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The question is “Replace the formula (p→(q→¬r))→(¬p→q) by an equivalent formula not involving ¬ or →”.

(p→(q→¬r))→(¬p→q)

≡(p→(¬q∨¬r))→(p∨q)

≡(p→¬(q∧r))→(p∨q)

≡(¬p∨¬(q∧r))→(p∨q)

≡¬(p∧(q∧r))→(p∨q)

≡(p∧(q∧r))∨(p∨q)

A second part to the question then asked, based on the statement in the first part and assuming p,q and r are all equally likely, what is the probability that the above statement is true?

Vladimir Nabokov
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1 Answers1

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The reduction is correct. The probability of $(P \land (Q \land R)) \lor (P \lor Q)$ being true is 3/4, because there are 8 possible assignments of truth-values to the propositional variables; and only 6 of those satisfy the formula.

Hunan Rostomyan
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