PHIL134” (which is in turn, the negation of “Theodore is
enrolled in PHIL134”) and of “Theodore doesn’t pass PHIL134” (which is in turn the negation of
“Theodore does pass PHIL134”). The consequent, which is, remember, “Theodore cannot go on to do
advanced logic subjects”, is the negation of “Theodore can go on to do advanced logic subjects”. Phew!
5. This is the conjunction of “I believe that you are either going to leave the party early or make a scene”
and “You know that you are either going to leave the party early or make a scene”. These both don’t go any
further. The first is an “I believe that …” statement. This is not a disjunction of “I believe that you are
going to leave the party early” and “I believe that you are going to make a scene”, as I might not believe
either disjunct of the disjunction, while still believing the whole disjunction. Same goes for the other
conjunct about knowledge.
Question {2.6}
1. Yukiko is not a linguist.
2. Yukiko is not a linguist or Pavlos is a logician.

3. Yukiko is not a linguist if and only if Christine is a lawyer.
4. If Yukiko is a linguist, then if Christine is a lawyer, Pavlos is a logician.
5. If Yukiko is a linguist and Pavlos is not a logician then Christine is a lawyer.
6. Yukiko is a linguist and Christine is a lawyer.
7. It’s not the case that both Yukiko is a linguist and Pavlos is a logician.
8. It’s not the case that Yukiko is a linguist if and only if Christine is a lawyer.
9. It’s not the case that Christine is not a lawyer.
10. Yukiko is a linguist if and only if either Christine is a lawyer or Pavlos is not a logician.
11. Yukiko is a linguist or Christine is a lawyer.
12. Yukiko is a linguist if and only if Pavlos is a logician.
13. If Yukiko is a linguist, Christine is a lawyer.
14. If Yukiko is a linguist only if Christine is a lawyer, then Pavlos is a logician.
15. I can’t get a good English translation for this. Can you?
Question {2.7}
1. ~ c
2. y & p
3. p ⊃ y
4. p ∨ ~ y
5. ~(p & c)
6. y ⊃ (p & ~ c)
7. (y ∨ ~ p) ⊃ c
8. y ≡ (p ∨ ~ c)
9. p ∨ (y ⊃ c)
10. (p ∨ y) ⊃ c
Question {2.8}

1, 4, 6, 9 and 10 are well-formed, the rest are not.

Chapter 3: Truth Tables
Basic
Question {3.1}
The asterisk is under the column of the main operator.
1:
y

~ y

0

1 0

1

0 1
*

2 and 12:
y p

~ y ∨ p

y ≡ p

0 0

1 0 1 0

0 1 0

0 1

1 0 1 1

0 0 1

1 0

0 1 0 0

1 0 0

1 1

0 1 1 1

1 1 1

*

*

3 and 8:
y c

~ y ≡ c

~ (y ≡ c)

0 0

1 0 0 0

0 0 1 0

0 1

1 0 1 1

1 0 0 1

1 0

0 1 1 0

1 1 0 0

1 1

0 1 0 1

0 1 1 1

*

*

4 and 14:
y c p

y ⊃ (c ⊃ p)

(y ⊃ c) ⊃ p

0 0 0

0 1 0 1 0

0 1 0 0 0

0 0 1

0 1 0 1 1

0 1 0 1 1

0 1 0

0 1 1 0 0

0 1 1 0 0

0 1 1

0 1 1 1 1

0 1 1 1 1

1 0 0

1 1 0 1 0

1 0 0 1 0

1 0 1

1 1 0 1 1

1 0 0 1 1

1 1 0

1 0 1 0 0

1 1 1 0 0

1 1 1

1 1 1 1 1

1 1 1 1 1

*

*

5 and 10:
y c p

(y & ~ p) ⊃ c

y ≡ (c ∨ ~ p)

0 0 0

0 0 1 0 1 0

0 0 0 1 1 0

0 0 1

0 0 0 1 1 0

0 1 0 0 0 1

0 1 0

0 0 1 0 1 1

0 0 1 1 1 0

0 1 1

0 0 0 1 1 1

0 0 1 1 0 1

1 0 0

1 1 1 0 0 0

1 1 0 1 1 0

1 0 1

1 0 0 1 1 0

1 0 0 0 0 1

1 1 0

1 1 1 0 1 1

1 1 1 1 1 0

1 1 1

1 0 0 1 1 1

1 1 1 1 0 1

*

*

6, 11 and 13:
y c

y & c

y ∨ c

y ⊃ c

0 0

0 0 0

0 0 0

0 1 0

0 1

0 0 1

0 1 1

0 1 1

1 0

1 0 0

1 1 0

1 0 0

1 1

1 1 1

1 1 1

1 1 1

*

*

*

7:
y p

~ (y & p)

0 0

1 0 0 0

0 1

1 0 0 1

1 0

1 1 0 0

1 1

0 1 1 1
*

9:
c

~ ~ c

0

0 1 0

1

1 0 1
*

15:
y c

(y ⊃ c) ≡ (c ⊃ y)

0 0

0 1 0 1 0 1 0

0 1

0 1 1 0 1 0 0

1 0

1 0 0 0 0 1 1

1 1

1 1 1 1 1 1 1
*

Question {3.2}
They are all tautologies except for the fourth one.
((p & q) ⊃ r) ⊃ (p ⊃ r)
1 0 0 1 0 0 1 0 0
Let p = 1, q = 0 and r = 0. That makes the formula false.
Question {3.3}
1:
p

~ ~ p

0

0 1 0

1

1 0 1

2:
p q

p & q

q & p

0 0

0 0 0

0 0 0

0 1

0 0 1

1 0 0

1 0

1 0 0

0 0 1

1 1

1 1 1

1 1 1

p q

p ⊃ q

~ p ∨ q

0 0

0 1 0

1 0 1 0

0 1

0 1 1

1 0 1 1

1 0

1 0 0

0 1 0 0

1 1

1 1 1

0 1 1 1

3:

4:

p q r

p ⊃ (q ⊃ r)

(p ⊃ q) ⊃ r

~ p ∨ ~ (q & ~ r)

0 0 0

0 1 0 1 0

0 1 0 0 0

1 0 1 1 0 0 1 0

0 0 1

0 1 0 1 1

0 1 0 1 1

1 0 1 1 0 0 0 1

0 1 0

0 1 1 0 0

0 1 1 0 0

1 0 1 0 1 1 1 0

0 1 1

0 1 1 1 1

0 1 1 1 1

1 0 1 1 1 0 0 1

1 0 0

1 1 0 1 0

1 0 0 1 0

0 1 1 1 0 0 1 0

1 0 1

1 1 0 1 1

1 0 0 1 1

0 1 1 1 0 0 0 1

1 1 0

1 0 1 0 0

1 1 1 0 0

0 1 0 0 1 1 1 0

1 1 1

1 1 1 1 1

1 1 1 1 1

0 1 1 1 1 0 0 1

The first and third propositions are equivalent to each other, and both are not equivalent to the second.
5. p & ~ p is equivalent to ~(q ⊃ q), as both are contradictions. r ∨ ~ r and s ⊃ s are equivalent as both are
tautologies.
Question {3.4}
1:
p q

p

p ⊃ q

q

0 02

0

0 1 0

0

0 1

0

0 1 1

1

1 0

1

1 0 0

0

1 1

1

1 1 1

1

The argument form is valid, as there is no row where the premises are true & the conclusion false.
2:
p q

p

q

p ≡ q

0 0

0

0

0 1 0

0 1

0

1

0 0 1

1 0

1

0

1 0 0

1 1

1

1

1 1 1

The argument form is valid as there is no row where the premises are both true & the conclusion false.
3:
p q

p & q

p ≡ q

0 0

0 0 0

0 1 0

0 1

0 0 1

0 0 1

1 0

1 0 0

1 0 0

1 1

1 1 1

1 1 1

The argument form is valid as there is no row where the premise is true & the conclusion false.
4:
p q

p

q ⊃ p

q

0 0

0

0 1 0

0

0 1

0

1 0 0

1

1 0

1

0 1 1

0 *

1 1

1

1 1 1

1

The argument form is invalid as there is a row where the premises are true & the conclusion false, marked
with the star.
5:
p q

p

q

p & q

0 0

0

0

0 0 0

0 1

0

1

0 0 1

1 0

1

0

1 0 0

1 1

1

1

1 1 1

The argument is valid as there is no row where the premises are both true & the conclusion false.
6:
p q

p

p ∨ q

0 0

0

0 0 0

0 1

0

0 1 1

1 0

1

1 1 0

1 1

1

1 1 1

The argument is valid as there is no row where the premise is true & the conclusion false.
7:
p q

p ≡ q

p ≡ ~ q

~ p

0 0

0 1 0

0 0 1 0

1 0

0 1

0 0 1

0 1 0 1

1 0

1 0

1 0 0

1 1 1 0

0 1

1 1

1 1 1

1 0 0 1

0 1

The argument is valid as there is no row where the premises are true & the conclusion false. (There is no
row where premises are true.)
8:
p q r

p ⊃ (q ⊃ r)

q ⊃ (p ⊃ r)

0 0 0

0 1 0 1 0

0 1 0 1 0

0 0 1

0 1 0 1 1

0 1 0 1 1

0 1 0

0 1 1 0 0

1 1 0 1 0

0 1 1

0 1 1 1 1

1 1 0 1 1

1 0 0

1 1 0 1 0

0 1 1 0 0

1 0 1

1 1 0 1 1

0 1 1 1 1

1 1 0

1 0 1 0 0

1 0 1 0 0

1 1 1

1 1 1 1 1

1 1 1 1 1

The argument is valid – there is no row where the premise is true and the conclusion false. They are true in
exactly the same rows, so they are equivalent.
9:
p

p ⊃ ~ p

~ p

0

0 1 1 0

1 0

1

1 0 0 1

0 1

The argument is valid, as there’s no row where the premise is true and the conclusion false. (Again, they’re
equivalent.)
10:
p

~ ~ p

p

0

0 1 0

0

1

1 0 1

1

The argument is valid. ~~p and p are equivalent, what’s more.
11:
p q r

p ⊃ q

(r ⊃ p) ⊃ (r ⊃ q)

0 0 0

0 1 0

0 1 0 1 0 1 0

0 0 1

0 1 0

1 0 0 1 1 0 0

0 1 0

0 1 1

0 1 0 1 0 1 1

0 1 1

0 1 1

1 0 0 1 1 1 1

1 0 0

1 0 0

0 1 1 1 0 1 0

1 0 1

1 0 0

1 1 1 0 1 0 0

1 1 0

1 1 1

0 1 1 1 0 1 1

1 1 1

1 1 1

1 1 1 1 1 1 1

The argument is valid – there is no row where the premise is true and the conclusion false. They are not
true in exactly the same rows, so they are not equivalent.
12:
p q

p ⊃ q

~ q ⊃ ~ p

0 0

0 1 0

1 0 1 1 0

0 1

0 1 1

0 1 1 1 0

1 0

1 0 0

1 0 0 0 1

1 1

1 1 1

0 1 1 0 1

The argument is valid. There’s no row with the premise true & conclusion false.
13:
p q

p

~ p ⊃ q

0 0

0

1 0 0 0

0 1

0

1 0 1 1

1 0

1

0 1 1 0

1 1

1

0 1 1 1

The argument is valid. There’s no row with the premise true & conclusion false.
14:
p q

p ⊃ (p ⊃ q)

p ⊃ q

0 0

0 1 0 1 0

0 1 0

0 1

0 1 0 1 1

0 1 1

1 0

1 0 1 0 0

1 0 0

1 1

1 1 1 1 1

1 1 1

The argument is valid. There’s no row with the premise true & conclusion false.
15:
p q

p

q ⊃ q

0 0

0

0 1 0

0 1

0

1 1 1

1 0

1

0 1 0

1 1

1

1 1 1

The argument is valid. There’s no row with the premise true & conclusion. In fact, the conclusion is a
tautology, and any argument like this is valid.
16:
p q r

p ∨ q

~ q ∨ r

p ∨ r

0 0 0

0 0 0

1 0 1 0

0 0 0

0 0 1

0 0 0

1 0 1 1

0 1 1

0 1 0

0 1 1

0 1 0 0

0 0 0

0 1 1

0 1 1

0 1 1 1

0 1 1

1 0 0

1 1 0

1 0 1 0

1 1 0

1 0 1

1 1 0

1 0 1 1

1 1 1

1 1 0

1 1 1

0 1 0 0

1 1 0

1 1 1

1 1 1

0 1 1 1

1 1 1

The argument is valid – there is no row where the premise is true and the conclusion false.
17:
p q

p ⊃ q

q ⊃ p

0 0

0 1 0

0 1 0

0 1

0 1 1

1 0 0 *

1 0

1 0 0

0 1 1

1 1

1 1 1

1 1 1

The argument is invalid. There’s a row with the premise true & conclusion false (with the star).
18:
p q r

p ⊃ (q ⊃ r)

