Theorem bj-hbalt

Description: Closed form of hbal . When in main part, prove hbal and hbald from it. (Contributed by BJ, 2-May-2019)

	Ref	Expression
Assertion	bj-hbalt	|- ( A. y ( ph -> A. x ph ) -> ( A. y ph -> A. x A. y ph ) )

Proof

Step	Hyp	Ref	Expression
1		alim	|- ( A. y ( ph -> A. x ph ) -> ( A. y ph -> A. y A. x ph ) )
2		ax-11	|- ( A. y A. x ph -> A. x A. y ph )
3	1 2	syl6	|- ( A. y ( ph -> A. x ph ) -> ( A. y ph -> A. x A. y ph ) )