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Solutions to Exercises A
Solution 1: Symbolic connectives
- (a) 68 ≠ (^34) (b) 1216 = 68 = 34 , that is ‘ 1216 = 68 and^68 = 34 ’. Only (b) is true.
- r ∧ s
- p is ‘ Two is a prime number ’ q is ‘ Two is an even number ’ p ∨ q
Solution 2: Arithmetic operators
- 3 + 1 + 4 or, more accurately, 3 + ( 1 + 4 ). Left operand = 3 and right operand = 5 so result = 3 + 5 = 8.
- ( 2 + 3 ) − ( 7 − 5 ) = 5 − 2 = 3 but ( 7 − 5 ) − ( 2 + 3 ) = 2 − 5 = −3, which is different.
Solution 3: Conjunction
- ‘ My cat is black but your cat is white ’ ‘ My cat is black ’ ∧ ‘ Your cat is white ’ = T F ∧ F = T F
- ‘ Shakespeare wrote both Hamlet and MacBeth’ ‘ Shakespeare wrote Hamlet ’ ∧ ‘ Shakespeare wrote MacBeth’ = T T ∧ T = T T
- q ∧ p = T T ∧ F = T F
- 2 × 7 = 27 ∧ 3 2 = 9 = T F ∧ T = T F
- 3 > 2 ∧ 3 > 2 = T T ∧ T = T T
Solution 4: Negation
- ‘ The moon is not made of blue cheese ’ = T ¬F = T T.
- ‘1234 is an not even number ’ = T ¬T = T F. Note that ‘1234 is an odd number ’ would be incorrect answer to this ques- tion as it depends upon the additional fact that an odd number is one which is not even.
Solution 5: Disjunction
- ‘ Either you broke the window or else I’m a Martian ’ = T F ∨ F = T F. Note that in general use this is just a more emphatic way of saying ‘you broke the window’.
- ‘ Either Shakespeare or Francis Bacon wrote Hamlet ’ = T T ∨ F = T T.
- r 1 ∨ r 2 = T F ∨ T = T T
- ‘ Either 2 is even or 3 is odd ’ = T T ∨ T = T T
Solution 6: Connective schemas
- (a) ‘ Rex does not have a wet nose ’ (b) 2 + 3 ≠ 8 (c) ¬ p (d) ¬ q (e) ¬ r 2 (f) ‘ Rex is black but Rover is white ’ (g) ‘ Either 2 + 3 = 8 or 3 2 = 9’ (h) p ∧ q (i) p 1 ∧ p 2 (j) r 2 ∨ q
- The following are possible answers. Note how some of the meaning of the original may be lost. (a) P ∧ Q where - P is interpreted as ‘ This book is long ’; - Q is interpreted as ‘ I read it quickly ’. (b) P ∨ Q where - P is ‘ There is a hole in the exhaust ’ - Q is ‘ A bracket has worked loose ’. (c) P ∨ Q where - P is ‘ Lunch will be served during the flight ’. - Q is ‘ Dinner will be served during the flight ’. (d) ¬P where P is 3 × 6 = 7
- ¬¬¬¬ ((p ∨ q) ∧ r )
- (( ¬ p) ∧ ( ¬ q)) ∨∨∨ ( ¬ r )
Solution 9: Compound propositions from parse trees
- q ∧ (p ∧ r )
- ¬¬ (p ∧ q)
Solution 10: Connective priorities
- (a) ¬ ( ¬ p) (b) ( ¬ p) ∧ q (c) q ∧ ( ¬ r ) (d) ( ¬ p) ∨ q (e) (p ∧ q) ∧ ( ¬ r ) (f) (p ∨ q) ∨ ( ¬ r ) (g) (q ∧ ( ¬ p)) ∨ q (h) (p 1 ∨ p 2 ) ∨ ( ¬ p 3 ) (i) 3 > 0 ∨ (( ¬ 1 + 1 = 2 ) ∧ 2 + 3 = 5 ) (j) ( ¬‘ Fido has three legs ’ ) ∨ ‘ Rex has four legs ’ (k) (p ∧ q) ∨ (( ¬ r ) ∧ p) (l) ( ¬ p) ∧ (( ¬ q) ∨ (p ∧ r )) (m) ¬ ( ¬ (( ¬ p 2 ) ∧ ( ¬ p 1 )))
