Theorem 3jaao 1460

Description: Inference conjoining and disjoining the antecedents of three implications. (Contributed by Jeff Hankins, 15-Aug-2009.) (Proof shortened by Andrew Salmon, 13-May-2011.) (Proof shortened by Garrett Katz, 16-Jun-2026.)

Hypotheses

Ref	Expression
3jaao.1	⊢ (𝜑 → (𝜓 → 𝜒))
3jaao.2	⊢ (𝜃 → (𝜏 → 𝜒))
3jaao.3	⊢ (𝜂 → (𝜁 → 𝜒))

Assertion

Ref	Expression
3jaao	⊢ ((𝜑 ∧ 𝜃 ∧ 𝜂) → ((𝜓 ∨ 𝜏 ∨ 𝜁) → 𝜒))

Proof of Theorem 3jaao

Step	Hyp	Ref	Expression
1		3jaao.1	. 2 ⊢ (𝜑 → (𝜓 → 𝜒))
2		3jaao.2	. 2 ⊢ (𝜃 → (𝜏 → 𝜒))
3		3jaao.3	. 2 ⊢ (𝜂 → (𝜁 → 𝜒))
4		3jao 1452	. 2 ⊢ (((𝜓 → 𝜒) ∧ (𝜏 → 𝜒) ∧ (𝜁 → 𝜒)) → ((𝜓 ∨ 𝜏 ∨ 𝜁) → 𝜒))
5	1, 2, 3, 4	syl3an 1178	1 ⊢ ((𝜑 ∧ 𝜃 ∧ 𝜂) → ((𝜓 ∨ 𝜏 ∨ 𝜁) → 𝜒))

This proof depends on syntax axioms:   → wi 4   ∨ w3o 1102   ∧ w3a 1103

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-an 402  df-or 862  df-3or 1104  df-3an 1105

This theorem is used by:  tpfo  14557  lpni  30862  3ornot23  45251