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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. $\sim(p \rightarrow p) \rightarrow \sim q$
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## 1 Answer

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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. :)

by Junior (869 points)

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