Theorem bicomdd

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)

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

Proof

Step	Hyp	Ref	Expression
1		bicomdd.1	⊢ ( 𝜑 → ( 𝜓 → ( 𝜒 ↔ 𝜃 ) ) )
2		bicom	⊢ ( ( 𝜒 ↔ 𝜃 ) ↔ ( 𝜃 ↔ 𝜒 ) )
3	1 2	syl6ib	⊢ ( 𝜑 → ( 𝜓 → ( 𝜃 ↔ 𝜒 ) ) )