(p ⊃ q) ⊃ r

0 0 0

0 1 0 1 0

0 1 0 0 0

0 0 1

0 1 0 1 1

0 1 0 1 1

0 1 0

0 1 1 0 0

0 1 1 0 0 *

0 1 1

0 1 1 1 1

0 1 1 1 1

1 0 0

1 1 0 1 0

1 0 0 1 0

1 0 1

1 1 0 1 1

1 0 0 1 1

1 1 0

1 0 1 0 0

1 1 1 0 0

1 1 1

1 1 1 1 1

1 1 1 1 1

The argument is invalid – there is a row where the premise is true and the conclusion false. (See the
asterisk.)
19:

p q r

(p & q) ⊃ r

p ⊃ (~ q ∨ r)

0 0 0

0 0 0 1 0

0 1 1 0 1 0

0 0 1

0 0 0 1 1

0 1 1 0 1 1

0 1 0

0 0 1 1 0

0 1 0 1 0 0

0 1 1

0 0 1 1 1

0 1 0 1 1 1

1 0 0

1 0 0 1 0

1 1 1 0 1 0

1 0 1

1 0 0 1 1

1 1 1 0 1 1

1 1 0

1 1 1 0 0

1 0 0 1 0 0

1 1 1

1 1 1 1 1

1 1 0 1 1 1

The argument is valid – there is no row where the premise is true and the conclusion false.
20:
p q r s

p ⊃ q

r ⊃ s

(q ⊃ r) ⊃ (p ⊃ s)

0 0 0 0

0 1 0

0 1 0

0 1 0 1 0 1 0

0 0 0 1

0 1 0

0 1 1

0 1 0 1 0 1 1

0 0 1 0

0 1 0

1 0 0

0 1 1 1 0 1 0

0 0 1 1

0 1 0

1 1 1

0 1 1 1 0 1 1

0 1 0 0

0 1 1

0 1 0

1 0 0 1 0 1 0

0 1 0 1

0 1 1

0 1 1

1 0 0 1 0 1 1

0 1 1 0

0 1 1

1 0 0

1 1 1 1 0 1 0

0 1 1 1

0 1 1

1 1 1

1 1 1 1 0 1 1

1 0 0 0

1 0 0

0 1 0

0 1 0 0 1 0 0

1 0 0 1

1 0 0

0 1 1

0 1 0 1 1 1 1

1 0 1 0

1 0 0

1 0 0

0 1 1 0 1 0 0

1 0 1 1

1 0 0

1 1 1

0 1 1 1 1 1 1

1 1 0 0

1 1 1

0 1 0

1 0 0 1 1 0 0

1 1 0 1

1 1 1

0 1 1

1 0 0 1 1 1 1

1 1 1 0

1 1 1

1 0 0

1 1 1 0 1 0 0

1 1 1 1

1 1 1

1 1 1

1 1 1 1 1 1 1

The argument form is valid. There’s no row where the premises are both true and the conclusion is false.
Question {3.5}
The argument form goes like this: (j & ~ b) ⊃ ~ j therefore j ⊃ b
This is a valid argument form:
j b

(j & ~ b) ⊃ ~ j

j ⊃ b

0 0

0 0 1 0 1 1 0

0 1 0

0 1

0 0 0 1 1 1 0

0 1 1

1 0

1 1 1 0 0 0 1

1 0 0

1 1

1 0 0 1 1 0 1

1 1 1

Question {3.6}
The argument form goes like this: ~(b & ~ m) therefore m ⊃ b It is invalid.
~ (b & ~ m)

m ⊃ b

1 0 0 0 1

1 0 0

Question {3.7}
The argument form is: e ⊃ c, c ⊃ p, p ⊃ ~ e therefore ~ e It is a valid form.
Question {3.8}
The argument form is: p ∨ c, p ⊃ m, c ⊃ s therefore ~(~ m &~ s) This is valid.
Question {3.9}
The argument form is: (r ⊃ ~ w) ⊃ r, therefore r. Surprisingly, this is valid. If the conclusion, r, is false,
then (r ⊃ ~ w) is true, and r is false, so the premise is also false.
Question {3.10}
(j ⊃ c) & (e ⊃ d) therefore (j ⊃ d) ∨ (e ⊃ c) Surprisingly, this is valid.
Question {3.11}
Here’s a truth table for exclusive disjunction (I’ll use the symbol x for it).
p q

p x q

0 0

0 0 0

0 1

0 1 1

1 0

1 1 0

1 1

1 0 1

The exclusive disjunction of p and q is true when exactly one of p and q is true.

Advanced:
Question {3.14}
Each formula here is a tautology. I think that this shows that ’ ⊃ ’ is not a good translation of ‘if’ in English.

Chapter 4: Trees Basic
Basic
Question {4.1}
You should get exactly the same answers as you got with truth tables. Here are the solutions for the
exercises in Questions {3.2} and {3.4}. First, the formulas from {3.2}
1.
~(p ⊃ ((p ⊃ q) ⊃ q))
|
p
~((p ⊃ q) ⊃ q)
|
p⊃q
~q
/

\

~p

q

X

X

The tree closes, so the formula is a tautology.
2.
~ (p ∨ (p ⊃ q))
|
~p
~ (p ⊃ q)
|
p
~q
X
The tree closes, so the formula is a tautology.
3.
~((p & (p ⊃ q)) ⊃ q)
|
p & (p ⊃ q)
~q

|
p
p⊃q
/

\

~p

q

X

X

The tree closes, so the formula is a tautology.
4.
~(((p & q) ⊃ r) ⊃ (p ⊃ r))
|
(p & q) ⊃ r
~(p ⊃ r)
|
p
~r
/

\

~(p & q)

r

/

\

X

~p

~q

X

^
open

The tree is open. A countermodel is as follows:
p=1q=0r=0
This makes the premise true and the conclusion false. The argument is invalid.
5.
~((p ⊃ q) ⊃ p) ⊃ p)
|
(p ⊃ q) ⊃ p
~p
/

\

~(p ⊃ q)

p

|

X

p
~q

X
The tree is closed, the formula is a tautology.
Now the argument forms from {3.4}
1.
p
p⊃q
~q
/

\

~p

q

X

X

The tree closes, the argument form is valid.
2.
p
q
~(p ≡ q)
/

\

p

~p

~q

q

X

X

The tree closes, the argument form is valid.
3.
p&q
~(p ≡ q)
|
p
q
/

\

p

~p

~q

q

X

X

The tree is closed. The argument form is valid.
4.

p
q⊃p
~q
/

\

~q

p

^

^

open

open

The tree remains open. A countermodel p = 1 and q = 0 makes the premises true and the conclusion false.
The argument form is invalid.
5.
p
q
~(p & q)
/

\

~p

~q

X

X

The tree is closed. The argument form is valid.
6.
p
~(p ∨ q)
|
~p
~q
X
The tree is closed. The argument form is valid.
7.
p≡q
p≡~q
~~ p
|
p
/
p

\
~p

q
/

~q
\

X

p

~p

~q

~~ q

X

X

The tree is closed, the argument form is valid.
8.
p ⊃ (q ⊃ r)
~(q ⊃ (p ⊃ r))
|
q
~(p ⊃ r)
|
p
~r
/

\
q⊃r

~p
X

/

\

~q

r

X

X

The tree closes, the argument form is valid.
9.
p⊃~p
~~ p
/

\

~p

~p

X

X

The tree closes, the argument form is valid.
10.
~~ p
~p
X
The tree closes, the argument form is valid.

11.
p⊃q
~((r ⊃ p) ⊃ (r ⊃ q))
|
r⊃p
~(r ⊃ q)
|
r
~q
/

\

~p

q

/

\

~r

p

X

X

X

The tree closes. The argument form is valid.
12.
p⊃q
~(~ q ⊃ ~ p)
|
~q
~~ p
/

\

~p

q

X

X

The tree closes, the argument form is valid.
13.
p
~(~ p ⊃ q)
|
~p
~q
X
The tree closes, the argument form is valid.
14.

p ⊃ (p ⊃ q)
~(p ⊃ q)
|
p
~q
/

\

~p

p⊃q

X

X

The tree closes, the argument form is valid.
Note that we can close the right branch using p ⊃ q and its negation. There is no need to resolve the formula
p ⊃ q.
15.
p
~(q ⊃ q)
|
q
~q
X
The tree closes, the argument form is valid.
16.
p∨q
~q∨r
~ (p ∨ r)
|
~p
~r
/

\

p

q

X

/

\

~q

r

X

X

The tree closes, the argument form is valid.
17.

p⊃q
~(q ⊃ p)
|
q
~p
/

\

~p

q

^

^

open

open

The tree is open.
The counterexample p = 0 and q = 1 makes the premises true, and the conclusion false.
So, the argument is invalid.
18.
p ⊃ (q ⊃ r)
~((p ⊃ q) ⊃ r)
|
p⊃q
~r
/

\

/

\

/

\
q⊃r

~p
/

\

/

~p

q

^

^

\

~p
/

q
\

/

\

~q

r

~q

r

^

X

X

X

The tree is open (see any of the branches marked with `^’). Any evaluation of the propositions with p and r
false (see the first branch) will make the premises true and the conclusion false. So:
p=0q=0r=0
p=0q=1r=0
are both countermodels.
The argument is invalid.

19.
(p & q) ⊃ r
~(p ⊃ (~ q ∨ r))
|
p
~(~ q ∨ r)
|
~~ q
~r
/

\

~ (p & q)

r

/

\

~p

~q

X

X

X

The tree closes, the argument form is valid.
20.
p⊃q
r⊃s
~((q ⊃ r) ⊃ (p ⊃ s))
|
q⊃r
~(p ⊃ s)
|
p
~s
/

\

~p

q

X

/

\

~q

r

X

/

\

~r

s

X

X

The tree is closed. The argument form is valid.
Question {4.2}

An exclusive disjunction A x B is true when A is true and B isn’t, or when B is true and A isn’t. So the rule
for an exclusive disjunction should be:
AxB
/

\

A

~A

~B

B

If you get a negated exclusive disjunction, you have a different rule. A x B is false when A and B are both
true, or when they are both false. So, you get:
~(A x B)
/

\

A

~A

B

~B

Question {4.3}
The Sheffer Stroke A | B is false when A and B are both true, and true otherwise. The negated rule is easiest.
~(A | B)
|
A
B
The formula A | B is true just when either A is false, or B is false. So you can use the rule:
A|B
/
~A

\
~B

Chapter 5: Vagueness and Bivalence
Basic
Question {5.1}
I will write ’ n ’ for ‘1/2’ as it’s smaller here. You can read it as “neither true nor false”. Each of the
following are not L3 tautologies:
p ∨ ~ p
n n n n
~ (p & ~ p)
n n n n n
(p & (p ⊃ q)) ⊃ q
n n n n 0 n 0
(p ⊃ ~ p) ⊃ ~ p
n 1 n n n n n
(p & ~ p) ⊃ q
n n n n n 0
((p ⊃ q) ⊃ p) ⊃ p
n n 0 1 n n n
All of the other formulas are tautologies.

Chapter 6: Conditionality
Basic
Question {6.1}
□ A & (A → B) ⊨ □ B, since if w ⊩ □ A and w ⊩ A → B, then for every v, v ⊩ A, and for every v, if v ⊩ A
then v ⊩ B. Therefore, for every v, v ⊩ B, and hence, w ⊩ □ B, as we wanted.
□ A & (A ⊃ B) ⊭ □ B, for we can have a model with two worlds w and v where A is true at both w and v, B
is true at w but not at v. Therefore, w ⊩ □ A, and w ⊩ A ⊃ B (since A and B are both true at w) but w does
not give us □ B, as B is not true at v.
~ □ A ⊨ □ ~ □ A, since if w ⊩ ~ □ A, then there is some world v where v ⊩ ~ A. Therefore in any world x, x
⊩ ~ □ A, so w ⊩ □ ~ □ A, as we wanted.
□( A & □ B) ⊨ □( A & B), since if w ⊨ □(A & □ B), it means that at any world v, v ⊩ A & □ B, so v ⊩ A, and
since v ⊩ □ B, at any world x, x ⊩ B. But if B is true at any world, and A is true at any world, A & B is true
at all worlds too, so w ⊩ □ ( A & B), as desired.
□ A & ~ □ B ⊨ ~ □ (~ A ∨ B), since if w ⊩ □ A &~ □ B, then there is some world v where v ⊩ ~ B, and v ⊩
A. Therefore v ⊩ ~(~ A ∨ B), and hence, w ⊩ ~ □(~ A ∨ B), as desired.
Question {6.2}
w ⊩ ◇ A if and only if for some v, v ⊩ A.
◇ A & ◇ B ⊭ ◇( A & B), since we have a model with two worlds, w and v. At w, A is true but B is false,
while at v, B is true and A false. Therefore ◇ A is true at both worlds, as is ◇ B, but ◇ ( A & B) is not true, as
there is no world where A and B are both true.
On the other hand, ◇ ( A ∨ B) ⊨ ◇ A ∨ ◇ B, since if w ⊩ ◇ ( A ∨ B), then there is a world v where v ⊩ A ∨ B,
and hence, either v ⊩ A, or v ⊩ B. In the first case, w ⊩ ◇ A (and hence w ⊩ ◇ A ∨ ◇ B) and in the second
case w ⊩ ◇ B (and hence w ⊩ ◇ A ∨ ◇ B). In either case, w ⊨ ◇ A ∨ ◇ B, as desired.
Questions {6.3}
Have a simple little situation in which there are four worlds, a, b, c and d. In world a, p and q are both true,
in world b, p is true and q false, in c, p is false and q true, and in world d, p and q are both false. r is true
only in world a. Therefore the premise (p & q) → r is true in every world. The truth of p & q means you’re
in world a, and in that world, r is true. The conclusion (p → r) ∨ (q → r) is false, since in world b, p is true
but r is false (so p → r is false everywhere), and in world c, q is true and r is false (so q → r is false
everywhere). Therefore, we can have the premise true and the conclusion false. This model has four
worlds, one for each possible state of the electrical circuitry. ( p & q) → r is true because in each state
where p & q is true, so is r. But p → r fails because in some state, p is true but r false (namely, when switch
one is flipped, but switch two is not — in world b) and q → r fails because in some other state, q is true but
r is false (in world c). If we used the material conditional, you need the same state to fail both p ⊃ r and q ⊃
r, and this cannot happen if (p & q) ⊃ r is true.

