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Show that p(double arrow) q and (pq)(¬p¬q) are logically equivalent without usin

ID: 1719742 • Letter: S

Question

Show that p(double arrow) q and (pq)(¬p¬q) are logically equivalent without using truth tables. To do this, either show that both are true, or that both are false, for exactly the same combination of truth values of the propositional variables in these expressions (whichever is easier). Show that p(double arrow) q and (pq)(¬p¬q) are logically equivalent without using truth tables. To do this, either show that both are true, or that both are false, for exactly the same combination of truth values of the propositional variables in these expressions (whichever is easier). Show that p(double arrow) q and (pq)(¬p¬q) are logically equivalent without using truth tables. To do this, either show that both are true, or that both are false, for exactly the same combination of truth values of the propositional variables in these expressions (whichever is easier). Show that p(double arrow) q and (pq)(¬p¬q) are logically equivalent without using truth tables. To do this, either show that both are true, or that both are false, for exactly the same combination of truth values of the propositional variables in these expressions (whichever is easier). Show that p(double arrow) q and (pq)(¬p¬q) are logically equivalent without using truth tables. To do this, either show that both are true, or that both are false, for exactly the same combination of truth values of the propositional variables in these expressions (whichever is easier). Show that p(double arrow) q and (pq)(¬p¬q) are logically equivalent without using truth tables. To do this, either show that both are true, or that both are false, for exactly the same combination of truth values of the propositional variables in these expressions (whichever is easier).

Explanation / Answer

Answer:

Consider the statement formula

p q (p q) (q p)

         ( ¬ p V q) ( ¬ q V p)    since p q ¬ p V q

         ( ¬ p ( ¬ q V p) ) V ( q ( ¬ q V p) ) Distributive law

         ( ¬ p ¬ q ) V ( ¬ p p ) V ( q ¬ q ) V ( q p) Distributive law

          ( ¬ p ¬ q ) V F V F V ( q p)     Since ( ¬ p p ) F and ( q ¬ q ) F

         ( ¬ p ¬ q ) V F V ( q p) since F V F F

         ( ¬ p ¬ q ) V ( q p)   Since ( ¬ p ¬ q ) V F ( ¬ p ¬ q )

            ( q p) V ( ¬ p ¬ q ) since pVq qVp ( commutative )

             ( p q) V ( ¬ p ¬ q )   since p q q p ( commutative )

Therefore , p q ( p q) V ( ¬ p ¬ q )

     

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