Theorem sseq1i 1524

Description: An equality inference for the subclass relationship.

Hypothesis

Ref	Expression
sseq1i.1	⊢ A = B

Assertion

Ref	Expression
sseq1i	⊢ (A ⊆ C ↔ B ⊆ C)

Proof of Theorem sseq1i

Step	Hyp	Ref	Expression
1		sseq1i.1	. 2 ⊢ A = B
2		sseq1 1521	. 2 ⊢ (A = B → (A ⊆ C ↔ B ⊆ C))
3	1, 2	ax-mp 6	1 ⊢ (A ⊆ C ↔ B ⊆ C)

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

This theorem is referenced by:  sseq12i 1526  eqsstr 1530  3sstr3 1538  syl5ss 1544  pwssun 1917  cotr 2625  cnvsym 2626  dffun2 2674  ordgt0ge1 3114  trcl 3489  rankr1 3518  cardne 3637  indpi 3828  shlesb1 5360  chsscon2 5385

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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