- The main connective is shown in bold. (a) ¬ (p 1 ∧ p 2 ) ∨∨∨ p 3 (b) ‘ Rex has four legs ’ ∨ ‘ Fido has three legs ’ ∨∨∨ ¬‘ Rover has a wet nose ’ (c) ¬¬¬¬ ¬ q (d) ¬ 1 + 1 = 2 ∨∨∨ ( 1 + 1 )^2 = 2 2 (e) ¬¬¬¬ (p ∨ q ∧ r ∨ ¬ s) (f) ¬ ( ¬ ( ¬ p 1 ∨ p 2 ∧ p 3 ) ∨ ¬ p 4 ) ∧∧∧ ¬ (p 5 ∧ p 6 ∨ ¬¬ p 7 )
Solution 11: Removing parentheses
- ¬ p ∨ q
- p 1 ∨ ¬ p 2 ∨ p 3
- p ∧ ¬ q ∨ ¬ r
- ¬ q 1 ∧ (q 1 ∨ q 2 ) ∨ q 1 ∧ q 2 , though perhaps ¬ q 1 ∧ (q 1 ∨ q 2 ) ∨ (q 1 ∧ q 2 ) is easier to read
- r ∧ ( ¬ p ∨ q)
Solution 12: Truth values of compound propositions
1. ¬ ( ¬T ) = T ¬F = T T
2. ¬T ∨ F = T F ∨ F = T F
3. ( F ∨ F ) ∨ F = T F ∨ F = T F
4. ¬¬ ( T ∨ ¬F ) ∧ ¬ ( F ∨ F )
= T ¬¬ ( T ∨ T ) ∧ ¬F
= T ¬¬T ∧ T
= T ¬F ∧ T
= T T ∧ T
= T T
5. ¬F ∨ T = T T ∨ T = T T
6. ¬F ∧ ( T ∨ ¬T ) = T T ∧ ( T ∨ F ) = T T ∧ T = T T
Solution 13: Compound propositions from propositional forms
- ‘ Rex has four legs and either Fido has three legs or Rover does not have a wet nose. ’
- (a) ‘ It is not cold but it is snowing. ’ (b) ‘ Either the water pump is not working or there is no anti-freeze in the radiator and last night was very cold. ’
3. Q ¬Q ¬¬Q ¬¬¬Q
T F T F
F T F T
4. P Q ¬Q P ∨ ¬Q ¬ ( P ∨ ¬Q )
T T F T F
T F T T F
F T F F T
F F T T F
5. P Q ¬Q P ∧ ¬Q
T T F F
T F T T
F T F F
F F T F
6. P 1 P 2 P 3 P 1 ∧ P 2 ( P 1 ∧ P 2 ) ∧ P 3
T T T T T
T T F T F
T F T F F
T F F F F
F T T F F
F T F F F
F F T F F
F F F F F
7. P Q R P ∧ R Q ∧ ( P ∧ R )
T T T T T
T T F F F
T F T T F
T F F F F
F T T F F
F T F F F
F F T F F
F F F F F
8. P Q R P ∨ Q ( P ∨ Q ) ∨ R
T T T T T
T T F T T
T F T T T
T F F T T
F T T T T
F T F T T
F F T F T
F F F F F
9. P Q R Q ∨ R P ∨ ( Q ∨ R )
T T T T T
T T F T T
T F T T T
T F F F T
F T T T T
F T F T T
F F T T T
F F F F F
10. P Q R P ∨ Q ( P ∨ Q ) ∧ R
T T T T T
T T F T F
T F T T T
T F F T F
F T T T T
F T F T F
F F T F F
F F F F F
11. P Q R P ∧ R Q ∧ R ( P ∧ R ) ∨ ( Q ∧ R ))
T T T T T T
T T F F F F
T F T T F T
T F F F F F
F T T F T T
F T F F F F
F F T F F F
F F F F F F
16. P Q R Q ∨ R P ∧ ( Q ∨ R ) ¬ ( P ∧ ( Q ∨ R ))
T T T T T F
T T F T T F
T F T T T F
T F F F F T
F T T T F T
F T F T F T
F F T T F T
F F F F F T
17. P 1 P 2 P 3 ¬P 2 ¬P 1 ¬P 2 ∧ ¬P 1 ¬P 3 ¬P 3 ∨ ( ¬P 2 ∧ ¬P 1 )
T T T F F F F F
T T F F F F T T
T F T T F F F F
T F F T F F T T
F T T F T F F F
F T F F T F T T
F F T T T T F T
F F F T T T T T
- See Figure A.
