Metamath Proof Explorer

Theorem predeq1

Description: Equality theorem for the predecessor class. (Contributed by Scott Fenton, 2-Feb-2011)

Ref Expression
Assertion predeq1 
|- ( R = S -> Pred ( R , A , X ) = Pred ( S , A , X ) )

Proof

Step Hyp Ref Expression
1 eqid 
 |-  A = A
2 eqid 
 |-  X = X
3 predeq123 
 |-  ( ( R = S /\ A = A /\ X = X ) -> Pred ( R , A , X ) = Pred ( S , A , X ) )
4 1 2 3 mp3an23 
 |-  ( R = S -> Pred ( R , A , X ) = Pred ( S , A , X ) )