Theorem cleqf 1536

Description: Establish equality between classes, using bound-variable hypotheses instead of distinct variable conditions.

Hypotheses

Ref	Expression
cleqf.1	|- (y e. A -> A.x y e. A)
cleqf.2	|- (y e. B -> A.x y e. B)

Assertion

Ref	Expression
cleqf	|- (A = B <-> A.x(x e. A <-> x e. B))

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

Proof of Theorem cleqf

Step	Hyp	Ref	Expression
1		dfcleq 1447	. 2 |- (A = B <-> A.y(y e. A <-> y e. B))
2		ax-17 1190	. . 3 |- ((x e. A <-> x e. B) -> A.y(x e. A <-> x e. B))
3		cleqf.1	. . . 4 |- (y e. A -> A.x y e. A)
4		cleqf.2	. . . 4 |- (y e. B -> A.x y e. B)
5	3, 4	hbbi 986	. . 3 |- ((y e. A <-> y e. B) -> A.x(y e. A <-> y e. B))
6		eleq1 1510	. . . 4 |- (x = y -> (x e. A <-> y e. A))
7		eleq1 1510	. . . 4 |- (x = y -> (x e. B <-> y e. B))
8	6, 7	bibi12d 627	. . 3 |- (x = y -> ((x e. A <-> x e. B) <-> (y e. A <-> y e. B)))
9	2, 5, 8	cbval 1148	. 2 |- (A.x(x e. A <-> x e. B) <-> A.y(y e. A <-> y e. B))
10	1, 9	bitr4 176	1 |- (A = B <-> A.x(x e. A <-> x e. B))

Syntax hints:   -> wi 3   <-> wb 146  A.wal 950   = wceq 1099   e. wcel 1105

This theorem is referenced by:  abeq2 1544  eq2ab 1549  cbvab 1880  ne0f 2258

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-4 951  ax-5 952  ax-6 953  ax-7 954  ax-gen 955  ax-8 1101  ax-9 1102  ax-12 1104  ax-17 1190  ax-ext 1436

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 957  df-cleq 1446  df-clel 1449

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