Theorem 3sstr3d 1542

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

Hypotheses

Ref	Expression
3sstr3d.1	⊢ (φ → A ⊆ B)
3sstr3d.2	⊢ (φ → A = C)
3sstr3d.3	⊢ (φ → B = D)

Assertion

Ref	Expression
3sstr3d	⊢ (φ → C ⊆ D)

Proof of Theorem 3sstr3d

Step	Hyp	Ref	Expression
1		3sstr3d.1	. 2 ⊢ (φ → A ⊆ B)
2		3sstr3d.2	. . 3 ⊢ (φ → A = C)
3		3sstr3d.3	. . 3 ⊢ (φ → B = D)
4	2, 3	sseq12d 1529	. 2 ⊢ (φ → (A ⊆ B ↔ C ⊆ D))
5	1, 4	mpbid 170	1 ⊢ (φ → C ⊆ D)

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

This theorem is referenced by:  3sstr4d 1543  shlej2t 5357

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

metamath.org