19. P Q R ¬P ¬Q ¬Q ∧ R ¬P ∨ ( ¬Q ∧ R )
T T T F F F F
T T F F F F F
T F T F T T T
T F F F T F F
F T T T F F T
F T F T F F T
F F T T T T T
F F F T T F T
( P ∨
( Q ∧ R
( P ∨
( Q ∧ R
∨^
P
Q
R
Q ∧ R
P ∨
( Q ∧ R
)^
( P ∨
( Q ∧ R
Q ∨ R
( Q ∨ R
)^
P ∧ ¬
( Q ∨ R
)^
( P ∧ ¬
( Q ∨ R
( P ∧ ¬
( Q ∨ R
T^
T^
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Figure A.1: Solution to Exercise 15, Question 18
Solution 21: Using laws of equivalence
1. ¬P ∧ P
= T P ∧ ¬P 〈 A ∧ B = T B ∧ A 〉
= T F 〈 A ∧ ¬A = T F 〉
2. ¬ ( P ∨ ¬Q ) ∨ ( P ∨ ¬Q )
= T ( P ∨ ¬Q ) ∨ ¬ ( P ∨ ¬Q ) 〈 A ∨ B = T B ∨ A 〉
= T T 〈 A ∨ ¬A = T T 〉
3. ( P ∧ Q ) ∨ P
= T P ∨ ( P ∧ Q ) 〈 A ∨ B = T B ∨ A 〉
= T P 〈 A ∨ A = T A 〉
4. ( P ∧ Q ) ∧ P
= T P ∧ ( P ∧ Q ) 〈 A ∧ B = T B ∧ A 〉
= T ( P ∧ P ) ∧ Q 〈 A ∧ ( B ∧ C ) = T ( A ∧ B ) ∧ C 〉
= T P ∧ Q 〈 A ∧ A = T A 〉
5. ( P ∨ ¬P ) ∧ ¬ ( Q ∧ ¬Q )
= T T ∧ ¬ ( Q ∧ ¬Q ) 〈 A ∨ ¬A = T T 〉
= T ¬ ( Q ∧ ¬Q ) ∧ T 〈 A ∧ B = T B ∧ A 〉
= T ¬ ( Q ∧ ¬Q ) 〈 A ∧ T = T A 〉
= T ¬Q ∨ ¬¬Q 〈 ¬ ( A ∧ B ) = T ¬A ∨ ¬B 〉
= T ¬Q ∨ Q 〈 ¬¬A = T A 〉
= T Q ∨ ¬Q 〈 A ∨ B = T B ∨ A 〉
= T T 〈 A ∨ ¬A = T T 〉
6. ¬T
= T ¬ ( P ∨ ¬P ) 〈 A ∨ ¬A = T T 〉
= T ¬P ∧ ¬¬P 〈 ¬ ( A ∨ B ) = T ¬A ∧ B 〉
= T ¬P ∧ P 〈 ¬¬A = T A 〉
= T P ∧ ¬P 〈 A ∧ B = T B ∧ A 〉
= T F 〈 A ∧ ¬A = T F 〉
7. F
= T ¬¬T 〈 Qu. 6 〉 = T T 〈 ¬¬A = T A 〉 Note that an earlier proof has been quoted in order to avoid repeating the steps used in that proof – the earlier proof can be viewed as a ‘subprogram’ called by the later proof.
Solution 22: Sets of propositional forms
- Set of two propositional forms, namely P ∧ Q and ¬Q.
