Theorem eqimss2 2107

Description: Equality implies the subclass relation.

Assertion

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

Proof of Theorem eqimss2

Step	Hyp	Ref	Expression
1		eqimss 2106	. 2 |- (A = B -> A (_ B)
2	1	eqcoms 1476	1 |- (B = A -> A (_ B)

Syntax hints:   -> wi 3   = wceq 955   (_ wss 2044

This theorem is referenced by:  vss 2304  suc11 3089  dmcoeq 3362  xp11 3472  fconst3 3845  oaass 4188  odi 4203  oen0 4206  zorn 4780  subgres 8081  hstoht 10115  dmdi2 10187

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 961  ax-gen 962  ax-8 963  ax-10 965  ax-12 967  ax-17 970  ax-4 972  ax-5o 974  ax-6o 977  ax-9o 1122  ax-10o 1139  ax-16 1209  ax-11o 1217  ax-ext 1458

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 980  df-sb 1171  df-clab 1463  df-cleq 1468  df-clel 1471  df-in 2048  df-ss 2050

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