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Metamath Proof Explorer

Theorem breqdi

Description: Equality deduction for a binary relation. (Contributed by Thierry Arnoux, 5-Oct-2020)

Ref Expression
Hypotheses breq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
breqdi.1 ⊢ ( 𝜑 → 𝐶 𝐴 𝐷 )
Assertion breqdi ( 𝜑 → 𝐶 𝐵 𝐷 )

Proof

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