Metamath Proof Explorer

Theorem reseq2d

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

Ref Expression
Hypothesis reseqd.1 
|- ( ph -> A = B )
Assertion reseq2d 
|- ( ph -> ( C |` A ) = ( C |` B ) )

Proof

Step Hyp Ref Expression
1 reseqd.1 
 |-  ( ph -> A = B )
2 reseq2 
 |-  ( A = B -> ( C |` A ) = ( C |` B ) )
3 1 2 syl 
 |-  ( ph -> ( C |` A ) = ( C |` B ) )