Proof by Cases/Formulation 2

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Theorem

$\vdash \left({\left({p \implies r}\right) \land \left({q \implies r}\right)}\right) \iff \left({\left({p \lor q}\right) \implies r}\right)$


This can be expressed as two separate theorems:

Forward Implication

$\vdash \left({\left({p \implies r}\right) \land \left({q \implies r}\right)}\right) \implies \left({\left({p \lor q}\right) \implies r}\right)$

Reverse Implication

$\vdash \left({\left({p \lor q}\right) \implies r}\right) \implies \left({\left({p \implies r}\right) \land \left({q \implies r}\right)}\right)$


Proof

By the tableau method of natural deduction:

$\vdash \left({\left({p \implies r}\right) \land \left({q \implies r}\right)}\right) \iff \left({\left({p \lor q}\right) \implies r}\right)$
Line Pool Formula Rule Depends upon Notes
1 $\left({\left({p \implies r}\right) \land \left({q \implies r}\right)}\right) \implies \left({\left({p \lor q}\right) \implies r}\right)$ Theorem Introduction (None) Proof by Cases: Forward Implication
2 $\left({\left({p \lor q}\right) \implies r}\right) \implies \left({\left({p \implies r}\right) \land \left({q \implies r}\right)}\right)$ Theorem Introduction (None) Proof by Cases: Reverse Implication
3 $\left({\left({p \implies r}\right) \land \left({q \implies r}\right)}\right) \iff \left({\left({p \lor q}\right) \implies r}\right)$ Biconditional Introduction: $\iff \mathcal I$ 1, 2

$\blacksquare$