Chapter 7: Natural Deduction
Basic
Question {7.1}
A⊃~B⊢B⊃~A:
1 B⊢B

Ax.

2 A⊃~B⊢A⊃~B

Ax.

3 A⊢A

Ax.

4 A ⊃ ~ B, A ⊢ ~ B

2,3 (⊃ E)

5 A ⊃ ~ B, A, B ⊢ \bot 1,4 (~ E)
6 A ⊃ ~ B, B ⊢ ~ A

5 (~ I)

7 A⊃~B⊢B⊃~A

6 (⊃ I)

~~~A⊢~A:
1 ~~~A⊢~~~A

Ax.

2 A⊢A

Ax.

3 ~A⊢~A

Ax.

4 A, ~ A ⊢ \bot

2,3 (~ E)

5 A ⊢ ~~ A

4 (~ I)

6 ~ ~ ~ A, A ⊢ \bot 1,5 (~ E)
7 ~~~A⊢~A

6 (~ I)

~ A ∨ ~ B ⊢ ~(A & B) :
1 ~A∨~B⊢~A∨~B

Ax.

2 A & B, ~ A ⊢ A & B

Ax.

3 A & B, ~ A ⊢ A

2 (& E)

4 A & B, ~ A ⊢ ~ A

Ax.

5 A & B, ~ A ⊢ \bot

3,4 (~ E)

6 A & B, ~ B ⊢ A & B
7 A & B, ~ B ⊢ B

Ax.
6 (& E)

8 A & B, ~ B ⊢ ~ B

Ax.

9 A & B, ~ B ⊢ \bot

7,8 (~ E)

10 ~ A ∨ ~ B, (A & B) ⊢ \bot

1,5,9 (∨
E)

11 ~ A ∨ ~ B ⊢ ~(A & B)

10 (~ I)

~(A ∨ B) ⊢ ~ A & ~ B :

1 ~(A ∨ B) ⊢ ~(A ∨ B)

Ax.

2 A⊢A

Ax.

3 A⊢A∨B
4 ~(A ∨ B), A ⊢ \bot

2 (∨ I)
1,3 (~ E)

5 ~(A ∨ B) ⊢ ~ A

4 (~ I)

6 B⊢B

Ax.

7 B⊢A∨B
8 ~(A ∨ B), B ⊢ \bot

6 (∨ I)
1,7 (~ E)

9 ~(A ∨ B) ⊢ ~ B

8 (~ I)

10 ~(A ∨ B) ⊢ ~ A & ~ B 5,9 (& I)
A & ~ B ⊢ ~(A ⊃ B) :
1 A&~B⊢A&~B

Ax.

2 A&~B⊢~B

1 (& E)

3 A&~B⊢A

1 (& E)

4 A⊃B⊢A⊃B
5 A & ~ B, A ⊃ B ⊢ B

Ax.
3,4 (⊃ E)

6 A & ~ B, A ⊃ B ⊢ \bot 2,5 (~ E)
7 A & ~ B ⊢ ~(A ⊃ B)

6 (~ I)

Question {7.2}
⊢ ((A ⊃ B) ⊃ A) ⊃ A :
1 ~A⊢~A

Ax.

2 A⊢A

Ax.

3 ~ A, A ⊢ \bot

1,2 (~ E)

4 ~ A, A ⊢ B

3 (\bot E)

5 ~A⊢A⊃B

4 (⊃ I)

6 (A ⊃ B) ⊃ A ⊢ (A ⊃ B) ⊃ A

Ax.

7 (A ⊃ B) ⊃ A, ~ A ⊢ A

5,6 (⊃ E)

8 (A ⊃ B) ⊃ A, ~ A ⊢ \bot

1,7 (~ E)

9 (A ⊃ B) ⊃ A ⊢ ~~ A

8 (~ I)

10 (A ⊃ B) ⊃ A ⊢ A

9 (DNE)

11 ⊢ ((A ⊃ B) ⊃ A) ⊃ A

10 (⊃ I)

~(A & B) ⊢ ~ A ∨ ~ B :
1 ~(~ A ∨ ~ B) ⊢ ~(~ A ∨ ~ B)

Ax.

2 ~A⊢~A

Ax.

3 ~A⊢~A∨~B

2 (∨ I)

4 ~(~ A ∨ ~ B), ~ A ⊢ \bot

1,3 (~ E)

5 ~(~ A ∨ ~ B) ⊢ ~~ A

4 (~ I)

6 ~(~ A ∨ ~ B) ⊢ A

5 (DNE)

7 ~B⊢~B

Ax.

8 ~B⊢~A∨~B

7 (∨ I)

9 ~(~ A ∨ ~ B), ~ B ⊢ \bot

1,8 (~ E)

10 ~(~ A ∨ ~ B) ⊢ ~~ B

9 (~ I)

11 ~(~ A ∨ ~ B) ⊢ B

10 (DNE)

12 ~(~ A ∨ ~ B) ⊢ A & B

6,11 (& I)

13 ~(A & B) ⊢ ~(A & B)

Ax.

14 ~(A & B), ~(~ A ∨ ~ B) ⊢ \bot 12,13 (~ E)
15 ~(A & B) ⊢ ~~(~ A ∨ ~ B)

14 (~ I)

16 ~(A & B) ⊢ ~ A ∨ ~ B

15 (DNE)

⊢ A ∨ (A ⊃ B) :
1 ~(A ∨ (A ⊃ B)) ⊢ ~(A ∨ (A ⊃ B))

Ax.

2 A⊢A

Ax.

3 ~A⊢~A

Ax.

4 A, ~ A ⊢ \bot

2,3 (~ E)

5 A, ~ A ⊢ B

4 (\bot E)

6 ~A⊢A⊃B

5 (⊃ I)

7 ~ A ⊢ A ∨ (A ⊃ B)

6 (∨ I)

8 ~(A ∨ (A ⊃ B)), ~ A ⊢ \bot

1,7 (~ E)

9 ~(A ∨ (A ⊃ B)) ⊢ ~~ A

8 (~ I)

10 ~(A ∨ (A ⊃ B)) ⊢ A

9 (DNE)

11 ~(A ∨ (A ⊃ B)) ⊢ A ∨ (A ⊃ B)

10 (∨ I)

12 ~(A ∨ (A ⊃ B)) ⊢ \bot

1,11 (~ I)

13 ⊢ ~~(A ∨ (A ⊃ B))

12 (~ I)

14 ⊢ A ∨ (A ⊃ B)

13 (DNE)

~A⊃~B⊢B⊃A:
1 ~A⊢~A

Ax.

2 ~A⊃~B⊢~A⊃~B

Ax.

3 ~ A ⊃ ~ B, ~ A ⊢ ~ B
4 B⊢B

1,2 (⊃ E)
Ax.

5 ~ A ⊃ ~ B, ~ A, B ⊢ \bot 3,4 (~ E)
6 ~ A ⊃ ~ B, B ⊢ ~~ A

5 (~ I)

7 ~ A ⊃ ~ B, B ⊢ A
8 ~A⊃~B⊢B⊃A

6 (DNE)
7 (⊃ I)

(A & B) ⊃ C ⊢ (A ⊃ C) ∨ (B ⊃ C) :
1 A⊢A

Ax.

2 B⊢B

Ax.

3 A, B ⊢ A & B
4 (A & B) ⊃ C ⊢ (A & B) ⊃ C
5 (A & B) ⊃ C, A, B ⊢ C

1,2 (& I)
Ax.
3,4 (⊃ E)

6 (A & B) ⊃ C, A ⊢ B ⊃ C

5 (⊃ I)

7 (A & B) ⊃ C, A ⊢ (A ⊃ C) ∨ (B ⊃ C)

6 (∨ I)

8 ~((A ⊃ C) ∨ (B ⊃ C)) ⊢ ~((A ⊃ C) ∨ (B ⊃ C))

Ax.

9 (A & B) ⊃ C, A, ~((A ⊃ C) ∨ (B ⊃ C)) ⊢ \bot

7,8 (~ E)

10 (A & B) ⊃ C, A, ~((A ⊃ C) ∨ (B ⊃ C)) ⊢ C

9 (\bot E)

11 (A & B) ⊃ C, ~((A ⊃ C) ∨ (B ⊃ C)) ⊢ A ⊃ C

10 (⊃ I)

12 (A & B) ⊃ C, ~((A ⊃ C) ∨ (B ⊃ C)) ⊢ (A ⊃ C) ∨ (B ⊃ C) 11 (∨ I)
13 (A & B) ⊃ C, ~((A ⊃ C) ∨ (B ⊃ C)) ⊢ \bot
14 (A & B) ⊃ C ⊢ ~~((A ⊃ C) ∨ (B ⊃ C))
15 (A & B) ⊃ C ⊢ (A ⊃ C) ∨ (B ⊃ C)

8,12 (~ E)
13 (~ I)
14 (DNE)

Chapter 8: Predicates, Names and Quantifiers
Basic
Question {8.1}
1. Anthea is a geologist.
2. Brian is not a hairdresser.
3. Either Anthea is a hairdresser or Brian is a geologist.
4. Both Anthea and Brian are hairdressers.
5. Brian is both a hairdresser and a geologist.
6. Brian is not a geologist.
7. If Brian is a geologist, he’s a hairdresser.
8. Anthea is larger than Brian.
9. Brian is not larger than Anthea.
10. All geologists are hairdressers.
11. Not all geologists are hairdressers.
12. All geologists are not hairdressers.
13. Some hairdressers are people.
14. All hairdressers are larger than Brian.
15. For any person there’s a person larger than that person.
16. For any person there’s a geologist larger than that person.
17. For any person, anything that that person is larger than is not larger than that person. (check errata)
18. Someone’s not a geologist.
19. There’s a person larger than any person.
20. Take any three people you like: If the first is larger than the second and the second is larger than the
third, then the first is larger than the third.
Question {8.2}

1. Pb
2. Hb ⊃ ~ Gb
3. ~ Gb ⊃ ~ Ga
4. ~(∃ x)(Px & Gx)
5. (∀ x)(Px ⊃ Gx)
6. (∀ x)(Gx ⊃ Px)
7. (∃ x)(Gx & Px)
8. (∃ x)(Px & Gx)
9. (∃ x)(Gx & Hx)
10. (∃ x)(Hx & ~ Px)
11. (∃ x)(~ Px & ~ Gx)
12. (∃ x)(Hx & (Gx & Px))
13. Ga ⊃ (∃ x)(Px & Lax)
14. (∀ x)(Gx ⊃ (∃ y)(Hy & Lxy))
15. (∀ x)((Gx & Hx) ⊃ ~ Px)
16. (∃ x)(Gx & (∀ y)(Hy ⊃ Lxy))
17. (∀ x)(Px ⊃ ~(Gx & Hx))
18. Which is it? (∀ x)(Px ⊃ (∃ y)(Gy & Lxy)) or (∀ x)(Gx ⊃ (∀ y)(Py ⊃ Lyx))?
19. (∀ x)(Gx ⊃ (∀ y)(Hy ⊃ Lxy))
20. (∀ x)((Hx & (∃ y)(Gy & Lxy)) ⊃ (∃ z)(Hz & Lxz))
21. (∃ x)(Gx & (∃ y)(Hy & Lxy))
22. (∃ x)(Gx & (∀ y)(Hy ⊃ Lxy))
23. (∀ x)(Hx ⊃ (Lxa & Lxb)
24. (∀ x)(Lxa ⊃ Lxb) ⊃ Lab
25. (∀ x)(Lxa ⊃ Gx) ⊃ (∀ x)(Hx ⊃ Lax)

26. (∃ x)(Hx & Lax) ⊃ (∃ x)(Gx & Lax)
27. (∃ x)(Px & (∀ y)(Hy ⊃ Lxy))
28. (∃ x)(Hx &~(∃ y)(Gy & Lxy))
29. (∀ x)(Px ⊃ (Gx ⊃ (Lxb & ~ Lxa)))
30. (∀ x)((Hx & (∃ y)(Gy & Lxy)) ⊃ (∀ z)(Pz ⊃ Lxz))
Question {8.3}
In these answers variables are bound by the quantifier that marked by the same type of *. If a variable has a
* above it, then it is bound by the quantifier with a * above it, similarly for variables with *‘s below.
10:
(∀ y) (G y ⊃ H y)
*

*

*

11:
~ (∀ x) (G x ⊃ H x)
*

*

*

12:
(∀ x) (G x ⊃ ~ H x)
*

*

*

13:
(∃ x) (H x & P x)
*

*

*

14:
(∀ x) (H x ⊃ L x b)
*

*

*

15:
*

*

*

(∀ x) (P x ⊃ (∃ y) (P y & L y x))
*

*

*

16:
*

*

*

(∀ x) (P x ⊃ (∃ y) (G y & L y x))
*

*

*

17:
*

*

*

(∀ x) (P x ⊃ (∀ y) (L x y ⊃ ~ L y x))
*

*

*

*

18:
(∃ y) (P y & ~ G y)
*

*

*

19:
*

*

*

(∃ x) (P x & (∀ y) (P y ⊃ L x y))
*

*

*

20:
*

*

*

(∀ x) (∀ y) (∀ z) (L x y ⊃ (L y z ⊃ L x z))
*

*
*

*
*

*

Question {8.4}
1. Lb, Rb, therefore Lb & Rb.
Lx = x studies linguistics, Rx = belongs to the rockclimber’s club, b = Brian.
2. (∃ x)Lx, (∃ x)Rx, therefore (∃ x)(Lx & Rx)
3. (∀ x)(Sx ⊃ (∃ y)(Ly & Oxy)) therefore (∃ y)(Ly & (∀ x)(Sx ⊃ Oxy))
Sx = x is a solid, Lx = x is a liquid, Oxy = x is soluble in y.
4. (∀ x)(Ex ⊃ (Sx ∨ Ax)), Ea therefore Sa & Aa
Ex = x is eligible for the clean desk prize. Sx = x is a secretary. Ax = x is an administrator. a = Ian.

5. (∀ x)Mx, therefore (~(∃ x)Mx ∨ ((∃ x)(Mx & Tx) & (∀ x)(Tx ⊃ Mx))) & ~(~(∃ x)Mx & ((∃ x)(Mx & Tx) &
(∀ x)(Tx ⊃ Mx))) {This was a hard one!}
Mx = x is material. Tx = x is mental.
6. (∃ x)(Mx & (∀ y)((My & ~ Syy) ⊃ Sxy)) therefore (∃ x)(Mx & Sxx)
Mx = x is a man in town. Sxy = x shaves y.
7. (∀ x)(Hx ⊃ Ax), therefore (∀ x)((∃ y)(Hy & Oxy) ⊃ (∃ y)(Ay & Oxy))
Hx = x is a horse. Ax = x is an animal. Oxy = x is a head of y.
8. (∀ x)(∀ y)((Rxy & Py) ⊃ Nx), ~(∃ x)(Nx & Fx), (∀ x)(∀ y)((Rxy & (Ny & ~ Py)) ⊃ ~ Fx), therefore (∀ x)(∀
y)((Rxy & Ny) ⊃ ~ Fx)
Rxy = x is a square root of y. Px = x is a perfect square. Nx = x is a natural number. Fx = x is a fraction.
9. ~(∃ x)(Px & Cx) ⊃ (∃ x)(Px & Dx), therefore (∃ x)(Px & (~ Cx ⊃ Dx))
Px = x is a person. Cx = x contributes to Oxfam. Dx = x dies of hunger.
10. ~(∃ x)(Ax & ~ Kx), (∀ x)(Gx ⊃ Ix), ~(∃ x)(Ix & Kx) therefore (∀ x)(Gx ⊃ ~ Ax).
{I have treated “no-one” as “nothing” here.}
Ax = x really appreciates Beethoven. Kx = x keeps silent while the Moonlight Sonata is played. Gx = x is a
guinea pig. Ix = x is hopelessly ignorant of music.

Advanced
Questions {8.5} and {8.6}:
Discuss these with your tutor.

Chapter 9: Models for Predicate Logic
Basic
Question {9.1}
1. True. (Each instance is true.)
2. True. (Each instance is true.)
3. False. It has a false instance, b for x and a for y, Tba ⊃ (Tab & ~ Fb)) is false.
4. True. It has a true instance (∀ y)((Tby & Fy) ⊃ (Gb & Tby))
5. False. It has a false instance (a for x and b for y) (∃ z)(Taz & Tab) ⊃ (∃ z)(Tbz & Tza)
The antecedent is true as it has a true instance (b for z). The consequent is false as it has no true instance.
