Kenneth Rosen Edition 7 Exercise 1.3 Question 9 (Page No. 35)

Question

Show that each of these conditional statements is a tautology by using truth tables.

1. $(p \wedge q) \rightarrow p $
2. $ p \rightarrow (p \vee q)$
3. $\sim p \rightarrow (p \rightarrow q)$
4. $(p\wedge q)\rightarrow (p \rightarrow q)$
5. $\sim (p \rightarrow q) \rightarrow p$
6. $\sim(p \rightarrow p) \rightarrow \sim q$

Answer 1

a)

p	q	p^q	~(p^q)	(p^q)->p=~(p^q) v p
0	0	0	1	1
0	1	0	1	1
1	0	0	1	1
1	1	1	0	1

 

b)

p	q	p v q	p->(p v q)= ~p v (p v q)
0	0	0	1
0	1	1	1
1	0	1	1
1	1	1	1

 

c)

p	q	p->q= ~p v q	~p->(p->q)=p v (p->q)
0	0	1	1
0	1	1	1
1	0	0	1
1	1	1	1

 

d)

p	q	p^q	p->q	(p^q)->(p->q)= ~(p^q) + (p->q)
0	0	0	1	1
0	1	0	1	1
1	0	0	0	1
1	1	1	1	1

 

e)

 

p	q	p->q	~(p->q)->p = (p->q) + p
0	0	1	1
0	1	1	1
1	0	0	1
1	1	1	1

 

f)

 

p	q	~(p->p)=~(~p v p)=0	~(p->p)-> ~q = (p->p) v ~q= 1 v q=1
0	0	0	1
0	1	0	1
1	0	0	1
1	1	0	1

 

Here are the answers. :)