- Set of two propositional forms, namely P ∧ Q and ¬Q.
- Set of one propositional form, namely ¬Q.
- The propositional form ¬Q.
- Set of one truth value, namely F.
- The truth value F.
- General expression for a set of two propositional forms, one of which has ∧ as the main connective and the other of which has ¬ as the main connective.
- General expression for a set of two propositional forms, one of which has ∧ as the main connective and the other of which has ¬ as the main connective.
- General expression for a set of two propositional forms, one of which has ∨ as the main connective with F as the right disjunct and the other of which has ∧ as the main connective with left conjunct T.
- A propositional form with ∧ as the main connective.
- Set of one propositional form, namely Q ∧ R.
- Does not represent anything meaningful in our notation – it is not a well formed expression.
- Set containing at least one propositional form.
- Set containing at least one propositional form; if A and B are different, then the set contains at least two propositional forms.
- Does not represent anything meaningful as A is not a set – it is not a well formed expression.
- A set with one propositional form, namely P 1 ∨ ¬P 2
Solution 23: Semantic entailment
- (a) P Q P ∧ Q P ∨ Q T T T T T F F T F T F T F F F F (b) P 1 P 2 ¬ ( P 1 ∨ P 2 ) ¬ ( P 2 ∧ P 1 ) T T F F T F F T F T F T F F T T
Solution 26: Validity of arguments
The characteristic argument form is given for each argument, from which it is possible to decide whether or not the argument is valid.
- P ∨ Q ∴ P ∧ Q (not valid).
- P ∧ Q ∴ P ∨ Q (valid).
- P ∧ Q ∴ P ∨ R (valid).
- P ∧ Q ∴ P (valid).
- P , Q ∴ P ∧ Q (valid).
- P , Q ∴ P ∨ Q (valid).
- P ∴ P ∨ Q (valid).
- P ∴ Q ∨ P (valid).
- P ∴ Q ∨ R (not valid).
- ¬¬ ( P ∧ ¬Q ) ∴ P ∧ ¬Q (valid). Note that it is also possible to use the fact that ¬¬P ∴ P is a valid argument form to show that the argument itself is valid.
Solution 27: Deduction rules
- In each case the instances of P and Q are given together with the corres- ponding inference form. (a) P : p , Q : q ∨ r , P , Q P ∧ Q. (b) P : p , Q : q ∨ r , P ∧ Q P. (c) P : p , Q : q ∨ r , P ∧ Q Q. (d) P : ¬¬¬ q , Q : p ∨ ¬¬ q , P ∧ Q Q. (e) P : ¬¬¬ q , Q : p ∨ ¬¬ q , P ∧ Q P. (f) P : p , Q : q ∧ r , P P ∨ Q. (g) P : ¬ q ∨ r , Q : ¬ p ∨ s , Q P ∨ Q. (h) P : ¬ p , ¬¬P P. (i) P : ¬¬ p , ¬¬P P. (j) P : (p ∧ (r ∨ q)) , ¬¬P P. (k) P : ¬¬ p , P ¬¬P. (l) P : ¬ p , P ¬¬P. (m) P : (p ∨ (r ∧ q)) , P ¬¬P.
- For each question, the appropriate rule of deduction is given followed by the necessary instantiations and then the inference form which constitutes that rule of deduction. (a) ∧I , P : ‘ The sky is blue ’, Q : ‘ Grass is green ’, P , Q P ∧ Q. (b) ∨I 1 , P : p 1 , Q : p 2 , P P ∨ Q. (c) ¬¬E , P : ‘ Fido has three legs ’, ¬¬P P. (d) ∧E 2 , P : ‘73 is prime ’, Q : ‘73 is odd ’, P ∧ Q Q.
(e) ¬¬I , P : ‘ The sky is blue ’, P ¬¬P. (f) ∨I 1 , P : ‘71 is prime ’, Q : ‘26 is prime ’, P P ∨ Q. (g) ∨I 2 , P : ‘ Rex has four legs ’, Q : ‘ Rex has a wet nose ’, Q P ∨ Q. (h) ∧E 1 , P : r , Q : s , P ∧ Q P. (i) ∨I 2 , P : 3 2 = 9 , Q : 1 + 1 = 2 ∧ 2 × 3 = 6 , Q P ∨ Q. (j) ¬¬E , P : ‘ Roses are red ’, Q : ‘ Violets are blue ’, ¬¬P P.
