Metamath Proof Explorer


Theorem 3sstr4g

Description: Substitution of equality into both sides of a subclass relationship. (Contributed by NM, 16-Aug-1994) (Proof shortened by Eric Schmidt, 26-Jan-2007)

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

Proof

Step Hyp Ref Expression
1 3sstr4g.1 ⊢ φ → A ⊆ B
2 3sstr4g.2 ⊢ C = A
3 3sstr4g.3 ⊢ D = B
4 2 1 eqsstrid ⊢ φ → C ⊆ B
5 4 3 sseqtrrdi ⊢ φ → C ⊆ D