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
|- ( ph -> A C_ B )
3sstr4g.2
|- C = A
3sstr4g.3
|- D = B
Assertion 3sstr4g
|- ( ph -> C C_ D )

Proof

Step Hyp Ref Expression
1 3sstr4g.1
 |-  ( ph -> A C_ B )
2 3sstr4g.2
 |-  C = A
3 3sstr4g.3
 |-  D = B
4 2 3 sseq12i
 |-  ( C C_ D <-> A C_ B )
5 1 4 sylibr
 |-  ( ph -> C C_ D )