Theorem sseq2i 1525

Description: An equality inference for the subclass relationship.

Hypothesis

Ref	Expression
sseq1i.1	⊢ A = B

Assertion

Ref	Expression
sseq2i	⊢ (C ⊆ A ↔ C ⊆ B)

Proof of Theorem sseq2i

Step	Hyp	Ref	Expression
1		sseq1i.1	. 2 ⊢ A = B
2		sseq2 1522	. 2 ⊢ (A = B → (C ⊆ A ↔ C ⊆ B))
3	1, 2	ax-mp 6	1 ⊢ (C ⊆ A ↔ C ⊆ B)

Syntax hints:   ↔ wb 127   = wceq 1091   ⊆ wss 1487

This theorem is referenced by:  sseq12i 1526  sseqtr 1532  3sstr3 1538  syl6ss 1546  ssindif0 1741  iunpwss 2039  dffun2 2674  tfrlem8 2956  iscard2 3660  alephislim 3688  cardaleph 3690  nnwo 4607  chsscon1 5384  hatomistic 5755  mdsymlem1 5776  mdsymlem3 5778  mdsymlem6 5781  mdsymlem8 5783

This theorem was proved from axioms:  ax-1 3  ax-2 4  ax-3 5  ax-mp 6  ax-4 673  ax-5 674  ax-6 675  ax-7 676  ax-gen 677  ax-8 798  ax-9 799  ax-10 800  ax-11 801  ax-12 802  ax-16 922  ax-17 925  ax-ext 1074

This theorem depends on definitions:  df-bi 128  df-an 198  df-ex 679  df-sb 853  df-clab 1093  df-cleq 1097  df-clel 1099  df-in 1491  df-ss 1492

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