Logic

1

Same as in-class exercise 1

2

$p$	$q$	$\neg p$	$\neg p \land q$	$p \lor (\neg p \land q)$	$p \lor q$	$p \lor (\neg p \land q) \to p \lor q$
T	T	F	F	T	T	T
T	F	F	F	T	T	T
F	T	T	T	T	T	T
F	F	T	F	F	F	T

3

$p$	$q$	$\neg p$	$\neg q$	$p \lor q$	$(p \lor q) \land \neg p$	$((p \lor q) \land \neg p) \land q$	$q \land \neg q$	$(((p \lor q) \land \neg p) \land q) \to q \land \neg q$
T	T	F	F	T	F	F	F	T
T	F	F	T	T	F	F	F	T
F	T	T	F	T	T	T	F	F
F	F	T	T	F	F	F	F	T

4

Same as in-class exercise 1

5

First attempt

(p \land q \land r) \lor (p \land \neg q \land r) \lor (\neg p \land \neg q \land r) \lor (\neg p \land q \land r) \equiv ((p \land r) \land (q \lor \neg q)) \lor (\neg p \land \neg q \land r) \lor (\neg p \land q \land r) \equiv ((p \land r) \land T) \lor (\neg p \land \neg q \land r) \lor (\neg p \land q \land r) \equiv (p \land r) \lor (\neg p \land \neg q \land r) \lor (\neg p \land q \land r) \equiv (r \land (p \lor (\neg p \land \neg q))) \lor (\neg p \land q \land r) \equiv (r \land ((p \lor \neg p) \land (p \lor \neg q))) \lor (\neg p \land q \land r) \equiv (r \land (T \land (p \lor \neg q))) \lor (\neg p \land q \land r) \equiv (r \land (p \lor \neg q)) \lor (\neg p \land q \land r) \equiv r \land ((p \lor \neg q) \lor (\neg p \land q)) \equiv r \land (p \lor \neg q \lor (\neg p \land q)) \equiv r \land (p \lor ((\neg q \lor \neg p) \land (\neg q \lor q))) \equiv r \land (p \lor ((\neg q \lor \neg p) \land T)) \equiv r \land (p \lor \neg q \lor \neg p) \equiv r \land ((p \lor \neg p) \lor \neg q) \equiv r \land (T \lor \neg q) \equiv r \land T \equiv r

Second attempt

(p \land q \land r) \lor (p \land \neg q \land r) \lor (\neg p \land \neg q \land r) \lor (\neg p \land q \land r) \equiv r \land ((p \land q) \lor (p \land \neg q) \lor (\neg p \land \neg q) \lor (\neg p \land q)) \equiv r \land ((p \land (q \lor \neg q)) \lor (\neg p \land (\neg q \lor q))) \equiv r \land ((p \land F) \lor (\neg p \land F)) \equiv r \land (F \lor F) \equiv r

6

p \to (q \to r) \equiv p \to (\neg q \lor r) \equiv \neg p \lor (\neg q \lor r) \equiv \neg p \lor \neg q \lor r \equiv \neg (p \land q) \lor r \equiv (p \land q) \to r

7

(p \to q) \to r \equiv (\neg p \lor q) \to r \equiv \neg (\neg p \lor q) \lor r \equiv (p \land \neg q) \lor r \equiv (p \lor r) \land (\neg q \lor r) \equiv (p \lor r) \land (q \to r)

8

(p \lor q) \to r \equiv \neg (p \lor q) \lor r \equiv (\neg p \land \neg q) \lor r \equiv (\neg p \lor r) \land (\neg q \lor r) \equiv (p \to r) \land (q \to r)

9

p \to (q \lor r) \equiv \neg p \lor (q \lor r) \equiv (\neg p \lor q) \land (\neg p \lor r) \equiv (p \to q) \land (p \to r)

10

(2 + 2 = 7) \leftrightarrow (5 + 3 = 4) \equiv F \leftrightarrow F \equiv T

11

Let the universe be $Z^{+}$

a

$\exists x (\forall y (x^{2} < y + 1))$

For $x = 1$ , $x^{2} < y + 1$ , since $y + 1 > 1$

True

b

$\forall x (\exists y (x^{2} + y^{2} < 12))$

For $x = 4$ , $x^{2} + y^{2} \geq 12$

False

c

$\forall x (\forall y (x^{2} + y^{2} > 0))$

For $x > 0 \land y > 0$ , $x^{2} + y^{2} \geq 1 + 1 = 2 > 0$

True

12

$p$	$q$	$p \to q$
T	T	T
T	F	F
F	T	T
F	F	T

when $p \to q$ , when $q$ is $T$ , $p$ can be $F$ , therefore $q$ does not implies $p$

13

Prove that $p \to \neg q$ and $q$ imply $\neg p$

p \to \neg q q \to \neg p q ∴ \neg p Premise 1 Contrapositive Premise 2, 3 Modus ponens 123

14

Prove that $p \to \neg q$ , $r \to q$ and $r$ imply $\neg p$

p \to \neg q r \to q q \to \neg p r \to \neg q r ∴ \neg q Premise Premise 1 Contrapositive 2, 3 Hypothetical syllogism Premise 4, 5 Modus ponens 12345

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Logic

1

2

3

4

5

6

7

8

9

10

11

a

b

c

12

13

14

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