Question {9.2}
Formulas 3, 4, 5, 6, 8, 9 and 10 are true. The rest are false.
Question {9.3}
2, 3, 4, 6, 7, 8 and 10 are tautologies. They are true in every two-element model (and in every model, in
fact!)
Number 1 is false in this model:
D = {a,b}
I(F) I(G)
a

1

0

b

1

1

(∀ x)(Fx ⊃ Gx) ∨ (∀ x)(Fx ⊃ ~ Gx) is false, as (∀ x)(Fx ⊃ Gx) is false (it has a false instance: Fa ⊃ Ga) and
(∀ x)(Fx ⊃ ~ Gx) is false (it has a false instance: Fa ⊃ ~ Ga)
The rest are false in this model:
D = {a,b}
I(F)
a

0

b

1

5: (∀ x)Fx ∨ (∀ x)~ Fx is false, as both (∀ x)Fx and (∀ x)~ Fx are false.

9: (∀ x)(Fx ⊃ (∀ y)Fy) is false, as it has a false instance Fb ⊃ (∀ y)Fy, as (∀ y)Fy is false, but Fb is true.
Question {9.4}
Arguments 3, 5, 6, 7 and 8 are valid. The rest are invalid.
1 has the following counterexample:
D = {a,b,c}
I(G) I(H)
a

1

1

b

0

1

c

0

0

(∀ x)(Gx ⊃ Hx) is true as every instance is true. (∀ x)(~ Gx ⊃ ~ Hx) is false as the instance ~ Gb ⊃ ~ Hb is
false.
2 has the following counterexample:
D = {a,b,c}
I(G) I(H)
a

1

1

b

1

0

c

0

0

(∀ x)Gx ⊃ (∀ x)Hx is true, as (∀ x)Gx and (∀ x)Hx are both false. However, (∀ x)(Gx ⊃ Hx) is not true, as it
has a false instance: Gb ⊃ Hb.
4 has the following counterexample:
D = {a,b,c}
I(G) I(H)
a

1

1

b

1

1

c

1

1

(∃ x)(Gx ≡ Hx) is true, it has a true instance with a for x. But ~(∀ x)(Gx ⊃ Hx) is false, as (∀ x)(Gx ⊃ Hx) is
true.
9 has the following counterexample:
D = {a,b,c}

I(G) I(H)
a

0

0

b

0

0

c

0

0

(∀ x)(Gx ⊃ ~ Hx) is true, as every instance is true. (Every instance of Gx is false, so the conditional is
always true!). However, no instance of (∃ x)(Gx & ~ Hx) is true (for the same reason: Gx is never true).
10 has the following counterexample:
D = {a,b,c}
I(G) I(H)
a

1

0

b

0

0

c

0

0

(∃ x)(Gx & ~ Hx) is now true. However ~(∀ x)(Gx ⊃ ~ Hx) is false, as (∀ x)(Gx ⊃ ~ Hx) is true.
Question {9.5}
Each of the three names can pick out any of the four objects. So for each name you have four choices. The
result is 4 x 4 x 4 = 16 x 4 = 64 different assignments of objects to names. In general, with n names and m
objects, you gave m x m x … x m (n times) = m to the power of n.
Question {9.6}
Given a three place predicate in a domain of 2 objects, you have a 2 x 2 x 2 table (think of a cube) to fill
with values of either 0 or 1. That is, you have 8 slots to fill in with either 0 or 1, and there are 2 to the
power of 8, namely 256, different ways of doing this. In general, for an n place predicate in a domain of m
objects, you have a staggering 2 to the power of (m to the power of n) ways of doing this. That’s a lot of
different combinations!
Question {9.7}
This is a hard one.
•
•
•

n names can be done k^n times (I write ``k to the n’’ as ``k^n’’.)
m monadic predicates can be done (2^k)^m times.
j dyadic predicates can be done (2^(k^2))^j times.

This gives you k^n x (2^k)^m x (2^(k^2))^j different models all up. That’s a lot!
Question {9.8}
Here is a formula which is satisfied only in a model with an infinite domain
(∀ x)~ Sxx & (∀ x)(∀ y)(∀ z)((Sxy & Syz) ⊃ Sxz) & (∀ x)(∃ y)Sxy

Why? Well, first show that it has a model with an infinite domain. In particular, it has a model with the
domain D = {0,1,2,3,4,…} of all the numbers. Interpret S like this: Sxy holds when x is smaller than y. So,
it has a table like this
I(S) 0 1 2 3 4 …
0

0 1 1 1 1 …

1

0 0 1 1 1 …

2

0 0 0 1 1 …

3

0 0 0 0 1 …

… … … … … … …
And the formula is true in this interpretation, as (1) no number is smaller than itself (2) If one number is
smaller than another, and that number is smaller than a third, then the first is smaller than the third. (3) For
any number, there’s a bigger number.
So, the whole formula is true in this model with an infinite domain.
Now we’ll show that in any finite model, this whole formula is false. In particular, we’ll show that if the
second and third conjuncts are true, then the first conjunct (∀ x)~ Sxx is false.
Suppose you’ve got a finite domain and you want to make (∀ x)(∃ y)Sxy true in it. This means that every
object in the domain is “smaller than” some object in the domain. Now suppose that (∀ x)(∀ y)(∀ z)((Sxy &
Syz) ⊃ Sxz) is true. This means that if x is “smaller than” y and y is “smaller than” z then x is “smaller than”
z.
Now look at what happens if these hold in a finite domain. Start with one object. It must be “smaller than”
some object. If it is “smaller than” itself, we’re done: we’ve made (∀ x)~ Sxx false. So, suppose it’s
“smaller than” something else. So, we’ve got a picture like this:
a —S→ b
Now b is “smaller than” some object too. If we have Sba, then since Sab and Sba, we must have Saa, which
conflicts with (∀ x)~ Sxx. If we have Sbb, this is bad too, so we must have a new object c, such that Sbc. So,
we’ve got a picture like this:
a —S→ b —S→ c
Now we have Sac too (since Sab and Sbc) so the picture is actually like this:
a —S→ b —S→ c
|

|
—

-S-

→

Now, there must be some object y where Scy: what could it be? If it were a, then we’d have Sac and Sca,
and so, Saa, which is bad. If it were b, we’d have Sbc and Scb, which gives Sbb, which is bad. If it were c,
we’d have Scc, which is bad too. So, we need a new object d where Scd.

This goes on forever. Every object requires a new object to be related to by S, because (∀ x)~ Sxx requires
that we can’t have S “looping back” to anywhere it’s already been. So, our formula can only be true in an
infinite domain. No finite list of objects will be enough.
So, we can answer our original question:
The formula
~[ (∀ x)~ Sxx & (∀ x)(∀ y)(∀ z)((Sxy & Syz) ⊃ Sxz) & (∀ x)(∃ y)Sxy ]
is true in every finite model, and it is false in some infinite models.

Chapter 10: Trees for Predicate Logic
Basic
Question {10.1}
1. A tautology. Here is my tree for it:
~((∃ x)Fx ≡ ~(∀ x)~ Fx)
/

\

(∃ x)Fx /a

~(∃ x)Fx \a

~~(∀ x)~ Fx

~(∀ x)~ Fx /a

|

|

(∀ x)~ Fx \a

~~ Fa

|

|

Fa

~ Fa

|

X

~ Fa
X
2. A tautology.
~((∃ x) Fx ∨ (∃ x)~ Fx)
|
~(∃ x)Fx \a
~(∃ x)~ Fx \a
|
~ Fa
|
~~ Fa
X
3. Not a tautology.
~((∀ x)(Fx ∨ Gx) ⊃ ((∀ x)Fx ∨ (Ax)Gx))
|
(∀ x)(Fx ∨ Gx) \ a, b
~((∀ x)Fx ∨ (∀ x)Gx)
|
~(∀ x)Fx / a
~(∀ x)Gx / b

|
~ Fa
|
~ Gb
|
Fa ∨ Ga
/

\

Fa

Ga

X

|
Fb ∨ Gb
/

\

Fb

Gb

^

X

This branch stays open. We have resolved every single-use formula, and the reusable formula (∀ x)(Fx ∨
Gx) has had every name substituted into it. So, we read off the open branch the following interpretation.
D = {a,b}
I(F) I(G)
a

0

1

b

1

0

And this indeed makes our original formula false. We have everything in the domain either F or G, yet it’s
not the case that everything is F and it’s not the case that everything is G.
4. Is a tautology.
~((∃ x)(Fx ⊃ p) ⊃ ((∀ x)Fx ⊃ p))
|
(∃ x)(Fx ⊃ p) \ a
~((∀ x)Fx ⊃ p)
|
(∀ x)Fx / a
~p
|
Fa ⊃ p
/

\

~ Fa

p

|

X

Fa
X

You need to keep an eye on the brackets in the original formula. On the left of the conditional, the (∃ x)
binds the whole (Fx ⊃ p). On the right, the (∀ x) binds just the Fx, and (∀ x)Fx is the antecedent of the
whole conditional (∀ x)Fx ⊃ p.
The question in the text book contains an error. The question should have been to test the formula (∃ x)(Fx
⊃ p) ≡ ((∀ x)Fx ⊃ p). This is also a tautology. The tree is:
~((∃ x)(Fx ⊃ p) ≡ ((∀ x)Fx ⊃ p))
/

\

/

\

(∃ x)(Fx ⊃ p) \ a

~(∃ x)(Fx ⊃ p) \ a,b

~((∀ x)Fx ⊃ p)

(∀ x)Fx ⊃ p

|

|

(∀ x)Fx / a

~(Fa ⊃ p)

~p

|

|

Fa

Fa ⊃ p

~p

/

\

/

\

~ Fa

p

~(∀ x)Fx / b

p

|

X

|

X

Fa

~ Fb

X

|
~(Fb ⊃ p)
|
Fb
~p
X

This one was quite difficult. You need to keep an eye on the brackets in the original formula. On the left of
the biconditional, the (∃ x) binds the whole (Fx ⊃ p). On the right, the (∀ x) binds just the Fx, and (∀ x)Fx is
the antecedent of the whole conditional (∀ x)Fx ⊃ p.
Then on the right branch we needed to substitute into the negated existential twice to get it to close.
We could have done it after processing (∀ x)Fx ⊃ p, and this would have resulted in a tree with fewer
substitutions. The right branch would have been:
~(∃ x)(Fx ⊃ p) \ a
(∀ x)Fx ⊃ p
/

\

~(∀ x)Fx / a

p

|

|

~ Fa

~(Fa ⊃ p)

|

|

~(Fa ⊃ p)

Fa

|

~p

Fa

