Metamath Proof Explorer


Theorem reseq1d

Description: Equality deduction for restrictions. (Contributed by NM, 21-Oct-2014)

Ref Expression
Hypothesis reseqd.1 ( 𝜑𝐴 = 𝐵 )
Assertion reseq1d ( 𝜑 → ( 𝐴𝐶 ) = ( 𝐵𝐶 ) )

Proof

Step Hyp Ref Expression
1 reseqd.1 ( 𝜑𝐴 = 𝐵 )
2 reseq1 ( 𝐴 = 𝐵 → ( 𝐴𝐶 ) = ( 𝐵𝐶 ) )
3 1 2 syl ( 𝜑 → ( 𝐴𝐶 ) = ( 𝐵𝐶 ) )