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Metamath Proof Explorer

Theorem brabg

Description: The law of concretion for a binary relation. (Contributed by NM, 16-Aug-1999) (Revised by Mario Carneiro, 19-Dec-2013)

Ref Expression
Hypotheses opelopabg.1 ⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
opelopabg.2 ⊢ ( 𝑦 = 𝐵 → ( 𝜓 ↔ 𝜒 ) )
brabg.5 ⊢ 𝑅 = { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝜑 }
Assertion brabg ( ( 𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷 ) → ( 𝐴 𝑅 𝐵 ↔ 𝜒 ) )

Proof

Step Hyp Ref Expression
1 opelopabg.1 ⊢ ( 𝑥 = 𝐴 → ( 𝜑 ↔ 𝜓 ) )
2 opelopabg.2 ⊢ ( 𝑦 = 𝐵 → ( 𝜓 ↔ 𝜒 ) )
3 brabg.5 ⊢ 𝑅 = { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝜑 }
4 1 2 sylan9bb ⊢ ( ( 𝑥 = 𝐴 ∧ 𝑦 = 𝐵 ) → ( 𝜑 ↔ 𝜒 ) )
5 4 3 brabga ⊢ ( ( 𝐴 ∈ 𝐶 ∧ 𝐵 ∈ 𝐷 ) → ( 𝐴 𝑅 𝐵 ↔ 𝜒 ) )