Metamath Proof Explorer


Theorem seqeq1d

Description: Equality deduction for the sequence builder operation. (Contributed by Mario Carneiro, 7-Sep-2013)

Ref Expression
Hypothesis seqeqd.1
|- ( ph -> A = B )
Assertion seqeq1d
|- ( ph -> seq A ( .+ , F ) = seq B ( .+ , F ) )

Proof

Step Hyp Ref Expression
1 seqeqd.1
 |-  ( ph -> A = B )
2 seqeq1
 |-  ( A = B -> seq A ( .+ , F ) = seq B ( .+ , F ) )
3 1 2 syl
 |-  ( ph -> seq A ( .+ , F ) = seq B ( .+ , F ) )