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

Theorem

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

Proof

 * align="right" | 4 ||
 * align="right" | 3
 * $p \implies q$
 * Sequent Introduction
 * 3
 * True Statement is Implied by Every Statement
 * align="right" | 5 ||
 * align="right" | 1, 3
 * $p \implies r$
 * $\implies \mathcal E$
 * 1, 4
 * align="right" | 6 ||
 * align="right" | 1, 2, 3
 * $r$
 * $\implies \mathcal E$
 * 5, 2
 * align="right" | 7 ||
 * align="right" | 1, 2
 * $q \implies r$
 * $\implies \mathcal I$
 * 3 - 6
 * align="right" | 8 ||
 * align="right" | 1
 * $p \implies \left({q \implies r}\right)$
 * $\implies \mathcal I$
 * 2 - 7
 * }
 * align="right" | 8 ||
 * align="right" | 1
 * $p \implies \left({q \implies r}\right)$
 * $\implies \mathcal I$
 * 2 - 7
 * }
 * 2 - 7
 * }
 * }