Praeclarum Theorema/Formulation 2

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Theorem

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


Proof

By the tableau method of natural deduction:

$\vdash \left({\left({p \implies q}\right) \land \left({r \implies s}\right)}\right) \implies \left({\left({p \land r}\right) \implies \left({q \land s}\right)}\right)$
Line Pool Formula Rule Depends upon Notes
1 1 $\left({\left({p \implies q}\right) \land \left({r \implies s}\right)}\right)$ Assumption (None)
2 1 $\left({p \land r}\right) \implies \left({q \land s}\right)$ Sequent Introduction 1 Praeclarum Theorema: Formulation 1
3 1 $\left({\left({p \implies q}\right) \land \left({r \implies s}\right)}\right) \implies \left({\left({p \land r}\right) \implies \left({q \land s}\right)}\right)$ Rule of Implication: $\implies \mathcal I$ 1 – 2 Assumption 1 has been discharged

$\blacksquare$


Sources