Theorem 3jaod 1246

Description: Disjunction of 3 antecedents (deduction). (Contributed by NM, 14-Oct-2005.)

Hypotheses

Ref	Expression
3jaod.1	⊢ (φ → (ψ → χ))
3jaod.2	⊢ (φ → (θ → χ))
3jaod.3	⊢ (φ → (τ → χ))

Assertion

Ref	Expression
3jaod	⊢ (φ → ((ψ ∨ θ ∨ τ) → χ))

Proof of Theorem 3jaod

Step	Hyp	Ref	Expression
1		3jaod.1	. 2 ⊢ (φ → (ψ → χ))
2		3jaod.2	. 2 ⊢ (φ → (θ → χ))
3		3jaod.3	. 2 ⊢ (φ → (τ → χ))
4		3jao 1243	. 2 ⊢ (((ψ → χ) ∧ (θ → χ) ∧ (τ → χ)) → ((ψ ∨ θ ∨ τ) → χ))
5	1, 2, 3, 4	syl3anc 1182	1 ⊢ (φ → ((ψ ∨ θ ∨ τ) → χ))

This proof depends on syntax axioms:   → wi 4   ∨ w3o 933

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

This proof depends on definitions:  df-bi 177  df-or 359  df-an 360  df-3or 935  df-3an 936

This theorem is used by:  3jaodan  1248  3jaao  1249  ltfintri  4467  nchoicelem1  6290  nchoicelem2  6291