Theorem sseqtrd 1536

Description: Substitution of equality into a subclass relationship.

Hypotheses

Ref	Expression
sseqtrd.1	⊢ (φ → A ⊆ B)
sseqtrd.2	⊢ (φ → B = C)

Assertion

Ref	Expression
sseqtrd	⊢ (φ → A ⊆ C)

Proof of Theorem sseqtrd

Step	Hyp	Ref	Expression
1		sseqtrd.1	. 2 ⊢ (φ → A ⊆ B)
2		sseqtrd.2	. . 3 ⊢ (φ → B = C)
3	2	sseq2d 1528	. 2 ⊢ (φ → (A ⊆ B ↔ A ⊆ C))
4	1, 3	mpbid 170	1 ⊢ (φ → A ⊆ C)

Syntax hints:   → wi 2   = wceq 1091   ⊆ wss 1487

This theorem is referenced by:  sseqtr4d 1537  nnaword2 3187  r1val1 3502  fodom 3613  shsub2t 5290  ococint 5298  spanssoc 5320  shub2t 5354  chub2t 5425  ssmd1 5734  atcvat3 5774  atcvat4 5775  mdsymlem1 5776  mdsymlem5 5780

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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