Show that each of these conditional statements is a tautology by using truth tables.
a)
p
q
p^q
~(p^q)
(p^q)->p=~(p^q) v p
0
1
b)
p v q
p->(p v q)= ~p v (p v q)
c)
p->q= ~p v q
~p->(p->q)=p v (p->q)
d)
p->q
(p^q)->(p->q)= ~(p^q) + (p->q)
e)
~(p->q)->p = (p->q) + p
f)
~(p->p)=~(~p v p)=0
~(p->p)-> ~q = (p->p) v ~q= 1 v q=1
Here are the answers. :)