Metamath Proof Explorer

Theorem qseq2d

Description: Equality theorem for quotient set, deduction form. (Contributed by Peter Mazsa, 27-May-2021)

Ref Expression
Hypothesis qseq2d.1 
|- ( ph -> A = B )
Assertion qseq2d 
|- ( ph -> ( C /. A ) = ( C /. B ) )

Proof

Step Hyp Ref Expression
1 qseq2d.1 
 |-  ( ph -> A = B )
2 qseq2 
 |-  ( A = B -> ( C /. A ) = ( C /. B ) )
3 1 2 syl 
 |-  ( ph -> ( C /. A ) = ( C /. B ) )