X

~p
X
There’s something tricky going on here. The name `a’ gets introduced in the right branch with the ~(∀
x)Fx. Then we substitute the name into the negated existential at the top of the tree here. The result, ~(Fa ⊃
p), gets placed in both branches below it. In one, `a’ is already there, and in the other, `a’ is new. This is
permissible, according to our rules.
5. Is a tautology.
~((∀ x)(Fx ⊃ Gx) ⊃ ((∀ x)Fx ⊃ (∀ x)Gx))
|
(∀ x)(Fx ⊃ Gx) \ a
~((∀ x)Fx ⊃ (∀ x)Gx)
|
(∀ x)Fx \ a
~(∀ x)Gx / a
|
~ Ga
|
Fa ⊃ Ga
/

\

~ Fa

Ga

|

X

Fa
X
6. Is a tautology.
~((∀ x)(Fx ∨ Gx) ⊃ ((∀ x) Fx ∨ (∃ x) Gx ))
|
(∀ x)(Fx ∨ Gx) \ a
~((∀ x)Fx ∨ (∃ x)Gx)
|
~(∀ x)Fx / a
~(∃ x)Gx \ a
|
~ Fa

|
~ Ga
|
Fa ∨ Ga
/

\

Fa

Ga

X

X

7. Is a tautology. The `p’ here is just a basic proposition.
~((∀ x)(Fx & p) ⊃ ((∀ x)Fx & p))
|
(∀ x)(Fx & p) \ a
~((∀ x)Fx & p)
/

\

~(∀ x)Fx / a

~p

|

|

~ Fa

Fa & p

|

|

Fa & p

Fa

|

p

Fa

X

p
X
8. Is a tautology.
~(∃ x)(Fx ⊃ (∀ y)Fy) \ a,b
|
~(Fa ⊃ (∀ y)Fy)
|
Fa
~(∀ y)Fy / b
|
~ Fb
|
~(Fb ⊃ (∀ y)Fy)
|
Fb
~(∀ y)Fy
X

9. Is a tautology.
~(∃ x)((∃ y)Fy ⊃ Fx) \ a,b
|
~((∃ y)Fy ⊃ Fa)
|
(∃ y)Fy / b
~ Fa
|
Fb
|
~((∃ y)Fy ⊃ Fb)
|
(∃ y)Fy
~ Fb
X
10. Is a tautology.
~(∃ x)(∀ y)(Fy ⊃ Fx) \ a,b
|
~(∀ y)(Fy ⊃ Fa) / b
|
~(Fb ⊃ Fa)
|
Fb
~ Fa
|
~(∀ y)(Fy ⊃ Fb) / c
|
~(Fc ⊃ Fb)
|
Fc
~ Fb
X
Question {10.2}
The arguments 1, 4 and 5 are valid. The rest are not.
1. Valid:

(∀ x)((Fx & Gx) ⊃ Hx) \ a
~(∀ x)(Fx ⊃ (~ Gx ∨ Hx)) / a
|
~(Fa ⊃ (~ Ga ∨ Ha))
|
Fa
~(~ Ga ∨ Ha)
|
~~ Ga
~ Ha
|
(Fa & Ga) ⊃ Ha
/

\

~(Fa & Ga)
/

Ha
\

~ Fa

X
~ Ga

X

X

2. Invalid.
(∀ x)(∀ y)(Lxy ⊃ Mxy) \ a,b
(∀ x)(∀ y)(Lxy ⊃ Lyx) \ a,b
~(∀ x)(∀ y)(Mxy ⊃ Myx) / a
|
~(∀ y)(May ⊃ Mya) / b
|
~(Mab ⊃ Mba)
|
Mab
~ Mba
|
(∀ y)(Lay ⊃ May) \ a,b
|
(∀ y)(Lby ⊃ Mby) \ a,b
|
(∀ y)(Lay ⊃ Lya) \ a,b
|
(∀ y)(Lby ⊃ Lyb) \ a,b
|
Laa ⊃ Maa

|
Lab ⊃ Mab
|
Lba ⊃ Mba
|
Lbb ⊃ Mbb
|
Laa ⊃ Laa
|
Lab ⊃ Lba
|
Lbb ⊃ Lbb
|
Lba ⊃ Lab
|
Lbb ⊃ Lbb
|
.
.
We’ve substituted every name into each universal quantifier, and the only formulae left to process have the
form Lcd ⊃ .., where c and d are each replaced by either a and b. Note that there is nothing in this branch
which compels us to take any formula of the form Lcd as true. The only results we have are that Mab is
true and Mba false. Each of the conditionals Lcd ⊃ … will be true if the antecedent is false. Therefore we
can construct the following interpretation:
D = {a,b}
I(L) a b
a

0 0

b

0 0

I(M) a b
a

0 1

b

0 0

The values for Maa and Mbb are arbitrary. Any will do. This makes the premises true and the conclusion
false, as you can see.
3. Invalid.
This tree is like the first infinite tree example we’ve seen. The premise (∀ x)(∃ y)Lxy on its own will
generate the infinite branch. We will therefore use the names a1, a2, a3 … to keep track of the pattern.

(∀ x)(∃ y)Lxy / a1,a2,a3,a4
~(∃ x)(∀ y)Lxy / a1,a2,a3,a4
|
(∃ y)La1y \ a2
|
~(∀ y)La1y \ a3
|
La1a2
|
~ La1a3
|
(∃ y)La2y \ a4
|
~(∀ y)La2y \ a5
|
La2a4
|
~ La2a5
|
(∃ y)La3y \ a6
|
~(∃ y)La3y \ a7
|
La3a6
|
~ La3a7
|
(∃ y)La4y \ a8
|
~(∀ y)La4y \ a9
|
La4a8
|
~ La4a9
|
.
.
.
You can see the pattern.

If we take each name to name one object only, we get the following domain D = {a1, a2, a3, a4, a5, a6, a7,
a8 …}
And L works like this. L relates object number n to object (2 x n), and it doesn't relate object n to object (2 x
n) + 1. So, a4 gets related to a8, and not to a9. a5 gets related to a10 and not to a11. And so on.
This will obviously make (∀ x)(∃ y)Lxy true (object n is related to object (2 x n). And (∃ x)(∀ y)Lxy is false.
There is no object which is related to everything. Object n is not related to object (2 x n) + 1. Here’s the
table, for those who like thinking in pictures:
I(L) a1 a2 a3 a4 a5 a6 a7 a8 a9 …
a1

1 0

a2

…
1 0

a3

…
1 0

a4
:

…
1 0 …

:

:

:

:

:

:

:

:

:

Now: Let’s see what we know about these objects, to see whether we can make a smaller interpretation.
a2 can’t be the same as a3, and a4 can’t be the same as a5, and a6 can’t be the same as a7, because of the
1s and 0s in each row. However, that is all that we can’t do. a1 can be the same as anything, as there’s
nothing in the a1 column, and the a1 row doesn’t clash with any row.
Now: a2 can be the same as a4, which can be the same as a6, etc… Similarly, a3 can be the same as a5,
which can be the same as a7, etc… So, let the odd-numbered names all name one object, and the evennumbered names name another.
Compressing the table, you get:
I(L) b c
b

0 1

c

1 0

As b is named by a1, a3, a5, etc… b can’t be related to itself (as a1 isn’t related to a3), but b is related to c,
as a1 is to a2. Similarly, c isn’t related to b (a2 isn’t related to a5) but c is related to itself (a2 is related to
a4).
This makes the premise true: (∀ x)(∃ y)Lxy — Everything is related to something (everything is related to
c). However, there isn’t something related to everything (nothing is related to b).
4. Is valid.
(∀ x)(∀ y)(Lxy ⊃ Lyx) \ a
(∀ x)(∀ y)(∀ z)((Lxy & Lyz) ⊃ Lxz) \ a
~(∀ x)((∃ y)Lxy ⊃ Lxx) / a
|
~((∃ y)Lay ⊃ Laa)

|
(∃ y)Lay / b
~ Laa
|
(*)

Lab
|
(∀ y)(Lay ⊃ Lya) \ b
|
Lab ⊃ Lba
/

\

~ Lab

Lba

X

|
(∀ y)(∀ z)((Lay & Lyz) ⊃ Laz) \ b
|
(∀ z)((Lab & Lbz) ⊃ Laz) \ a
|
(Lab & Lba) ⊃ Laa
/

\

~(Lab & Lba)

Laa

/

\

~ Lab

X
~ Lba

X

X

The tree is very much longer if you don’t do the substitutions I did here (or ones very much like them). To
do this, you need to think about how the reasoning goes. Down to the line I marked (*), it is completely
automatic, using the edict “do particular rules before general ones”. Then you must make the first
substitution into a universal quantifier. I picked the formula (∀ x)(∀ y)(Lxy ⊃ Lyx) because I saw that I
could substitute a for x and b for y to get Lab ⊃ Lba. I knew I had Lab, and this would give me new and
interesting information: Lba. Then our facts are Lab, Lba and ~ Laa. This then is inconsistent with the other
premise: (∀ x)(∀ y)(∀ z)((Lxy & Lyz) ⊃ Lxz) when you substitute a, b and a for x, y, and z, respectively. So,
they were the substitutions I made, and that closed the tree.
If you just did any substitutions you liked, in any order, the tree would still close (if you applied the rules
correctly.) It would just have been much longer.
5. Is valid too.
(∀ x)(∀ y)(Lxy ∨ Lyx) / a,b
(∀ x)(∀ y)(∀ z)((Lxy & Lyz) ⊃ Lxz)
~(∀ x)(∀ y)(∃ z)(Lxz & Lyz) / a
|
~(∀ y)(∃ z)(Laz & Lyz) / b
|

~(∃ z)(Laz & Lbz) / a,b
|
~(Laa & Lba)
|
~(Lab & Lbb)
|
(∀ y)(Lay ∨ Lya) / a,b
|
Laa ∨ Laa
|
Lab ∨ Lba
|
(∀ y)(Lby ∨ Lyb) / b
|
Lbb ∨ Lbb
/

\

/

\

/

\

~ Laa
/

~ Lba
\

/

Laa

Laa

X

X

\

Lab
/

Lba
\

~ Lab

X
~ Lbb

X

/

\

Lbb

Lbb

X

X

Notice that the premise (∀ x)(∀ y)(∀ z)((Lxy & Lyz) ⊃ Lxz) is completely irrelevant to the proof. We got the
tree to close without it. I didn’t substitute a into (∀ y)(Lby ∨ Lyb) in this proof as it would just have given
us Lba ∨ Lab, and we already had Lab ∨ Lba on an earlier line. Similarly, I didn’t branch the disjunctions
until I knew I could use them to close one disjunct. That way the tree remained reasonably manageable.
Question {10.3}
Arguments 1, 6, 7, 8, 9 and 10 are valid, in the way that I formalised them in the solutions to {8.4}. The
rest are not. I will present trees for the valid ones and just small interpretations for those which are not.
1. Lb, Rb, therefore Lb & Rb. That’s obviously valid.
Lb
Rb
~(Lb & Rb)
/

\

~ Lb

~ Rb

X

X

2. (∃ x)Lx, (∃ x)Rx, therefore (∃ x)(Lx & Rx) . Invalid.
D = {a, b}
I(L) I(R)
a

1

0

b

0

1

3. (∀ x)(Sx ⊃ (∃ y)(Ly & Oxy)) therefore (∃ y)(Ly & (∀ x)(Sx ⊃ Oxy))
Invalid.
D = {a, b, c, d}
I(S) I(L)
a

1

0

b 1

0

c

0

1

d 0

1

I(O) a b c d
a

0 0 1 0

b

0 0 0 1

c

0 0 0 0

d

0 0 0 0

a and b are solids, c and d are liquids, a dissolves in c, and b in d. So each solid dissolves in some liquid.
However, there’s no liquid such that each solid dissolves in it.
4. (∀ x)(Ex ⊃ (Sx ∨ Ax)), Ea therefore Sa & Aa
Invalid.
D = {a}
I(E) I(S) I(A)
a

1

1

0

Ian is a secretary but not and administrator. He’s eligible for the clean desk prize.
5. (∀ x)Mx, therefore (~(∃ x)Mx ∨ ((∃ x)(Mx & Tx) & (∀ x)(Tx ⊃ Mx))) & ~(~(∃ x)Mx & ((∃ x)(Mx & Tx) &
