Theorem eqssd 1518

Description: Equality deduction from two subclass relationships. Compare Theorem 4 of [Suppes] p. 22.

Hypotheses

Ref	Expression
eqssd.1	|- (ph -> A (_ B)
eqssd.2	|- (ph -> B (_ A)

Assertion

Ref	Expression
eqssd	|- (ph -> A = B)

Proof of Theorem eqssd

Step	Hyp	Ref	Expression
1		eqssd.1	. . 3 |- (ph -> A (_ B)
2		eqssd.2	. . 3 |- (ph -> B (_ A)
3	1, 2	jca 236	. 2 |- (ph -> (A (_ B /\ B (_ A))
4		eqss 1516	. 2 |- (A = B <-> (A (_ B /\ B (_ A))
5	3, 4	sylibr 175	1 |- (ph -> A = B)

Syntax hints:   -> wi 2   /\ wa 196   = wceq 1091   (_ wss 1487

This theorem is referenced by:  int0el 1985  tz7.7 2224  onint 2261  oaass 3163  mapenlem2 3385  r1val1 3502  rankr1 3518  rankr1id 3539  oncard 3636  distrpr 3926  ltexpri 3943  reclem4pr 3953  infxpidmlem7 4939  pjoml 5271  ococint 5298  chsupsn 5313  chabs1t 5432  spansncv 5542  atcvatlem 5770  atcvat3 5774  sumdmdlem 5786

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-in 1491  df-ss 1492

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