Theorem 3sstr4g 1541

Description: Substitution of equality into both sides of a subclass relationship.

Hypotheses

Ref	Expression
3sstr4g.1	⊢ (φ → A ⊆ B)
3sstr4g.2	⊢ C = A
3sstr4g.3	⊢ D = B

Assertion

Ref	Expression
3sstr4g	⊢ (φ → C ⊆ D)

Proof of Theorem 3sstr4g

Step	Hyp	Ref	Expression
1		3sstr4g.1	. 2 ⊢ (φ → A ⊆ B)
2		3sstr4g.2	. . 3 ⊢ C = A
3	2	cleqcomi 1105	. 2 ⊢ A = C
4		3sstr4g.3	. . 3 ⊢ D = B
5	4	cleqcomi 1105	. 2 ⊢ B = D
6	1, 3, 5	3sstr3g 1540	1 ⊢ (φ → C ⊆ D)

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

This theorem is referenced by:  unss2 1629  sslin 1662  uniss 1936  iunss1 2002  ssopab2 2119  cnvss 2512  rnss 2558  ssres 2589  ssres2 2590  imass1 2621  imass2 2622  shlej2 5350  chpssat 5756

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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