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Metamath Proof Explorer

Theorem breq12d

Description: Equality deduction for a binary relation. (Contributed by NM, 8-Feb-1996) (Proof shortened by Andrew Salmon, 9-Jul-2011)

Ref Expression
Hypotheses breq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
breq12d.2 ⊢ ( 𝜑 → 𝐶 = 𝐷 )
Assertion breq12d ( 𝜑 → ( 𝐴 𝑅 𝐶 ↔ 𝐵 𝑅 𝐷 ) )

Proof

Step Hyp Ref Expression
1 breq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 breq12d.2 ⊢ ( 𝜑 → 𝐶 = 𝐷 )
3 breq12 ⊢ ( ( 𝐴 = 𝐵 ∧ 𝐶 = 𝐷 ) → ( 𝐴 𝑅 𝐶 ↔ 𝐵 𝑅 𝐷 ) )
4 1 2 3 syl2anc ⊢ ( 𝜑 → ( 𝐴 𝑅 𝐶 ↔ 𝐵 𝑅 𝐷 ) )