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Metamath Proof Explorer

Theorem bj-cbv1hv

Description: Version of cbv1h with a disjoint variable condition, which does not require ax-13 . (Contributed by BJ, 16-Jun-2019) (Proof modification is discouraged.)

		Ref	Expression
	Hypotheses	bj-cbv1hv.1	\|- ( ph -> ( ps -> A. y ps ) )
		bj-cbv1hv.2	\|- ( ph -> ( ch -> A. x ch ) )
		bj-cbv1hv.3	\|- ( ph -> ( x = y -> ( ps -> ch ) ) )
	Assertion	bj-cbv1hv	\|- ( A. x A. y ph -> ( A. x ps -> A. y ch ) )

Proof

Step	Hyp	Ref	Expression
1		bj-cbv1hv.1	\|- ( ph -> ( ps -> A. y ps ) )
2		bj-cbv1hv.2	\|- ( ph -> ( ch -> A. x ch ) )
3		bj-cbv1hv.3	\|- ( ph -> ( x = y -> ( ps -> ch ) ) )
4		nfa1	\|- F/ x A. x A. y ph
5		nfa2	\|- F/ y A. x A. y ph
6		2sp	\|- ( A. x A. y ph -> ph )
7	6 1	syl	\|- ( A. x A. y ph -> ( ps -> A. y ps ) )
8	5 7	nf5d	\|- ( A. x A. y ph -> F/ y ps )
9	6 2	syl	\|- ( A. x A. y ph -> ( ch -> A. x ch ) )
10	4 9	nf5d	\|- ( A. x A. y ph -> F/ x ch )
11	6 3	syl	\|- ( A. x A. y ph -> ( x = y -> ( ps -> ch ) ) )
12	4 5 8 10 11	cbv1v	\|- ( A. x A. y ph -> ( A. x ps -> A. y ch ) )