Metamath Proof Explorer


Theorem ssbrd

Description: Deduction from a subclass relationship of binary relations. (Contributed by NM, 30-Apr-2004)

Ref Expression
Hypothesis ssbrd.1 ( 𝜑𝐴𝐵 )
Assertion ssbrd ( 𝜑 → ( 𝐶 𝐴 𝐷𝐶 𝐵 𝐷 ) )

Proof

Step Hyp Ref Expression
1 ssbrd.1 ( 𝜑𝐴𝐵 )
2 1 sseld ( 𝜑 → ( ⟨ 𝐶 , 𝐷 ⟩ ∈ 𝐴 → ⟨ 𝐶 , 𝐷 ⟩ ∈ 𝐵 ) )
3 df-br ( 𝐶 𝐴 𝐷 ↔ ⟨ 𝐶 , 𝐷 ⟩ ∈ 𝐴 )
4 df-br ( 𝐶 𝐵 𝐷 ↔ ⟨ 𝐶 , 𝐷 ⟩ ∈ 𝐵 )
5 2 3 4 3imtr4g ( 𝜑 → ( 𝐶 𝐴 𝐷𝐶 𝐵 𝐷 ) )