Implication is Left Distributive over Disjunction/Formulation 1/Reverse Implication

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Theorem

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


Proof

By the tableau method of natural deduction:

$\left({p \implies q}\right) \lor \left({p \implies r}\right) \vdash p \implies \left({q \lor r}\right)$
Line Pool Formula Rule Depends upon Notes
1 1 $\left({p \implies q}\right) \lor \left({p \implies r}\right)$ Assumption (None)
2 2 $p \implies q$ Assumption (None)
3 3 $p$ Assumption (None)
4 2, 3 $q$ Modus Ponendo Ponens: $\implies \mathcal E$ 2, 3
5 2, 3 $q \lor r$ Rule of Addition: $\lor \mathcal I_1$ 4
6 2 $p \implies \left({q \lor r}\right)$ Rule of Implication: $\implies \mathcal I$ 3 – 5 Assumption 3 has been discharged
7 7 $p \implies r$ Assumption (None)
8 8 $p$ Assumption (None)
9 7, 8 $r$ Modus Ponendo Ponens: $\implies \mathcal E$ 7, 8
10 7, 8 $q \lor r$ Rule of Addition: $\lor \mathcal I_2$ 9
11 7 $p \implies \left({q \lor r}\right)$ Rule of Implication: $\implies \mathcal I$ 8 – 10 Assumption 8 has been discharged
12 1 $p \implies \left({q \lor r}\right)$ Proof by Cases: $\text{PBC}$ 1, 2 – 6, 7 – 11 Assumptions 2 and 7 have been discharged

$\blacksquare$