Proof by Cases/Formulation 1

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Theorem

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


This can be expressed as two separate theorems:

Forward Implication

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

Reverse Implication

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


Proof

We apply the Method of Truth Tables.

As can be seen by inspection, the truth values under the main connectives match for all boolean interpretations.


$\begin{array}{|ccccccc||ccccc|} \hline (p & \implies & r) & \land & (q & \implies & r) & (p & \lor & q) & \implies & r \\ \hline F & T & F & T & F & T & F & F & F & F & T & F \\ F & T & T & T & F & T & T & F & F & F & T & T \\ F & T & F & F & T & F & F & F & T & T & F & F \\ F & T & T & T & T & T & T & F & T & T & T & T \\ T & F & F & F & F & T & F & T & T & F & F & F \\ T & T & T & T & F & T & T & T & T & F & T & T \\ T & F & F & F & T & F & F & T & T & T & F & F \\ T & T & T & T & T & T & T & T & T & T & T & T \\ \hline \end{array}$

$\blacksquare$