Theorem 3jao 1243

Description: Disjunction of 3 antecedents. (Contributed by NM, 8-Apr-1994.)

Assertion

Ref	Expression
3jao	⊢ (((φ → ψ) ∧ (χ → ψ) ∧ (θ → ψ)) → ((φ ∨ χ ∨ θ) → ψ))

Proof of Theorem 3jao

Step	Hyp	Ref	Expression
1		df-3or 935	. 2 ⊢ ((φ ∨ χ ∨ θ) ↔ ((φ ∨ χ) ∨ θ))
2		jao 498	. . . 4 ⊢ ((φ → ψ) → ((χ → ψ) → ((φ ∨ χ) → ψ)))
3		jao 498	. . . 4 ⊢ (((φ ∨ χ) → ψ) → ((θ → ψ) → (((φ ∨ χ) ∨ θ) → ψ)))
4	2, 3	syl6 29	. . 3 ⊢ ((φ → ψ) → ((χ → ψ) → ((θ → ψ) → (((φ ∨ χ) ∨ θ) → ψ))))
5	4	3imp 1145	. 2 ⊢ (((φ → ψ) ∧ (χ → ψ) ∧ (θ → ψ)) → (((φ ∨ χ) ∨ θ) → ψ))
6	1, 5	syl5bi 208	1 ⊢ (((φ → ψ) ∧ (χ → ψ) ∧ (θ → ψ)) → ((φ ∨ χ ∨ θ) → ψ))

Syntax hints:   → wi 4   ∨ wo 357   ∨ w3o 933   ∧ w3a 934

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 177  df-or 359  df-an 360  df-3or 935  df-3an 936

This theorem is referenced by:  3jaob  1244  3jaoi  1245  3jaod  1246