Theorem bicomdd 39669

Description: Commute two sides of a biconditional in a deduction. (Contributed by Rodolfo Medina, 19-Oct-2010.) (Proof shortened by Andrew Salmon, 29-Jun-2011.)

Hypothesis

Ref	Expression
bicomdd.1	⊢ (𝜑 → (𝜓 → (𝜒 ↔ 𝜃)))

Assertion

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

Proof of Theorem bicomdd

Step	Hyp	Ref	Expression
1		bicomdd.1	. 2 ⊢ (𝜑 → (𝜓 → (𝜒 ↔ 𝜃)))
2		bicom 225	. 2 ⊢ ((𝜒 ↔ 𝜃) ↔ (𝜃 ↔ 𝜒))
3	1, 2	imbitrdi 254	1 ⊢ (𝜑 → (𝜓 → (𝜃 ↔ 𝜒)))

This proof depends on syntax axioms:   → wi 4   ↔ wb 209

This proof depends on axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-3 8

This proof depends on definitions:  df-bi 210

This theorem is used by:  ibdr  39673