This is an inofficial mirror of http://metamath.tirix.org for personal testing of a visualizer extension only.

Metamath Proof Explorer

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	imbitrdi	⊢ ( 𝜑 → ( 𝜓 → ( 𝜃 ↔ 𝜒 ) ) )