# Proof by Cases/Formulation 2

## Theorem

$\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \iff \paren {\paren {p \lor q} \implies r}$

This can be expressed as two separate theorems:

### Forward Implication

$\vdash \paren {\paren {p \implies r} \land \paren {q \implies r} } \implies \paren {\paren {p \lor q} \implies r}$

### Reverse Implication

$\vdash \paren {\paren {p \lor q} \implies r} \implies \paren {\paren {p \implies r} \land \paren {q \implies r} }$

## Proof

By the tableau method of natural deduction:

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

$\blacksquare$