Question

Prove [(p->q )^p] ->q is a tautology​

Answer 1

Step-by-step explanation:

p->q = ~p v q

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

= (~p^p) v(q^p) = F v (q^p) = q^p

~p^p = F ,implies its false for all the combinations of t and f

and False or any statement = statement

((p-> q)^p)-> q

(q^p) ->q

~(q^p)vq

(~qv~p) v q

(~qvq) v(~pvq)

T v (~pvq)

~q v q= True in all cases

T, true or any statement = True

T v(~pvq) = T for all cases

as its true for all the combinations, it is tautology