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Exercise 503Logic, sets and proofs › Proof techniquesHigh schoolAdded 3 Nov 2026

A tautology

#propositional logic#tautology#truth table

Statement

Show that (P⇒Q)⇔(¬P∨Q)(P\Rightarrow Q)\Leftrightarrow(\neg P\lor Q) is a tautology.

Hint

Build the truth table.

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Solution

PQP⇒Q¬P∨Q(P⇒Q)⇔(¬P∨Q)TTTTTTFFFTFTTTTFFTTT\begin{array}{|c|c|c|c|c|} \hline P & Q & P \Rightarrow Q & \neg P \lor Q & (P \Rightarrow Q) \Leftrightarrow (\neg P \lor Q) \\ \hline \text{T} & \text{T} & \text{T} & \text{T} & \text{T} \\ \text{T} & \text{F} & \text{F} & \text{F} & \text{T} \\ \text{F} & \text{T} & \text{T} & \text{T} & \text{T} \\ \text{F} & \text{F} & \text{T} & \text{T} & \text{T} \\ \hline \end{array}

All entries are T.

Conclusion. The formula is a tautology; implication is definable via disjunction.

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