Theorem hblem 1567

Description: Lemma for hbeq 1568 and hbel 1569.

Hypothesis

Ref	Expression
hblem.1	|- (y e. A -> A.x y e. A)

Assertion

Ref	Expression
hblem	|- (z e. A -> A.x z e. A)

Distinct variable groups:   y,A   x,y,z

Proof of Theorem hblem

Step	Hyp	Ref	Expression
1		eleq1 1537	. . 3 |- (y = z -> (y e. A <-> z e. A))
2	1	albidv 1280	. . 3 |- (y = z -> (A.x y e. A <-> A.x z e. A))
3	1, 2	imbi12d 628	. 2 |- (y = z -> ((y e. A -> A.x y e. A) <-> (z e. A -> A.x z e. A)))
4		hblem.1	. 2 |- (y e. A -> A.x y e. A)
5	3, 4	chvarv 1329	1 |- (z e. A -> A.x z e. A)

Syntax hints:   -> wi 3  A.wal 956   = wceq 958   e. wcel 960

This theorem is referenced by:  hbeq 1568  hbel 1569  isumvaltf 7193

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 965  ax-17 973  ax-4 975  ax-5o 977  ax-9o 1125  ax-ext 1462

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 983  df-cleq 1472  df-clel 1475

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