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
|- ( ph -> A C_ B )
3sstr4d.2
|- ( ph -> C = A )
3sstr4d.3
|- ( ph -> D = B )
Assertion 3sstr4d
|- ( ph -> C C_ D )

Proof

Step Hyp Ref Expression
1 3sstr4d.1
 |-  ( ph -> A C_ B )
2 3sstr4d.2
 |-  ( ph -> C = A )
3 3sstr4d.3
 |-  ( ph -> D = B )
4 2 3 sseq12d
 |-  ( ph -> ( C C_ D <-> A C_ B ) )
5 1 4 mpbird
 |-  ( ph -> C C_ D )