Metamath Proof Explorer


Theorem 3sstr4d

Description: Substitution of equality into both sides of a subclass relationship. (Contributed by NM, 30-Nov-1995) (Proof shortened by Eric Schmidt, 26-Jan-2007)

Ref Expression
Hypotheses 3sstr4d.1 ⊢ φ → A ⊆ B
3sstr4d.2 ⊢ φ → C = A
3sstr4d.3 ⊢ φ → D = B
Assertion 3sstr4d ⊢ φ → C ⊆ D

Proof

Step Hyp Ref Expression
1 3sstr4d.1 ⊢ φ → A ⊆ B
2 3sstr4d.2 ⊢ φ → C = A
3 3sstr4d.3 ⊢ φ → D = B
4 2 1 eqsstrd ⊢ φ → C ⊆ B
5 4 3 sseqtrrd ⊢ φ → C ⊆ D