Theorem eqimss2 2106

Description: Equality implies the subclass relation.

Assertion

Ref	Expression
eqimss2	|- (B = A -> A (_ B)

Proof of Theorem eqimss2

Step	Hyp	Ref	Expression
1		eqimss 2105	. 2 |- (A = B -> A (_ B)
2	1	eqcoms 1475	1 |- (B = A -> A (_ B)

Syntax hints:   -> wi 3   = wceq 954   (_ wss 2043

This theorem is referenced by:  vss 2303  suc11 3088  dmcoeq 3358  xp11 3468  fconst3 3841  oaass 4185  odi 4200  oen0 4203  zorn 4777  subgres 8069  hstoht 10097  dmdi2 10169

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 960  ax-gen 961  ax-8 962  ax-10 964  ax-12 966  ax-17 969  ax-4 971  ax-5o 973  ax-6o 976  ax-9o 1121  ax-10o 1138  ax-16 1208  ax-11o 1216  ax-ext 1457

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 979  df-sb 1170  df-clab 1462  df-cleq 1467  df-clel 1470  df-in 2047  df-ss 2049

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