Praeclarum Theorema/Formulation 1/Proof 1

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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 4 $p \land r$ Assumption (None)
5 4 $p$ Rule of Simplification: $\land \mathcal E_1$ 4
6 4 $r$ Rule of Simplification: $\land \mathcal E_2$ 4
7 1, 4 $q$ Modus Ponendo Ponens: $\implies \mathcal E$ 2, 5
8 1, 4 $s$ Modus Ponendo Ponens: $\implies \mathcal E$ 3, 6
9 1, 4 $q \land s$ Rule of Conjunction: $\land \mathcal I$ 7, 8
10 1 $\left({p \land r}\right) \implies \left({q \land s}\right)$ Rule of Implication: $\implies \mathcal I$ 4 – 9 Assumption 4 has been discharged

$\blacksquare$