Self-Distributive Law for Conditional/Reverse Implication/Formulation 1

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Theorem

$\paren {p \implies q} \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r}$


Proof

By the tableau method of natural deduction:

$\paren {p \implies q} \implies \paren {p \implies r} \vdash p \implies \paren {q \implies r} $
Line Pool Formula Rule Depends upon Notes
1 1 $\paren {p \implies q} \implies \paren {p \implies r}$ Premise (None)
2 2 $p$ Assumption (None)
3 3 $q$ Assumption (None)
4 3 $p \implies q$ Sequent Introduction 3 True Statement is implied by Every Statement
5 1, 3 $p \implies r$ Modus Ponendo Ponens: $\implies \mathcal E$ 1, 4
6 1, 2, 3 $r$ Modus Ponendo Ponens: $\implies \mathcal E$ 5, 2
7 1, 2 $q \implies r$ Rule of Implication: $\implies \II$ 3 – 6 Assumption 3 has been discharged
8 1 $p \implies \paren {q \implies r}$ Rule of Implication: $\implies \II$ 2 – 7 Assumption 2 has been discharged

$\blacksquare$