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Metamath Proof Explorer

Theorem eqvreleqd

Description: Equality theorem for equivalence relation, deduction version. (Contributed by Peter Mazsa, 23-Sep-2021)

Ref Expression
Hypothesis eqvreleqd.1 ⊢ ( 𝜑 → 𝑅 = 𝑆 )
Assertion eqvreleqd ( 𝜑 → ( EqvRel 𝑅 ↔ EqvRel 𝑆 ) )

Proof

Step Hyp Ref Expression
1 eqvreleqd.1 ⊢ ( 𝜑 → 𝑅 = 𝑆 )
2 eqvreleq ⊢ ( 𝑅 = 𝑆 → ( EqvRel 𝑅 ↔ EqvRel 𝑆 ) )
3 1 2 syl ⊢ ( 𝜑 → ( EqvRel 𝑅 ↔ EqvRel 𝑆 ) )