Praeclarum Theorema/Formulation 1/Proof 3

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Theorem

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


Proof

By the tableau method of natural deduction:

$\left({p \implies q}\right) \land \left({r \implies s}\right) \vdash \left({p \land r}\right) \implies \left({q \land s}\right)$
Line Pool Formula Rule Depends upon Notes
1 1 $\left({p \implies q}\right) \land \left({r \implies s}\right)$ Premise (None)
2 1 $p \implies q$ Rule of Simplification: $\land \mathcal E_1$ 1
3 1 $r \implies s$ Rule of Simplification: $\land \mathcal E_2$ 1
4 1 $p \land r \implies q \land r$ Sequent Introduction 2 Factor Principles/Conjunction on Right/Formulation 1/Proof 2
5 1 $q \land r \implies q \land s$ Sequent Introduction 3 Factor Principles/Conjunction on Left/Formulation 1/Proof 2
6 1 $p \land r \implies q \land s$ Sequent Introduction 4,5 Hypothetical Syllogism

$\blacksquare$