Theorem 3orel2 1515

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.) (Proof shortened by Eric Schmidt, 8-Oct-2025.)

Assertion

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

Proof of Theorem 3orel2

Step	Hyp	Ref	Expression
1		3orcoma 1109	. 2 ⊢ ((𝜑 ∨ 𝜓 ∨ 𝜒) ↔ (𝜓 ∨ 𝜑 ∨ 𝜒))
2		3orel1 1107	. 2 ⊢ (¬ 𝜓 → ((𝜓 ∨ 𝜑 ∨ 𝜒) → (𝜑 ∨ 𝜒)))
3	1, 2	biimtrid 245	1 ⊢ (¬ 𝜓 → ((𝜑 ∨ 𝜓 ∨ 𝜒) → (𝜑 ∨ 𝜒)))

This proof depends on syntax axioms:  ¬ wn 3   → wi 4   ∨ wo 861   ∨ w3o 1102

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210  df-or 862  df-3or 1104

This theorem is used by:  nogesgn1o  28023  nosep1o  28031  nosupbnd1lem5  28062