Theorem impbidd 210

Description: Deduce an equivalence from two implications. Double deduction associated with impbi 208 and impbii 209. Deduction associated with impbid 212. (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		impbi 208	. 2 ⊢ ((𝜒 → 𝜃) → ((𝜃 → 𝜒) → (𝜒 ↔ 𝜃)))
4	1, 2, 3	syl6c 70	1 ⊢ (𝜑 → (𝜓 → (𝜒 ↔ 𝜃)))

Syntax hints:   → wi 4   ↔ wb 206

This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This theorem depends on definitions:  df-bi 207

This theorem is referenced by:  pm5.74  270  elabgtOLD  3657  seglecgr12  36053  disjlem18  38742  prtlem18  38819