Theorem jaob 328

Description: Disjunction of antecedents. Compare Theorem *4.77 of [WhiteheadRussell] p. 121.

Assertion

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

Proof of Theorem jaob

Step	Hyp	Ref	Expression
1		orc 225	. . . 4 ⊢ (φ → (φ ∨ χ))
2	1	syl4 19	. . 3 ⊢ (((φ ∨ χ) → ψ) → (φ → ψ))
3		olc 224	. . . 4 ⊢ (χ → (φ ∨ χ))
4	3	syl4 19	. . 3 ⊢ (((φ ∨ χ) → ψ) → (χ → ψ))
5	2, 4	jca 236	. 2 ⊢ (((φ ∨ χ) → ψ) → ((φ → ψ) ∧ (χ → ψ)))
6		jao 274	. . 3 ⊢ ((φ → ψ) → ((χ → ψ) → ((φ ∨ χ) → ψ)))
7	6	imp 277	. 2 ⊢ (((φ → ψ) ∧ (χ → ψ)) → ((φ ∨ χ) → ψ))
8	5, 7	impbi 139	1 ⊢ (((φ ∨ χ) → ψ) ↔ ((φ → ψ) ∧ (χ → ψ)))

Syntax hints:   → wi 2   ↔ wb 127   ∨ wo 195   ∧ wa 196

This theorem is referenced by:  unss 1632  prsspw 1858  intun 1989  intpr 1990  ordsseleq 2227

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6

This theorem depends on definitions:  df-bi 128  df-or 197  df-an 198

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