Principle of Dilemma/Formulation 1/Forward Implication/Proof 1

Theorem

 * $\left({p \implies q}\right) \land \left({\neg p \implies q}\right) \vdash q$

Proof

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 * align="right" | 1
 * $p \lor \neg p \implies q \lor q$
 * Sequent Introduction
 * 2, 3
 * Constructive Dilemma
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 * $p \lor \neg p$
 * LEM
 * (None)
 * LEM
 * (None)


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 * $q$
 * Sequent Introduction
 * 6
 * Rule of Idempotence
 * Rule of Idempotence