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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. $\sim(p \rightarrow p) \rightarrow \sim q$
asked ago in Mathematical Logic by Loyal (8.8k points) | 9 views

1 Answer

0 votes

a)

p

q

p^q

~(p^q)

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

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b)

p

q

p v q

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

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c)

p

q

p->q= ~p v q

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

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d)

p

q

p^q

p->q

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

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e)

 

p

q

p->q

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

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f)

 

p

q

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

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

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Here are the answers. :)

answered ago by Junior (547 points)

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