Metamath Proof Explorer


Theorem recseq

Description: Equality theorem for recs . (Contributed by Stefan O'Rear, 18-Jan-2015)

Ref Expression
Assertion recseq
|- ( F = G -> recs ( F ) = recs ( G ) )

Proof

Step Hyp Ref Expression
1 wrecseq3
 |-  ( F = G -> wrecs ( _E , On , F ) = wrecs ( _E , On , G ) )
2 df-recs
 |-  recs ( F ) = wrecs ( _E , On , F )
3 df-recs
 |-  recs ( G ) = wrecs ( _E , On , G )
4 1 2 3 3eqtr4g
 |-  ( F = G -> recs ( F ) = recs ( G ) )