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Theorem bj-sseq 15229
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 473 . 2  |-  ( ph  ->  ( ( ps  /\  ch )  <->  ( A  C_  B  /\  B  C_  A
) ) )
4 eqss 3194 . 2  |-  ( A  =  B  <->  ( A  C_  B  /\  B  C_  A ) )
53, 4bitr4di 198 1  |-  ( ph  ->  ( ( ps  /\  ch )  <->  A  =  B
) )
Colors of variables: wff set class
Syntax hints:    -> wi 4    /\ wa 104    <-> wb 105    = wceq 1364    C_ wss 3153
This theorem was proved from axioms:  ax-mp 5  ax-1 6  ax-2 7  ax-ia1 106  ax-ia2 107  ax-ia3 108  ax-5 1458  ax-7 1459  ax-gen 1460  ax-ie1 1504  ax-ie2 1505  ax-8 1515  ax-11 1517  ax-4 1521  ax-17 1537  ax-i9 1541  ax-ial 1545  ax-i5r 1546  ax-ext 2175
This theorem depends on definitions:  df-bi 117  df-nf 1472  df-sb 1774  df-clab 2180  df-cleq 2186  df-clel 2189  df-in 3159  df-ss 3166
This theorem is referenced by: (None)
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