Implication is Left Distributive over Conjunction/Reverse Implication/Formulation 2/Proof

Theorem

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

Proof

Let us use the following abbreviations

 $\displaystyle \phi$ $\text{ for }$ $\displaystyle \left({p \implies q}\right) \land \left({p \implies r}\right)$ $\displaystyle \psi$ $\text{ for }$ $\displaystyle p \implies \left({q \land r}\right)$

By the tableau method of natural deduction:

$\left({\left({p \implies q}\right) \land \left({p \implies r}\right)}\right) \implies \left({p \implies \left({q \land r}\right)}\right)$
Line Pool Formula Rule Depends upon Notes
1 1 $\phi$ Assumption (None)
2 1 $\psi$ Sequent Introduction 1 Implication is Left Distributive over Conjunction: Formulation 1
3 $\phi \implies \psi$ Rule of Implication: $\implies \mathcal I$ 1 – 2 Assumption 1 has been discharged

Expanding the abbreviations leads us back to:

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

$\blacksquare$