Question

Discrete math question

Prove that

¬(q→p)∧(p∧q∧s→r)∧p is a contradiction without using truth table

Answer 1

Given ¬(q   p) (pqs   r) p   Contradiction

Take L.H.S ¬(q   p) (pqs   r) p

¬(¬q V p) [¬(pqs) V r] p    { Law of Implies (PQ) = ¬PVQ }

[¬(¬q) ¬p] [(¬p V ¬q V ¬s) V r] p    {By De Morgan's law ¬ (P Q) = ¬P V ¬Q || ¬ (P V Q) = ¬P ¬Q }

[q ¬p] [(¬p V ¬q V ¬s) V r] p    {By De Morgan's law ¬ (¬P) = P }

[q ¬p] [(¬q V ¬p V ¬s) V r] p   {Commutative law (P V Q) = (Q V P)}

[q ¬q] [(¬p V ¬p) V ¬s V r] p   {Commutative law (P Q) = (Q P)}

[q ¬q] [¬p V (¬s V r)] p    { we know that A V A = A }

[q ¬q] [¬p V p] (¬s V r)   {Commutative law (P Q) = (Q P)}

[F] [T] (¬s V r)  { we know that P ¬P = F & P V ¬P = T }

[F] ¬s (T V r)   {Commutative law (P Q) = (Q P)}

[F ¬s] (T V r)   {Associative law (P Q) R = P (Q R)}

[F] (T) { we know that F ¬P = F & A V T = T }

[F T ]  {Associative law (P Q) R = P (Q R)}

[F]   { we know that T F = F }

Contradiction

R.H.S

L.H.S R.H.S

So ¬(q   p) (pqs   r) p  is a Contradiction