Theorem impbida 805

Description: Deduce an equivalence from two implications. (Contributed by NM, 17-Feb-2007.)

Hypotheses

Ref	Expression
impbida.1	⊢ ((φ ∧ ψ) → χ)
impbida.2	⊢ ((φ ∧ χ) → ψ)

Assertion

Ref	Expression
impbida	⊢ (φ → (ψ ↔ χ))

Proof of Theorem impbida

Step	Hyp	Ref	Expression
1		impbida.1	. . 3 ⊢ ((φ ∧ ψ) → χ)
2	1	ex 423	. 2 ⊢ (φ → (ψ → χ))
3		impbida.2	. . 3 ⊢ ((φ ∧ χ) → ψ)
4	3	ex 423	. 2 ⊢ (φ → (χ → ψ))
5	2, 4	impbid 183	1 ⊢ (φ → (ψ ↔ χ))

This proof depends on syntax axioms:   → wi 4   ↔ wb 176   ∧ 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-an 360

This theorem is used by:  eqrdav  2352  funfvbrb  5402  f1o2d  5728  ersymb  5954  erth  5969