Verify-that-p-q-p-q-is-tautology-using-laws-of-algebra-

Question Number 191735 by Spillover last updated on 29/Apr/23

V e r i f y t h a t

⌝ (p \to q) \to (p \land^{⌝} q) i s t a u t o l o g y u s i n g l a w s o f

a l g e b r a

Answered by manxsol last updated on 30/Apr/23

\sim (\sim p q) \to (p \sim q)

(\sim p q) (p \sim q)

(\sim p q) p} (\sim p q \sim q)

{V q} {V q)

V V = V t a u t o l o g i a

p \to q =\sim p q

d i s t r i b u t i v a

p (q r) = p q) (p r)

Answered by aba last updated on 30/Apr/23

\begin{array}{ccccc} p & q & ⌉ (p \to q) & p \land ⌉ q & ⌉ (p \to q) \to (p \land ⌉ q) \\ 0 & 0 & 0 & 0 & 1 \\ 0 & 1 & 0 & 0 & 1 \\ 1 & 0 & 1 & 1 & 1 \\ 1 & 1 & 0 & 0 & 1 \end{array}

Commented by Spillover last updated on 30/Apr/23

r e a d t h e q u e s t i o n c a r e f u l l . t h e q u e s t i o n s a y s

u s e t h e l a w s o f a l g e b r a . T h a n k s f o r y o u r s o l u t i o n

Commented by aba last updated on 30/Apr/23

i know

Answered by Spillover last updated on 30/Apr/23

\sim (p \to q) \to (p \land \sim q)

\sim [(\sim p \lor q) \lor \sim (p \land \sim q)

p \land \sim q \lor \sim p \lor q

p \lor \sim p \sim q \lor q

T \lor T

T