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Metamath Proof Explorer

Theorem dmqseqeq1i

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

Ref Expression
Hypothesis dmqseqeq1i.1 ⊢ 𝑅 = 𝑆
Assertion dmqseqeq1i ( ( dom 𝑅 / 𝑅 ) = 𝐴 ↔ ( dom 𝑆 / 𝑆 ) = 𝐴 )

Proof

Step Hyp Ref Expression
1 dmqseqeq1i.1 ⊢ 𝑅 = 𝑆
2 dmqseqeq1 ⊢ ( 𝑅 = 𝑆 → ( ( dom 𝑅 / 𝑅 ) = 𝐴 ↔ ( dom 𝑆 / 𝑆 ) = 𝐴 ) )
3 1 2 ax-mp ⊢ ( ( dom 𝑅 / 𝑅 ) = 𝐴 ↔ ( dom 𝑆 / 𝑆 ) = 𝐴 )