Theorem eqssi 1517

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

Hypotheses

Ref	Expression
eqssi.1	|- A (_ B
eqssi.2	|- B (_ A

Assertion

Ref	Expression
eqssi	|- A = B

Proof of Theorem eqssi

Step	Hyp	Ref	Expression
1		eqssi.1	. . 3 |- A (_ B
2		eqssi.2	. . 3 |- B (_ A
3	1, 2	pm3.2i 234	. 2 |- (A (_ B /\ B (_ A)
4		eqss 1516	. 2 |- (A = B <-> (A (_ B /\ B (_ A))
5	3, 4	mpbir 165	1 |- A = B

Syntax hints:   /\ wa 196   = wceq 1091   (_ wss 1487

This theorem is referenced by:  inv 1723  unv 1724  unipw 1960  find 2396  dmv 2546  ecopoprdm 3245  dfom3 3477  rankval3 3525  rankuni 3533  rankun 3535  ranklon 3540  cfom 3710  dmaddpq 3853  dmmulpq 3855  dmaddsr 3988  dmmulsr 3989  chcmh 5148  omlsi 5250  choc1 5292  shsidm 5358  shsumval2 5361  chm1 5378  chdmm1 5398  chj1 5410  chm0 5411  shjshs 5412  span0 5448  spanun 5450  sshhococ 5451  spansn 5462  pjoml4 5497

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

metamath.org