Theorem hblem 1564

Description: Lemma for hbeq 1565 and hbel 1566.

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 1534	. . 3 |- (y = z -> (y e. A <-> z e. A))
2	1	albidv 1278	. . 3 |- (y = z -> (A.x y e. A <-> A.x z e. A))
3	1, 2	imbi12d 626	. 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 1327	1 |- (z e. A -> A.x z e. A)

Syntax hints:   -> wi 3  A.wal 954   = wceq 956   e. wcel 958

This theorem is referenced by:  hbeq 1565  hbel 1566  isumvaltf 7193

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-gen 963  ax-17 971  ax-4 973  ax-5o 975  ax-9o 1123  ax-ext 1459

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 981  df-cleq 1469  df-clel 1472

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