Theorem eqimssi 2111

Description: Infer subclass relationship from equality.

Hypothesis

Ref	Expression
eqimssi.1	|- A = B

Assertion

Ref	Expression
eqimssi	|- A (_ B

Proof of Theorem eqimssi

Step	Hyp	Ref	Expression
1		ssid 2080	. 2 |- A (_ A
2		eqimssi.1	. 2 |- A = B
3	1, 2	sseqtr 2093	1 |- A (_ B

Syntax hints:   = wceq 956   (_ wss 2047

This theorem is referenced by:  funi 3545  tz7.48-2 3957  trcl 4645  zorn2lem4 4791  om2uzf1o 6301  idcn 7766  cncfmet1 7906  1alg 10654  0alg 10689

This theorem was proved from axioms:  ax-1 4  ax-2 5  ax-3 6  ax-mp 7  ax-7 962  ax-gen 963  ax-8 964  ax-10 966  ax-12 968  ax-17 971  ax-4 973  ax-5o 975  ax-6o 978  ax-9o 1123  ax-10o 1140  ax-16 1210  ax-11o 1218  ax-ext 1459

This theorem depends on definitions:  df-bi 147  df-an 225  df-ex 981  df-sb 1172  df-clab 1464  df-cleq 1469  df-clel 1472  df-in 2051  df-ss 2053

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