Answer to Question #298738 in Discrete Mathematics for ASIF

Question

Explain, without using a truth table, why (p ∨ q ∨ r) ∧

(¬p ∨ ¬q ∨ ¬r) is true when at least one of p, q, and r

is true and at least one is false, but is false when all three

variables have the same truth value.

Answer 1

Let us explain, without using a truth table, why $(p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r)$ is true when at least one of $p,$ $q,$ and $r$ is true and at least one is false, but is false when all three variables have the same truth value.

If at least one of $p,$$q,$ and $r$ is true then the disjunction $(p ∨ q ∨ r)$ is true. If at least one of $p,$$q,$ and $r$ is false then the disjunction $(¬p ∨ ¬q ∨ ¬r)$ is true. Therefore, in this case the conjunction $(p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r)$ is true.

If all three variables have the same truth value equal to true then the value of disjunction $(¬p ∨ ¬q ∨ ¬r)$ is false, and hence the conjunction $(p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r)$ is false. In the case when all three variables have the same truth value equal to false then the value of disjunction $(p ∨ q ∨ r)$ is false, and hence the conjunction $(p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r)$ is false.