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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 1 eqsstrd 
 |-  ( ph -> C C_ B )
5 4 3 sseqtrrd 
 |-  ( ph -> C C_ D )