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Archangel Macsika

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. Expert's answer Let us explain, without using a truth table, why (p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r)(p∨q∨r)∧(¬p∨¬q∨¬r) is true when at least one of p,p, q,q, and rr 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,p,q,q, and rr is true then the disjunction (p ∨ q ∨ r)(p∨q∨r) is true. If at least one of p,p,q,q, and rr is false then the disjunction (¬p ∨ ¬q ∨ ¬r)(¬p∨¬q∨¬r) is true. Therefore, in this case the conjunction (p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r)(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)(¬p∨¬q∨¬r) is false, and hence the conjunction (p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r)(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)(p∨q∨r) is false, and hence the conjunction (p ∨ q ∨ r) ∧ (¬p ∨ ¬q ∨ ¬r)(p∨q∨r)∧(¬p∨¬q∨¬r) is false.

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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.

Expert's answer

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.


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Question ID: mtid-5-stid-8-sqid-190-qpid-78