Theorem impbidd 125

Description: Deduce an equivalence from two implications. (Contributed by Rodolfo Medina, 12-Oct-2010.)

Hypotheses

Ref	Expression
impbidd.1	⊢ (𝜑 → (𝜓 → (𝜒 → 𝜃)))
impbidd.2	⊢ (𝜑 → (𝜓 → (𝜃 → 𝜒)))

Assertion

Ref	Expression
impbidd	⊢ (𝜑 → (𝜓 → (𝜒 ↔ 𝜃)))

Proof of Theorem impbidd

Step	Hyp	Ref	Expression
1		impbidd.1	. 2 ⊢ (𝜑 → (𝜓 → (𝜒 → 𝜃)))
2		impbidd.2	. 2 ⊢ (𝜑 → (𝜓 → (𝜃 → 𝜒)))
3		bi3 117	. 2 ⊢ ((𝜒 → 𝜃) → ((𝜃 → 𝜒) → (𝜒 ↔ 𝜃)))
4	1, 2, 3	syl6c 65	1 ⊢ (𝜑 → (𝜓 → (𝜒 ↔ 𝜃)))

Syntax hints:   → wi 4   ↔ wb 103

This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia2 105  ax-ia3 106

This theorem depends on definitions:  df-bi 115

This theorem is referenced by:  impbid21d  126  pm5.74  177  con1biimdc  800  pclem6  1305