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Metamath Proof Explorer

Theorem releqi

Description: Equality inference for the relation predicate. (Contributed by NM, 8-Dec-2006)

Ref Expression
Hypothesis releqi.1 ⊢ 𝐴 = 𝐵
Assertion releqi ( Rel 𝐴 ↔ Rel 𝐵 )

Proof

Step Hyp Ref Expression
1 releqi.1 ⊢ 𝐴 = 𝐵
2 releq ⊢ ( 𝐴 = 𝐵 → ( Rel 𝐴 ↔ Rel 𝐵 ) )
3 1 2 ax-mp ⊢ ( Rel 𝐴 ↔ Rel 𝐵 )