Self-Distributive Law for Conditional/Reverse Implication/Formulation 2/Proof
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Theorem
- $\vdash \paren {\paren {p \implies q} \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} }$
Proof
By the tableau method of natural deduction:
Line | Pool | Formula | Rule | Depends upon | Notes | |
---|---|---|---|---|---|---|
1 | 1 | $\paren {p \implies q} \implies \paren {p \implies r}$ | Assumption | (None) | ||
2 | 1 | $p \implies \paren {q \implies r}$ | Sequent Introduction | 1 | Self-Distributive Law for Conditional: Formulation 1 | |
3 | $\paren {\paren {p \implies q} \implies \paren {p \implies r} } \implies \paren {p \implies \paren {q \implies r} }$ | Rule of Implication: $\implies \II$ | 1 – 2 | Assumption 1 has been discharged |
$\blacksquare$