Solution 28: Tabular derivations
1. 1 P ∧ Q P ∧ E 1
2 P ¬¬P ¬¬I
3 P ∧ Q ¬¬P 1 , 2
2. 1 ¬¬P P ¬¬E
2 P , Q P ∧ Q ∧ I
3 ¬¬P , Q P ∧ Q 1 , 2
3. 1 ¬¬ ( P ∨ Q ) P ∨ Q ¬¬E
2 P ∨ Q ( P ∨ Q ) ∨ ¬¬ ( P ∧ Q ) ∨ I 1
3 ¬¬ ( P ∨ Q ) ( P ∨ Q ) ∨ ¬¬ ( P ∧ Q ) 1 , 2
4. 1 ¬¬A A ¬¬E
2 A , B A ∧ B ∧ I
3 ¬¬A , B A ∧ B 1 , 2
4 ¬¬B B ¬¬E
5 ¬¬A , ¬¬B A ∧ B 4 , 3
Solution 29: Deduction trees
- (a) P ∧ Q P
∧E 1
¬¬P
¬¬I
(b) ¬¬¬Q ∧ ( P ∨ ¬¬Q ) ¬¬¬Q
∧E 1
¬Q
¬¬E
(c) P ∧ Q P
∧E 1
P ∨ Q
∨I 1
(d) ¬¬P ∧ Q ¬¬P
∧E 1
P
¬¬E
¬¬P ∧ Q
Q
∧E 2
P ∧ Q
∧I
(e) ¬¬P ∧ ¬¬Q ¬¬P
∧E 1
P
¬¬E
¬¬P ∧ ¬¬Q
¬¬Q
∧E 2
Q
¬¬E
P ∧ Q
∧I
(f) ¬¬ ( P ∧ Q ) P ∧ Q
¬¬E
Q
∧E 2
P ∨ Q
∨I 2
¬¬ ( P ∨ Q )
¬¬I
or, alternatively,
¬¬ ( P ∧ Q ) P ∧ Q
¬¬E
P
∧E 1
P ∨ Q
∨I 1
¬¬ ( P ∨ Q )
¬¬I
(g) ¬¬ ( P ∧ Q ) P ∧ Q
¬¬E
P
∧E 1
¬¬P
¬¬I
¬¬P ∨ ¬¬Q
∨I 1
or, alternatively,
¬¬ ( P ∧ Q ) P ∧ Q
¬¬E
Q
∧E 2
¬¬Q
¬¬I
¬¬P ∨ ¬¬Q
∨I 2
(h) ¬¬ ( P ∧ Q ) P ∧ Q
¬¬E
P
∧E 1
¬¬P
¬¬I
¬¬ ( P ∧ Q )
P ∧ Q
¬¬E
Q
∧E 2
¬¬Q
¬¬I
¬¬P ∧ ¬¬Q
∧I
(i) ¬¬P ∧ ¬¬Q ¬¬P
∧E 1
P
¬¬E
¬¬P ∧ ¬¬Q
¬¬Q
∧E 2
Q
¬¬E
P ∧ Q
∧I
¬¬ ( P ∧ Q )
¬¬I
- (a) P ∧ Q P
∧E 1
Q ∧ P
∨I 2
(b) P ∧ ¬Q ¬Q
∧E 2
P ∧ ¬Q
P
∧E 1
¬Q ∧ P
∧I
(c) P Q P ∧ Q
∧I
R ∨ P ∧ Q
∨I 2
Solution 30: Using ¬I and ∨E
P ∧ Q
Q
∧E 2
¬Q
Q ∧ ¬Q
∧I
¬ ( P ∧ Q )
∗¬I
P ∧ Q
P
∧E 1
¬P ∧ R
¬P
∧E 1
P ∧ ¬P
∧I
¬ ( ¬P ∧ R )
∗¬I
