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Metamath Proof Explorer

Theorem pm5.21nd

Description: Eliminate an antecedent implied by each side of a biconditional. Variant of pm5.21ndd . (Contributed by NM, 20-Nov-2005) (Proof shortened by Wolf Lammen, 4-Nov-2013)

	Ref	Expression
Hypotheses	pm5.21nd.1	\|- ( ( ph /\ ps ) -> th )
	pm5.21nd.2	\|- ( ( ph /\ ch ) -> th )
	pm5.21nd.3	\|- ( th -> ( ps <-> ch ) )
Assertion	pm5.21nd	\|- ( ph -> ( ps <-> ch ) )

Proof

Step	Hyp	Ref	Expression
1		pm5.21nd.1	\|- ( ( ph /\ ps ) -> th )
2		pm5.21nd.2	\|- ( ( ph /\ ch ) -> th )
3		pm5.21nd.3	\|- ( th -> ( ps <-> ch ) )
4	1	ex	\|- ( ph -> ( ps -> th ) )
5	2	ex	\|- ( ph -> ( ch -> th ) )
6	3	a1i	\|- ( ph -> ( th -> ( ps <-> ch ) ) )
7	4 5 6	pm5.21ndd	\|- ( ph -> ( ps <-> ch ) )