Theorem dfss3f 1500

Description: Equivalence for subclass relation requiring only that x not be free in A and B (but not necessarily absent from them).

Hypotheses

Ref	Expression
dfss2f.1	|- (y e. A -> A.x y e. A)
dfss2f.2	|- (y e. B -> A.x y e. B)

Assertion

Ref	Expression
dfss3f	|- (A (_ B <-> A.x e. A x e. B)

Distinct variable group(s):   y,A   y,B   x,y

Proof of Theorem dfss3f

Step	Hyp	Ref	Expression
1		dfss2f.1	. . 3 |- (y e. A -> A.x y e. A)
2		dfss2f.2	. . 3 |- (y e. B -> A.x y e. B)
3	1, 2	dfss2f 1499	. 2 |- (A (_ B <-> A.x(x e. A -> x e. B))
4		df-ral 1205	. 2 |- (A.x e. A x e. B <-> A.x(x e. A -> x e. B))
5	3, 4	bitr4 154	1 |- (A (_ B <-> A.x e. A x e. B)

Syntax hints:   -> wi 2   <-> wb 127  A.wal 672   e. wcel 1092  A.wral 1201   (_ wss 1487

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-ral 1205  df-in 1491  df-ss 1492

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