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Metamath Proof Explorer

Theorem dmqseqeq1d

Description: Equality theorem for domain quotient set, deduction version. (Contributed by Peter Mazsa, 26-Sep-2021)

Ref Expression
Hypothesis dmqseqeq1d.1 
|- ( ph -> R = S )
Assertion dmqseqeq1d 
|- ( ph -> ( ( dom R /. R ) = A <-> ( dom S /. S ) = A ) )

Proof

Step Hyp Ref Expression
1 dmqseqeq1d.1 
 |-  ( ph -> R = S )
2 dmqseqeq1 
 |-  ( R = S -> ( ( dom R /. R ) = A <-> ( dom S /. S ) = A ) )
3 1 2 syl 
 |-  ( ph -> ( ( dom R /. R ) = A <-> ( dom S /. S ) = A ) )