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 ) )