Theorem 3orel2 1484

Description: Partial elimination of a triple disjunction by denial of a disjunct. (Contributed by Scott Fenton, 26-Mar-2011.) (Proof shortened by Andrew Salmon, 25-May-2011.)

Assertion

Ref	Expression
3orel2	⊢ (¬ 𝜓 → ((𝜑 ∨ 𝜓 ∨ 𝜒) → (𝜑 ∨ 𝜒)))

Proof of Theorem 3orel2

Step	Hyp	Ref	Expression
1		3orrot 1091	. 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜓 ∨ 𝜒 ∨ 𝜑))
2		3orel1 1090	. . 3 ⊢ (¬ 𝜓 → ((𝜓 ∨ 𝜒 ∨ 𝜑) → (𝜒 ∨ 𝜑)))
3		orcom 867	. . 3 ⊢ ((𝜒 ∨ 𝜑) ↔ (𝜑 ∨ 𝜒))
4	2, 3	syl6ib 250	. 2 ⊢ (¬ 𝜓 → ((𝜓 ∨ 𝜒 ∨ 𝜑) → (𝜑 ∨ 𝜒)))
5	1, 4	syl5bi 241	1 ⊢ (¬ 𝜓 → ((𝜑 ∨ 𝜓 ∨ 𝜒) → (𝜑 ∨ 𝜒)))

Syntax hints:  ¬ wn 3   → wi 4   ∨ wo 844   ∨ w3o 1085

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 206  df-or 845  df-3or 1087

This theorem is referenced by:  nogesgn1o  33876  nosep1o  33884  nosupbnd1lem5  33915