(∀ x)(Tx ⊃ Mx)))

D = {a}
Invalid.
I(M) I(T)
a

1

0

(∀ x)Mx is obviously true. However, the conclusion is false as ~(∃ x)Mx is false, and so is (∃ x)(Mx & Tx).
6. (∃ x)(Mx & (∀ y)((My & ~ Syy) ⊃ Sxy)) therefore (∃ x)(Mx & Sxx)
This is valid:
(∃ x)(Mx & (∀ y)((My & ~ Syy) ⊃ Sxy)) / a
~(∃ x)(Mx & Sxx) / a
|
Ma & (∀ y)((My & ~ Syy) ⊃ Say)
|
Ma
(∀ y)((My & ~ Syy) ⊃ Say) / a
|
(Ma & ~ Saa) ⊃ Saa
|
~(Ma & Saa)
/

\

~ Ma

~ Saa

X

/

\

~(Ma & ~ Saa)

Saa

/

\

X

~ Ma

~~ Saa

X

X

7. (∀ x)(Hx ⊃ Ax), therefore (∀ x)((∃ y)(Hy & Oxy) ⊃ (∃ y)(Ay & Oxy))
This is valid:
(∀ x)(Hx ⊃ Ax) \ b
~(∀ x)((∃ y)(Hy & Oxy) ⊃ (∃ y)(Ay & Oxy)) / a
|
~((∃ y)(Hy & Oay) ⊃ (∃ y)(Ay & Oay))
|
(∃ y)(Hy & Oay) / b
~(∃ y)(Ay & Oay) \ b

|
Hb & Oab
|
Hb
Oab
|
~(Ab & Oab)
/

\

~ Ab

~ Oab

|

X

Hb ⊃ Ab
/

\

~ Hb

Ab

X

X

8. (∀ x)(∀ y)((Rxy & Py) ⊃ Nx), ~(∃ x)(Nx & Fx), (∀ x)(∀ y)((Rxy & (Ny & ~ Py)) ⊃ ~ Fx), therefore (∀ x)(∀
y)((Rxy & Ny) ⊃ ~ Fx)
This is valid:
(∀ x)(∀ y)((Rxy & Py) ⊃ Nx) \ a
~(∃ x)(Nx & Fx) \ a
(∀ x)(∀ y)((Rxy & (Ny & ~ Py))
⊃ ~ Fx) \ a
~(∀ x)(∀ y)((Rxy & Ny) ⊃ ~ Fx) /
a
|
~(∀ y)((Ray & Ny) ⊃ ~ Fa) / b
|
~((Rab & Nb) ⊃ ~ Fa)
|
Rab & Nb
~~ Fa
|
Rab
Nb
|
(∀ y)((Ray & (Ny & ~ Py)) ⊃ ~
Fa) \ b
|
(Rab & (Nb & ~ Pb)) ⊃ ~ Fa)
/

\

~( Rab & (Nb &
~ Pb))

~ Fa

/

\

X

~
Rab

~(Nb & ~ Pb)

X

/

\

~ Nb

~~ Pb

X

|
(∀ y)((Ray & Py) ⊃
Na) \ b
|
(Rab & Pb) ⊃ Na
/

\

~(Rab & Pb)
/

Na
\

~ Rab

|
~(Na &
Fa)

~ Pb

X

X

/
~ Na

~
Fa

X

X

9. ~(∃ x)(Px & Cx) ⊃ (∃ x)(Px & Dx), therefore (∃ x)(Px & (~ Cx ⊃ Dx))
This is valid, surprisingly!
~(∃ x)(Px & Cx) ⊃ (∃
x)(Px & Dx)
~(∃ x)(Px & (~ Cx ⊃ Dx))
\a
/
/

\

\
\

~~(∃ x)(Px &
Cx)

(∃ x)(Px & Dx) /
a

|

|

(∃ x)(Px & Cx) /
a

Pa & Da

|

|

Pa & Ca

Pa

|

|

Pa

Da

|

|

Ca

~(Pa & (~ Ca ⊃

Da))
|

/

~(Pa & (~ Ca ⊃
Da))
/

\

\

~
Pa

~(~ Ca ⊃
Da)

X

|

~
Pa

~(~ Ca ⊃
Da)

~ Ca

X

|

~ Da

~ Ca

X

~ Da
X
Why is it valid? Well, if ~(∃ x)(Px & Cx) ⊃ (∃ x)(Px & Dx) is true, then either someone contributes to
Oxfam or someone dies. If someone contributes to Oxfam then that person is someone who will die if she
doesn’t. (Why is this true? Because of the material conditional! It is true if the antecedent is false.) On the
other hand, if someone dies, then that person is someone who will die if she doesn’t contribute to Oxfam.
(Why is this true? Because of the material conditional! It is true if the consequent is true.)
10. ~(∃ x)(Ax & ~ Kx), (∀ x)(Gx ⊃ Ix), ~(∃ x)(Ix & Kx) therefore (∀ x)(Gx ⊃ ~ Ax).
{I have treated “no-one” as “nothing” here.}
This is valid.
~(∃ x)(Ax & ~ Kx) \a
(∀ x)(Gx ⊃ Ix) \a
~(∃ x)(Ix & Kx) \a
~(∀ x)(Gx ⊃ ~ Ax) / a
|
~(Ga ⊃ ~ Aa)
|
Ga
~~ Aa
|
Ga ⊃ Ia
/

\

~ Ga

Ia

X

|
~(Ia & Ka)
/
~ Ia
X

\
~ Ka
|
~(Aa & ~ Ka)

/

\

~ Aa

~~ Ka

X

X

Questions {10.4}
1.
a = PS
Sx = x is an axiomatic system.
Txy = x is a theorem of y.
Ox = x is a tautology.
Vx = x is a variable.
Cx = x is post-consistent.
Sa, (∀ x)(Txa ⊃ Ox), ~(∃ x)(Vx & Ox), (∀ x)((Sx & ~(∃ y)(Vy & Tyx)) ⊃ Cx), therefore Ca
This is valid.
Sa
(∀ x)(Txa ⊃ Ox) \b
~(∃ x)(Vx & Ox) \b
(∀ x)((Sx & ~(∃ y)(Vy & Tyx)) ⊃ Cx)
~ Ca
|
(Sa & ~(∃ y)(Vy & Tya)) ⊃ Ca
/

\
Ca

~(Sa & ~(∃ y)(Vy & Tya))
/

\

X

~ Sa

~~(∃ y)(Vy & Tya)

X

|
(∃ y)(Vy & Tya) / b
|
Vb & Tba
|
Vb
Tba
|
Tba ⊃ Ob

/

\

~ Tba

Ob

X

|
~(Vb & Ob)
/
~ Vb
X

\
~ Ob
X

With this argument, (as with the others) it is helpful to keep in mind what each name stands for. The name
’a’ names PS. It’s a system. The name ’b’ names a propositional variable which is a theorem of PS. So that
is the sensible thing to substitute into the second and third premises, not a.
2.
p = Every decision is to be referred to the central authority.
Px = x is a program.
Dx = x is delayed.
Ax = x receives approval.
p ⊃ (∃ x)(Px & Dx), (∀ x)(Px ⊃ Ax), ~(∃ x)(Dx & Ax), therefore ~ p.
It is possible to expand p as follows:
Cx = x is a decision
Rx = x is to be referred to the central authority,
and p is then (∀ x)(Cx ⊃ Rx).
But this plays no role in the argument.
It is valid either way. Here is the tree for when we use ’p’.
p ⊃ (∃ x)(Px & Dx)
(∀ x)(Px ⊃ Ax) \a
~(∃ x)(Dx & Ax) \a
~~ p
/

\

~p

(∃ x)(Px & Dx) / a

X

|
Pa & Da
|
Pa

Da
|
Pa ⊃ Aa
/

\

~ Pa

Aa

X

|
~(Da & Aa)
/

\

~ Da

~ Aa

X

X

3.
Tx = x is true.
Fx = x is false.
Kxy = x knows y.
Px = x is a person.
j = John.
(∀ x)(Tx ⊃ ~ Fx) & (∀ x)(~ Fx ⊃ Tx), (∀ x)(Fx ⊃ ~(∃ y)(Py & Kyx)), (∃ x)(Kjx) & Pj Therefore (∃ x)Tx.
This is valid. Consider the following tree:
(∀ x)(Tx ⊃ ~ Fx)
(∀ x)(~ Fx ⊃ Tx) \a
(∀ x)(Fx ⊃ ~(∃ y)(Py & Kyx)) \a
(∃ x)Kjx & Pj
~(∃ x)Tx \a
|
(∃ x)Kjx / a
Pj
|
Kja
|
~ Ta
|
~ Fa ⊃ Ta
/
~~ Fa

\
Ta

|

X

Fa
|
Fa ⊃ ~(∃ y)(Py & Kya)
/

\

~ Fa

~(∃ y)(Py & Kya) \j

X

|
~(Pj & Kja)
/
~ Pj

\
~ Kja

X

X

The tree is quite short, when you know what substitutions to make, to make things close.
Question {10.5}
For each of these we need to find a counterexample to the arguments. One way to do this is to do a tree for
each and read off the counterexample
Here are some suggestions:
(1) To show that reflexivity and symmetry does not produce transitivity:
Let the domain be {a,b,c}
and R is interpreted as:
I(R) a b c
a

1 1 0

b

1 1 1

c

0 1 1

In this interpretation the premises are true and the conclusion is false.
See if you can think of a way of determining whether a relation is reflexive, symmetry or transitive at a
mere glance of its table.
(2) To show that reflexivity and transitivity does not produce symmetry:
Let the domain be {a,b}
and R is interpreted as:
I(R) a b
a

1 1

b

0 1

In this interpretation the premises are true and the conclusion is false.
(3) To show that symmetry and transitivity does not produce reflexivity:
Let the domain be {a}
and R is interpreted as:
I(R) a
a

0

In this interpretation the premises are true and the conclusion is false.
Question {10.6}
This is the tree for the argument (∀ x)(∀ y)(Rxy ⊃ Ryx), (∀ x)(∀ y)(∀ z)((Rxy & Ryz) ⊃ Rxz), (∀ x)(∃ y)Rxy
therefore (∀ x)Rxx.
(∀ x)(∀ y)(Rxy ⊃ Ryx) \a
(∀ x)(∀ y)(∀ z)((Rxy & Ryz) ⊃ Rxz) \a
(∀ x)(∃ y)Rxy \a
~(∀ x)Rxx / a
|
~ Raa
|
(∃ y)Ray / b
|
Rab
|
(∀ y)(Ray ⊃ Rya) \b
|
Rab ⊃ Rba
/

\

~ Rab

Rba

X

|
(∀ y)(∀ z)((Ray & Ryz) ⊃ Raz) \b
|
(∀ z)((Rab & bz) ⊃ Raz) \a
|
(Rab & Rba) ⊃ Raa
/

\

~(Rab & Rba)
/
~ Rab
X

Raa
\

X
~ Rba
X

Chapter 11: Identity and Functions
Basic
Question {11.1}
1. Alphonzo is Bernadette.
2. Alphonzo is Bernadette and Alphonzo isn’t Candide.
3. There’s someone (other than Candide) who is rich.
4. Everyone (other than Candide) is rich.
5. Everyone (other than Bernadette) is smaller than Bernadette.
6. No-one (other than Bernadette) is smaller than Candide.
7. Everyone works for someone.
8. Everyone works for someone other than themselves.
9. Someone works for someone.
10. Someone works for someone other than themselves.
Question {11.2}
1. Ra & ~ Rc
2. (∃ x)(Px & Rx & Sxc)
3. (∃ x)(Px & x ≠ c & Sxa)
4. (∀ x)(Px ⊃ (∃ y)(Py & Sxy))
5. (∀ x)(Px ⊃ (∃ y)(Py & x ≠ y & Sxy))
6. (∀ x)((Px & x ≠ a) ⊃ (∃ y)(Py & Wxy))
7. ~(∃ x)((Px & x ≠ b) & Wxa)
8. (∃ x)(∃ y)(x ≠ y & Px & Py)
9. (∀ x)(∀ y)(∀ z)(∀ w)((Px & Py & Pz & Pw) ⊃ (x=y ∨ x=z ∨ x=w ∨ y=z ∨ y=w ∨ z=w))
10. (∃ x)((Px & Wxa) & (∀ y)((Py & Way) ⊃ x=y))
Question {11.3}

The first four are tautologies, and the last is not.
1. Tautology.
~(∀ x)(x=x) / a
|
a≠a
X
2. Tautology
~(∀ x)(∀ y)(x=y ⊃ y=x) / a
|
~(∀ x)(a=y ⊃ y=a) / b
|
~(a=b ⊃ b=a)
|
a=b
b≠a
|
b≠b
X
3. Tautology
~(∀ x)(∃ y)x=y / a
|
~(∃ y)a=y \a
|
a≠a
X
4. Tautology
~(∀ x)(∀ y)(∀ z)((x=y & y=z) ⊃ x=z) / a
|
~(∀ y)(∀ z)((a=y & y=z) ⊃ a=z) / b
|
~(∀ z)((a=b & b=z) ⊃ a=z) / c
|
~((a=b & b=c) ⊃ a=c)
|

a=b & b=c
a≠c
|
a=b
b=c
|
b≠c
|
c≠c
X
5. Not a tautology
~(∀ x)(∀ y)(∀ z)(∀ w)((Rxy & Rzw) ⊃ (y ≠ z ∨ Rxw)) / a
|
~(∀ y)(∀ z)(∀ w)((Ray & Rzw) ⊃ (y ≠ z ∨ Raw)) / b
|
~(∀ z)(∀ w)((Rab & Rzw) ⊃ (b ≠ z ∨ Raw)) / c
|
~(∀ w)((Rab & Rcw) ⊃ (b ≠ c ∨ Raw)) / d
|
~((Rab & Rcd) ⊃ (b ≠ c ∨ Rad))
|
Rab & Rcd
~(b ≠ c ∨ Rad)
|
Rab
Rcd
|
~(b ≠ c)
~ Rad
|
b=c
|
Rac
This branch remains open. The branch contains the names a, b, c, and d, and we know that b=c. This means
we have a domain D = {a,c,d}, and R works at least like this:
I(R) a c d
a

