Metamath Proof Explorer


Theorem reseq2d

Description: Equality deduction for restrictions. (Contributed by Paul Chapman, 22-Jun-2011)

Ref Expression
Hypothesis reseqd.1 ( 𝜑𝐴 = 𝐵 )
Assertion reseq2d ( 𝜑 → ( 𝐶𝐴 ) = ( 𝐶𝐵 ) )

Proof

Step Hyp Ref Expression
1 reseqd.1 ( 𝜑𝐴 = 𝐵 )
2 reseq2 ( 𝐴 = 𝐵 → ( 𝐶𝐴 ) = ( 𝐶𝐵 ) )
3 1 2 syl ( 𝜑 → ( 𝐶𝐴 ) = ( 𝐶𝐵 ) )