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 ( 𝜑𝐴 = 𝐵 )
Assertion seqeq1d ( 𝜑 → seq 𝐴 ( + , 𝐹 ) = seq 𝐵 ( + , 𝐹 ) )

Proof

Step Hyp Ref Expression
1 seqeqd.1 ( 𝜑𝐴 = 𝐵 )
2 seqeq1 ( 𝐴 = 𝐵 → seq 𝐴 ( + , 𝐹 ) = seq 𝐵 ( + , 𝐹 ) )
3 1 2 syl ( 𝜑 → seq 𝐴 ( + , 𝐹 ) = seq 𝐵 ( + , 𝐹 ) )