1 0

c

1

d
Any interpretation which has this form will make our original formula false. Here is an instance of the
quantified formula:
(Rac & Rcd) ⊃ (c ≠ c ∨ Rad)
which is false. So the original formula is false. So, here is one interpretation in which the formula is false.
I(R) a c d
a

1 1 0

c

1 1 1

d

1 1 1

Question {11.4}
Argument 1 is invalid, and argument 2 is valid.
1.
(∀ x)(Fx ⊃ Gx) \a
(∀ x)(Fx ⊃ x=a) \a
~ Ga
|
Fa ⊃ Ga
Fa ⊃ a=a
/

\

~ Fa
/

Ga
\

~ Fa

X
a=a

Both branches on the left here are open. They give rise to this interpretation
I(F) I(G)
a

0

0

The premises are true, but the conclusion is false.
2.
(∀ x)(∀ y)x=y \a
~((∀ x)Gx ∨ (∀ x)~ Gx)
|

~(∀ x)Gx / a
~(∀ x)~ Gx / b
|
~ Ga
|
~~ Gb
|
(∀ y)a=y \b
|
a=b
|
~ Gb
X
The tree closes, the argument is valid.
Question {11.5}
1. At most two frogs are blue, therefore at most three frogs are blue. (∀ x)(∀ y)(∀ z)(((Fx & Bx) & (Fy &
By) & (Fz & Bz)) ⊃ (x=y ∨ y=z ∨ x=z)) therefore,
(∀ x)(∀ y)(∀ z)(∀ w)(((Fx & Bx) & (Fy & By) & (Fz & Bz) & (Fw & Bw)) ⊃ (x=y ∨ x=z ∨ x=w ∨ y=z ∨ y=w
∨ z=w))
This is a valid argument. I will write the tree in shorthand, skipping obvious steps or running things
together (such as the long disjunctions or conjunctions).
(∀ x)(∀ y)(∀ z)(((Fx & Bx) & (Fy & By) & (Fz &
Bz)) ⊃ (x=y ∨ y=z ∨ x=z))
~(∀ x)(∀ y)(∀ z)(∀ w)(((Fx & Bx) & (Fy & By) &
(Fz & Bz) & (Fw & Bw)) ⊃ (x=y ∨ x=z ∨ x=w ∨
y=z ∨ y=w ∨ z=w))
|
~(((Fa & Ba) & (Fb & Bb) & (Fc & Bc) & (Fd &
Bd)) ⊃ (a=b ∨ a=c ∨ a=d ∨ b=c ∨ b=d ∨ c=d))
|
Fa & Ba
|
Fb & Bb
|
Fc & Bc
|
Fd & Bd
|

~(a=b ∨ a=c ∨ a=d ∨ b=c ∨ b=d ∨ c=d)
|
a≠b
a≠c
a≠d
b≠c
b≠d
c≠d
|
((Fa & Ba) & (Fb & Bb) & (Fc & Bc)) ⊃ (a=b ∨
a=c ∨ b=c)
/

\
a=b ∨
a=c ∨
b=c

~((Fa & Ba) &
(Fb & Bb) & (Fc
& Bc))
/

|

\

/

|

\

~(Fa &
Ba)

~(Fb & Bb)

~(Fc &
Bc)

a=b

a=c

b=c

X

X

X

X

X

X

2. At least two frogs are blue, therefore at least three frogs are blue. (∃ x)(∃ y)((Bx & Fx) & (By & Fy) & x
≠ y) therefore (∃ x)(∃ y)(∃ z)((Bx & Fx) & (By & Fy) & (Bz & Fz) & x ≠ y & x ≠ z & y ≠ z)
(∃ x)(∃ y)((Bx & Fx) & (By & Fy) & x ≠ y) / a,b
~(∃ x)(∃ y)(∃ z)((Bx & Fx) & (By & Fy) & (Bz & Fz) & x ≠ y & x ≠ z & y ≠ z) / a,b
|
Ba & Fa
|
Bb & Fb
|
a≠b
|
~(∃ z)((Ba & Fa) & (Bb & Fb) & (Bz & Fz) & a ≠ b & a ≠ z & b ≠ z)
|
:
:
This tree remains open. There’s nothing that we can do to close it. Here’s an interpretation read off the tree
which makes the premise true and the conclusion false. D = {a,b}
I(B) I(F)
a

1

1

b

1

1

Question {11.6}
(You don’t need to know how to do this one.)
Part 1:
a=b
f(a) ≠ f(b)
|
f(b) ≠ f(b)
X
Part 2:
f(a) = f(b) ⊬ a=b
Consider the counterexample:
D={a,b}
I(f)
a a
b a

Chapter 12: Definite Descriptions
Basic:
Question {12.1}
1. (Ix)(Vx,Fx)
2. (Ix)(Lx,Hx)
3. (Ix)(Fx & Hx,Sxb)
4. La
5. (Ix)(Lx,x=a)
6. (Ix)(Hx & ~ Fx, Scx)
7. (Ix)(Fx & Hx, (Iy)(Hy & ~ Fy,Sxy))
8. (∃ y)(Hy & ~ Fy & (Ix)(Fx & Hx, Sxy))
9. (∀ x)((Iy)(Fy & Hy, x ≠ y) ⊃ (Iy)(Fy & Hy, Sxy))
10. (Ix)(Vx, Sax) & (Ix)(Fx & Hx, Sbx)
Question {12.2}
1. (∃ x)(Vx & (∀ y)(Vy ⊃ x=y) & Fx)
2. (∃ x)(Lx & (∀ y)(Ly ⊃ x=y) & Hx)
3. (∃ x)(Fx & Hx & (∀ y)((Fy & Hy) ⊃ x=y) & Sxb)
4. La
5. (∃ x)(Lx & (∀ y)(Ly ⊃ x=y) & x=a)
6. (∃ x)(Hx & ~ Fx & (∀ y)((Hy & ~ Fy) ⊃ x=y) & Scx)
7. (∃ x)(Fx & Hx & (∀ y)((Fy & Hy) ⊃ x=y) & ((∃ y)(Hy & ~ Fy & (∀ z)((Hz & ~ Fz) ⊃ y=z) & Sxy)))
8. (∃ y)(Hy & ~ Fy & (∃ x)(Fx & Hx & (∀ z)((Fz & Hz) ⊃ x=z) & Sxy))
9. (∀ x)((∃ y)(Fy & Hy & (∀ z)((Fz & Hz) ⊃ y=z) & x ≠ y) ⊃ (∃ y)(Fy & Hy & (∀ z)((Fz & Hz) ⊃ y=z) &
Sxy))
10. (∃ x)(Vx & (∀ y)(Vy ⊃ x=y) & Sax) & (∃ x)(Fx & Hx & (∀ y)((Fy & Hy) ⊃ x=y) & Sbx)

Question {12.3}
1. (Ix)(Fx, Gx) & (Ix)(Fx, Hx), therefore (Ix)(Fx, Gx & Hx) is valid.
(Ix)(Fx, Gx) & (Ix)(Fx, Hx)
~(Ix)(Fx, Gx & Hx) / b
|
(Ix)(Fx, Gx) / a
(Ix)(Fx, Hx) / b
|
Fa
(∀ x)(Fx ⊃ x=a) \b
Ga
|
Fb
(∀ x)(Fx ⊃ x=b) \c
Hb
|
Fb ⊃ b=a
/

\

~ Fb

b=a

X

/

|

\

~(∀ x)(Fx ⊃ b=x) / c

~ Fb
X

~(Gb & Hb)

|

/

\

~(Fc ⊃ b=c)

~ Gb

~ Hb

|

|

X

Fc

~ Ga

b≠c

X

|
Fc ⊃ c=b
/

\

~ Fc

c=b

X

|
b≠b
X

2.
(Ix)(Fx & Gx, Hx), therefore (Ix)(Fx, Hx). This is invalid.
(Ix)(Fx & Gx, Hx) / a

~(Ix)(Fx, Hx) \a
|
Fa & Ga
(∀ x)((Fx & Gx) ⊃ x=a) \b
Ha
|
Fa
Ga
/

|

\

~ Fa

~(∀ x)(Fx ⊃ x=a) / b

~ Ha

X

|

X

~(Fb ⊃ b=a)
|
Fb
b≠a
|
(Fb & Gb) ⊃ b=a
/

\

~(Fb & Gb)
/

b=a
\

~ Fb

X
~ Gb

X
The remaining branch is open. We could substitute a in the universal quantifier, but that would just give us
(Fa & Ga) ⊃ a=a, which is a tautology anyway.
The open branch gives us a model D = {a,b}
I(F) I(G) I(H)
a

1

1

b

1

0

1

There’s nothing to tell us whether b has property H or not. So let’s just say Hb is false.
I(F) I(G) I(H)
a

1

1

1

b

1

0

0

3. (Ix)(Fx, Gx), therefore (∀ x)(~ Gx ⊃ ~ Fx) is valid.
(Ix)(Fx, Gx) / a
~(∀ x)(~ Gx ⊃ ~ Fx) / b

|
Fa
(∀ x)(Fx ⊃ x=a) \b
Ga
|
~(~ Gb ⊃ ~ Fb)
|
~ Gb
~~ Fb
|
Fb ⊃ b=a
/

\

~ Fb

b=a

X

|
~ Ga
X

4. (Ix)(Fx, ~ Gx), therefore ~(Ix)(Fx, Gx), is valid.
(Ix)(Fx, ~ Gx) / a
~~(Ix)(Fx, Gx)
|
(Ix)(Fx, Gx) / b
|
Fa
(∀ x)(Fx ⊃ x=a) \b
~ Ga
|
Fb
(∀ x)(Fx ⊃ x=b)
Gb
|
Fb ⊃ b=a
/

\

~ Fb

b=a

X

|
Ga
X

5. Fa, (Ix)(Fx, Gx), therefore (Ix)(Fx, x=a), is valid.

Fa
(Ix)(Fx, Gx) / b
~(Ix)(Fx, x=a) / b
|
Fb
(∀ x)(Fx ⊃ x=b) \a
Gb
|
Fa ⊃ a=b
/

\

~ Fa

a=b

X

/
~ Fb
X

|

\

~(∀ x)(Fx ⊃ x=b)

b≠a

X

|
b≠b
X

Question {12.4}
(Ix)(Fx, Fx) is “the F is an F”, which means, there’s exactly one thing with property F.

Chapter 13: Some Things do not Exist
Basic
Question {13.1}
1. The tree closes, the argument is valid.
(∃ x)Fx / a
~(∃ x)(E!x & Fx) \a
|
E!a
Fa
/

\

~ E!a

~(E!a & Fa)

X

/

\

~ E!a

~ Fa

X

X

2. The argument is invalid
(∀ x)Fx \a
~(∃ x)Fx / a
/

\

/

\

~ E!a
/

Fa
\

~ E!a

/
~ Fa

\

~ E!a

~ Fa

*

X

The other branches are open. We get an interpretation
D = {a}
I(F) I(E!)
a

1

0

from the branch with the asterisk. (Or Fa being false would be equally as appropriate, had we chosen a
different branch)
The premise is true, as we have
E!a ⊃ Fa

0 1 1
And the conclusion is false, as we have
E!a & Fa
0

0 1

Therefore, the argument is invalid.
3. The argument is invalid.
(∀ x)(Gx ⊃ E!x) \a
E!a
~~ Fa
Fa
/

\
Ga ⊃ E!a

~ E!a
X

/
~ Ga

\
E!a

The tree is open as the rightmost branch remains open. The interpretation is:
D = {a}
I(F) I(G) I(E!)
a

1

0

0

makes the premises true, as we have
E!a ⊃ (Fa ⊃ E!a)
0

1 1

0

0

and
E!a
1
and the conclusion
~ Fa
0 1
is false, showing the argument to be invalid.

There is a mistake in the question the correct version should be:
(∀ x)(Fx ⊃ E!x), ~ E!a, therefore ~ Fa (see errata)
In this case the argument is invalid, the tree is
(∀ x)(Fx ⊃ E!x) \a
~ E!a
~~ Fa
Fa
/

\
Fa ⊃ E!a

~ E!a
/
~ Fa
X

\
E!a
X

We get the following counterexample from the leftmost branch:
D = {a}
I(F) I(E!)
a

1

0

Question {13.2}
In any interpretation (Ax)(E!x ⊃ Fx) is true if and only if every instance (E!a ⊃ Fa) is true (for each a in the
domain). But this is true iff Fa is true for each a in the inner domain, and that happens iff (∀ x)Fx is true.
Similarly, (Sx)(E!x & Fx) is true if and only if some instance (E!a & Fa) is true (for some a in the domain),
and this is true iff Fa is true for some a in the inner domain, which happens iff (∃ x)Fx is true.

Chapter 14: What is a Predicate?
Basic
Question {14.1}
I think that the narrow predications appear in numbers 1, 3, 4, 5 and 7.
For 2, Superman might be famous for running very fast, but Clark Kent isn’t.
For 6, I might be sick and tired of that guy who’s been emailing me a lot, but I might not be sick and tired
of you, even though you are that person.
For 8, what I can distinguish depends on my capacity for description or whatever. I might be able to tell
Superman apart from Jimmy Olsen (a photographer for the Daily Planet), but not Clark Kent from Jimmy
Olsen.
For 9, I might know Superman but not Clark Kent.
Question {14.2}
I might think that something bit me, without there being something such that I think that it bit me. I might
think I’ve been bitten, but I might not know what bit me.

Logic Text Errata
This page contains the list of known errors in the book ’’Logic’’ (Greg Restall, published by Routledge,
2006).
Like any text, errors creep in at various stages of the process, and these can detract when reading it. So, this
