Theorem jaob 758

Description: Disjunction of antecedents. Compare Theorem *4.77 of [WhiteheadRussell] p. 121. (Contributed by NM, 30-May-1994.) (Proof shortened by Wolf Lammen, 9-Dec-2012.)

Assertion

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

Proof of Theorem jaob

Step	Hyp	Ref	Expression
1		pm2.67-2 391	. . 3 ⊢ (((φ ∨ χ) → ψ) → (φ → ψ))
2		olc 373	. . . 4 ⊢ (χ → (φ ∨ χ))
3	2	imim1i 54	. . 3 ⊢ (((φ ∨ χ) → ψ) → (χ → ψ))
4	1, 3	jca 518	. 2 ⊢ (((φ ∨ χ) → ψ) → ((φ → ψ) ∧ (χ → ψ)))
5		pm3.44 497	. 2 ⊢ (((φ → ψ) ∧ (χ → ψ)) → ((φ ∨ χ) → ψ))
6	4, 5	impbii 180	1 ⊢ (((φ ∨ χ) → ψ) ↔ ((φ → ψ) ∧ (χ → ψ)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∨ wo 357   ∧ wa 358

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

This theorem is used by:  pm4.77  762  pm5.53  771  pm4.83  895  unss  3438  ralunb  3445  intun  3959  intpr  3960