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Theorem bj-sseq 11336
Description: If two converse inclusions are characterized each by a formula, then equality is characterized by the conjunction of these formulas. (Contributed by BJ, 30-Nov-2019.)
Hypotheses
Ref Expression
bj-sseq.1  |-  ( ph  ->  ( ps  <->  A  C_  B
) )
bj-sseq.2  |-  ( ph  ->  ( ch  <->  B  C_  A
) )
Assertion
Ref Expression
bj-sseq  |-  ( ph  ->  ( ( ps  /\  ch )  <->  A  =  B
) )

Proof of Theorem bj-sseq
StepHypRef Expression
1 bj-sseq.1 . . 3  |-  ( ph  ->  ( ps  <->  A  C_  B
) )
2 bj-sseq.2 . . 3  |-  ( ph  ->  ( ch  <->  B  C_  A
) )
31, 2anbi12d 457 . 2  |-  ( ph  ->  ( ( ps  /\  ch )  <->  ( A  C_  B  /\  B  C_  A
) ) )
4 eqss 3038 . 2  |-  ( A  =  B  <->  ( A  C_  B  /\  B  C_  A ) )
53, 4syl6bbr 196 1  |-  ( ph  ->  ( ( ps  /\  ch )  <->  A  =  B
) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    /\ wa 102    <-> wb 103    = wceq 1289    C_ wss 2997
This theorem was proved from axioms:  ax-1 5  ax-2 6  ax-mp 7  ax-ia1 104  ax-ia2 105  ax-ia3 106  ax-5 1381  ax-7 1382  ax-gen 1383  ax-ie1 1427  ax-ie2 1428  ax-8 1440  ax-11 1442  ax-4 1445  ax-17 1464  ax-i9 1468  ax-ial 1472  ax-i5r 1473  ax-ext 2070
This theorem depends on definitions:  df-bi 115  df-nf 1395  df-sb 1693  df-clab 2075  df-cleq 2081  df-clel 2084  df-in 3003  df-ss 3010
This theorem is referenced by: (None)
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