page contains the list of known errors in the text. Page references are from the first edition, from 2006 (the
only so far).
Please add any errors you find in the book by editing this page. (Just click this link to do it: type in the box,
sign your changes and press “save.”)
•

•
•
•

•

Use the format If you’re not ’‘sure’’ that you’ve found an error, please add a
comment with a page reference, prefixed by a question mark, like this: “? Page
63: …”
Check the Sand Box page for tips with editing pages. Don’t worry if you make a
mistake. You can correct your errors, or someone else can.
It’s polite to ‘sign’ your work by adding a username in the box near the ‘save’
button on the form for editing the page.
Quick editing tips.
o Connectives: & (conjunction), ∨ (disjunction), ~ (negation), ⊃
(conditional), ≡ (biconditional).
o Turnstiles: ⊢ (syntactic), ⊨ (semantic), ⊬ (negated syntactic), ⊭ (negated
semantic)
o Quantifiers: ∀ (universal), ∃ (existential)
o In many of these cases, you need to have a space after the symbol for it to
be decoded properly: p ∨ q works, but /p/vv/q/ doesn’t.
If the symbols do not show up propertly in your browser, a stopgap measure is to
look at the source of the page to see what you are missing. (I do know that the
symbols work correctly using Safari v2 on Mac OS X, and in all versions of
Firefox 1.5 that I have tried.)

Thanks! Greg.

Introduction
Chapter 1: Propositions and arguments
•
•

Page 11, eight lines from the bottom. “in order to to give” should be “in order to
give”.
Page 18, Exercise {1.3}, part 1. “Critical Thinking” should be “PHIL137”.

Chapter 2: Connectives and argument forms
•

Page 29: Paragraph two, sentence one should read “The first, p~, is not a formula,
since a negation always attaches to a formula on its right” - not “on its left”.

•
•

Page 34: Exercise {2.6}, part 5: Need a space between the “ ⊃ “ and the “c”.
Page 34: Exercise {2.6}, part 14: This should be “(y ⊃ c) ⊃ p” and not “y ⊃ (c ⊃
p)” (which is a repeat of part 4).

Chapter 3: Truth tables
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Page 50: Exercise 3.2: the last formula has an extra closing bracket on the end.
Delete it.
Page 51: Exercise (3.4), part 15: There is a missing atomic proposition q. This
part should read p, q therefore q ⊃ q.

Chapter 4: Trees
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•
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Page 62: The rule for “Negated Disjunction” is misprinted. The A and B should
both be negated. The rule should read: from ~(A ∨ B) to derive ~ A and ~ B.
Page 62: The rule for “Negated Conditional” is misprinted. The B should be
negated, not the A. The rule should read: from ~(A ⊃ B) to infer A and ~ B.
*Page 67: There are two mistakes in the tree in Box 4.2. First, the first appearance
of “~ s” on the right side of the tree should be “~ q”. (To put it another way, the
mistaken “~ s” is the one that appears immediately below “~ p”). Second, in the
fourth branch (ending in r ⊃ p and a closure), the “~ q” immediately before the “r
⊃ p” should just be “q”.
Page 68: There are three mistakes in the tree in Box 4.3. First, the leftmost branch
should be closed. Second, at the end of the second branch from the left, “~q”
should be “q”. Third, at the end of the third branch from the left, “q” should be
“~q”.
Page 69: paragraph before “Proof of fact 1”. All of the capital ‘X’s should be in
italics.

Chapter 5: Vagueness and bivalence
Chapter 6: Conditionality
•

Page 89: First paragraph: “…from John Howard is Prime Minister of Australia to
if John Howard is loses the election, John Howard is Prime Minister of
Australia.” should probably be something like “if John Howard loses the
election,...”.

Chapter 7: Natural deduction
•

Pages 103-4: The description of ∨-elimination on p. 104 doesn’t match the
sequent formulation on p. 103. To match Y,B ⊢ C should be X,B ⊢ C. In the
description on p. 104 the phrase “… provided you’ve got both X too)” is more
than a little odd. It might make better sense to have X,A ⊢ C, Y,B ⊢ C and Z ⊢ A
vv B above the line and X,Y Z ⊢ C below the line on p. 103, then make

•

appropriate changes in the first paragraph on p. 104, including changing “…
provided you’ve got both X too)” to “… provided you’ve got both X and Y too)”.
(The thought behind the restriction on axiom sequents mooted on p. 109 would
suggest this form.)
Page 109: Second paragraph: All GBP symbols (£) ought to be turnstiles: ⊢ .

? Page 109, second paragraph. “Then there is no way to deduce A ⊢ B → A since the B was not used in the
deduction of A.” What’s wrong with this deduction (in the style of Lemmon’s Beginning Logic in which
the Rule of Assumptions is, in effect, the axiom A ⊢ A)?
1 A ⊢ A Axiom 2 B ⊢ B Axiom 3 A,B ⊢ A & B &-I 4 A,B ⊢ A &-E 5 A ⊢ B → A ->-I
BTW, why are the deductions in this chapter given using A, B, C, D rather than p, q, r, s?

Chapter 8: Predicates, names and quantifiers
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•

Page 115: “…is a multiple of ten…” should be “…is a multiple of ten”.
Page 116: 6 lines from the bottom: “p is the predicate ‘is a philosopher’” should
be “P is the predicate ‘is a philosopher’”.
Page 119: Third example formula should be (∃x)(Px & (∃y)(Ly & Kxy)).

? Page 120: Add clause: any atomic formula is a formula. Cf. p. 28. Without it a lot hangs on the
italicization of formula in atomic formula.
? Page 120: The Latinate ‘arity’ naturally goes along with unary, binary, ternary, …. With the Greek
monadic, dyadic, triadic, … shoudln’t it be ‘adicity’?
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•

Page 123: Second para. “Table 8.1 gives are some more examples…” should be
“Table 8.1 gives some more examples …”
Page 124: Ex. 8.2 17 should be (∀x)(Px ⊃ (∀y)Lxy ⊃ ~ Lyx)

Chapter 9: Models for predicate logic
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•

Page 129: Square brackets in [Descartes, Kant] should be curly brackets:
{Descartes, Kant}.
Page 143, Box 9.2 The second formula in the tree should be “∃ x ~Gx”, not “∀ x
~Gx”.
Page 145, Summary, point 5. “∀ x” should be “∃ x”
Page 146, Ex. 9.1 “(∀ x)( Rxa ⊃ (∃ y)(Py & Rxy))” between the specification of
the model and “evaluate the following formulas:” should be deleted.
Page 146, Ex. 9.1 question 1 should be “(∀ x)(Tax ⊃ Gx)”, not “(∀ x)(Tx ⊃ Gx)”,
where I(a)=a.

Chapter 10: Trees for predicate logic
•

Page 165, Ex 10.1 (4) The solutions on this site report that the main connective of
exercise 10.1 (4) is a biconditional. In the book it is a material conditional.

•

Page 165, Ex 10.1 (5) The question should be (∀ x)(Fx ⊃ Gx) ⊃ ((∀ x)Fx ⊃ (∀
x)Gx), not (∀ x)Fx(Fx ⊃ Gx) ⊃ ((∀ x)Fx ⊃ (∀ x)Gx)

Chapter 11: Identity and functions
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•

•
•
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Page 176, bottom: the interpretation given is g(b,c) = c. It should be g(b,c) = a.
Page 177, last tree: The formulas at the top of the tree should be centred on the
formulas themselves (with the tags for names substituted in hanging off to the
right, as with the other trees).
Page 177, end: The displayed tree should end “tree continued on next page.”
Page 181, Exercise {11.10} in 5 and 7 the variable y should be bound by a
universal quantifier.
Page 182, Exercise {11.10}, part 5. It should read “RA ⊢ (∀ x)(∀ y)(x + y = y +
x)” and not “RA < …”

Chapter 12: Definite descriptions
•

Page 187, the middle branch for the tree rule for “~I(Fa,Ga)” should be “~(∀
y)(Fy ⊃ y=a)” not “(∀ y)(Fy ⊃ y=a)”.

Chapter 13: Some things do not exist
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Page 199, under the heading “Universal” the rule should take “(∀ x)A” to “~ E!a”
or “A(x:=a) (any a)”. Under the heading “Negated Universal” the rule should take
“~(∀ x)A” to “E!a” and “~ A(x:=a) (a new)”.
Page 201, Tree in Box 13.1: The branch ending “~E!a” should end in a closure.
Page 203, Question 13.1.3 should be (∀ x)(Fx ⊃ E!x), ~ E!a, therefore ~ Fa.

Chapter 14: What is a predicate?
Chapter 15: What is logic?