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Metamath Proof Explorer

Theorem baibr

Description: Move conjunction outside of biconditional. (Contributed by NM, 11-Jul-1994)

		Ref	Expression
	Hypothesis	baib.1	$⊢ φ \leftrightarrow (ψ \land χ)$
	Assertion	baibr	$⊢ ψ \to (χ \leftrightarrow φ)$

Step	Hyp	Ref	Expression
1		baib.1	$⊢ φ \leftrightarrow (ψ \land χ)$
2	1	baib	$⊢ ψ \to (φ \leftrightarrow χ)$
3	2	bicomd	$⊢ ψ \to (χ \leftrightarrow φ)$