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I was following the solution of one particular propositional logic puzzle where I encountered that the initial well-formed formula was reduced to:

((Q → ((A → B) ∧ (¬A → ¬B))) ∧ (¬Q → ((¬A → B) ∧ (A → ¬B))))

The example then used the definition of Logical Equivalence(<=>) and the below identity:

((¬p ↔ q) ↔ ¬(p ↔ q))

to reduce the formula to

Q ↔ (A ↔ B)

Can anyone please help me understand how the final concise form was derived using the identity?

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    Next time you can use "LaTeX" for mathematical notation, so you don't have to copy so much. – 2017-02-09

1 Answers 1

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Another equivalence you want to use here is:

$P \leftrightarrow Q \Leftrightarrow (P \to Q) \land (\neg P \to \neg Q)$

Thus, the $(A \to B) \land (\neg A \to \neg B)$ part in your original expression is equivalent to $A \leftrightarrow B$.

Likewise, the $(\neg A \to B) \land (A \to \neg B)$ part is equivalent to $A \leftrightarrow \neg B$

Hence, you get:

$(Q \to (A \leftrightarrow B)) \land (\neg Q \to (A \leftrightarrow \neg B))$

So by the provided equivalence, that is the same as:

$(Q \to (A \leftrightarrow B)) \land (\neg Q \to \neg (A \leftrightarrow B))$

And using the earlier mentioned equivalence yet again, that is equivalent to:

$Q \leftrightarrow (A \leftrightarrow B)$

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    Thanks. Got it. This compound statement: ((¬A → B) ∧ (A → ¬B)) actually troubled me. – 2017-02-09
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    @NINCOMPOOP Yeah, it is not *exactly* of the right format, since that would be $(A \to \neg B) \land (\neg A \to \neg \neg B)$, but the difference is a simple double negation and a commutation. – 2017-02-09