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 ⊢ φ → A ⊆ B
Assertion ssbrd ⊢ φ → C A D → C B D

Proof

Step Hyp Ref Expression
1 ssbrd.1 ⊢ φ → A ⊆ B
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 D → C B D