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

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Theorem

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


Proof

By the tableau method of natural deduction:

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

$\blacksquare$


Sources