Metamath Proof Explorer


Theorem breq123d

Description: Equality deduction for a binary relation. (Contributed by NM, 29-Oct-2011)

Ref Expression
Hypotheses breq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
breq123d.2 ⊢ ( 𝜑 → 𝑅 = 𝑆 )
breq123d.3 ⊢ ( 𝜑 → 𝐶 = 𝐷 )
Assertion breq123d ( 𝜑 → ( 𝐴 𝑅 𝐶 ↔ 𝐵 𝑆 𝐷 ) )

Proof

Step Hyp Ref Expression
1 breq1d.1 ⊢ ( 𝜑 → 𝐴 = 𝐵 )
2 breq123d.2 ⊢ ( 𝜑 → 𝑅 = 𝑆 )
3 breq123d.3 ⊢ ( 𝜑 → 𝐶 = 𝐷 )
4 1 3 breq12d ⊢ ( 𝜑 → ( 𝐴 𝑅 𝐶 ↔ 𝐵 𝑅 𝐷 ) )
5 2 breqd ⊢ ( 𝜑 → ( 𝐵 𝑅 𝐷 ↔ 𝐵 𝑆 𝐷 ) )
6 4 5 bitrd ⊢ ( 𝜑 → ( 𝐴 𝑅 𝐶 ↔ 𝐵 𝑆 𝐷 ) )