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Metamath Proof Explorer

Theorem breq1dd

Description: Equality deduction for a binary relation. (Contributed by Thierry Arnoux, 10-Jan-2026)

Ref Expression
Hypotheses breq1dd.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
breq1dd.2 ⊢ ( 𝜑 → 𝐴 𝑅 𝐶 )
Assertion breq1dd ( 𝜑 → 𝐵 𝑅 𝐶 )

Proof

Step Hyp Ref Expression
1 breq1dd.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 breq1dd.2 ⊢ ( 𝜑 → 𝐴 𝑅 𝐶 )
3 1 breq1d ⊢ ( 𝜑 → ( 𝐴 𝑅 𝐶 ↔ 𝐵 𝑅 𝐶 ) )
4 2 3 mpbid ⊢ ( 𝜑 → 𝐵 𝑅 𝐶 )