Theorem 3jaob 1244

Description: Disjunction of 3 antecedents. (Contributed by NM, 13-Sep-2011.)

Assertion

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

Proof of Theorem 3jaob

Step	Hyp	Ref	Expression
1		3mix1 1124	. . . 4 ⊢ (φ → (φ ∨ χ ∨ θ))
2	1	imim1i 54	. . 3 ⊢ (((φ ∨ χ ∨ θ) → ψ) → (φ → ψ))
3		3mix2 1125	. . . 4 ⊢ (χ → (φ ∨ χ ∨ θ))
4	3	imim1i 54	. . 3 ⊢ (((φ ∨ χ ∨ θ) → ψ) → (χ → ψ))
5		3mix3 1126	. . . 4 ⊢ (θ → (φ ∨ χ ∨ θ))
6	5	imim1i 54	. . 3 ⊢ (((φ ∨ χ ∨ θ) → ψ) → (θ → ψ))
7	2, 4, 6	3jca 1132	. 2 ⊢ (((φ ∨ χ ∨ θ) → ψ) → ((φ → ψ) ∧ (χ → ψ) ∧ (θ → ψ)))
8		3jao 1243	. 2 ⊢ (((φ → ψ) ∧ (χ → ψ) ∧ (θ → ψ)) → ((φ ∨ χ ∨ θ) → ψ))
9	7, 8	impbii 180	1 ⊢ (((φ ∨ χ ∨ θ) → ψ) ↔ ((φ → ψ) ∧ (χ → ψ) ∧ (θ → ψ)))

Syntax hints:   → wi 4   ↔ wb 176   ∨ 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: (None)