Theorem jao 274

Description: Disjunction of antecedents. Compare Theorem *3.44 of [WhiteheadRussell] p. 113.

Assertion

Ref	Expression
jao	⊢ ((φ → ψ) → ((χ → ψ) → ((φ ∨ χ) → ψ)))

Proof of Theorem jao

Step	Hyp	Ref	Expression
1		con3 86	. 2 ⊢ ((φ → ψ) → (¬ ψ → ¬ φ))
2		pm3.43i 235	. . . . 5 ⊢ ((¬ ψ → ¬ φ) → ((¬ ψ → ¬ χ) → (¬ ψ → (¬ φ ∧ ¬ χ))))
3		con1 84	. . . . 5 ⊢ ((¬ ψ → (¬ φ ∧ ¬ χ)) → (¬ (¬ φ ∧ ¬ χ) → ψ))
4	2, 3	syl6 23	. . . 4 ⊢ ((¬ ψ → ¬ φ) → ((¬ ψ → ¬ χ) → (¬ (¬ φ ∧ ¬ χ) → ψ)))
5		oran 255	. . . 4 ⊢ ((φ ∨ χ) ↔ ¬ (¬ φ ∧ ¬ χ))
6	4, 5	bisyl7 189	. . 3 ⊢ ((¬ ψ → ¬ φ) → ((¬ ψ → ¬ χ) → ((φ ∨ χ) → ψ)))
7		con3 86	. . 3 ⊢ ((χ → ψ) → (¬ ψ → ¬ χ))
8	6, 7	syl5 22	. 2 ⊢ ((¬ ψ → ¬ φ) → ((χ → ψ) → ((φ ∨ χ) → ψ)))
9	1, 8	syl 12	1 ⊢ ((φ → ψ) → ((χ → ψ) → ((φ ∨ χ) → ψ)))

Syntax hints:  ¬ wn 1   → wi 2   ∨ wo 195   ∧ wa 196

This theorem is referenced by:  jaoi 275  jaob 328  jaod 329  3jao 632  indpi 3828

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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