Metamath Proof Explorer


Theorem relss

Description: Subclass theorem for relation predicate. Theorem 2 of Suppes p. 58. (Contributed by NM, 15-Aug-1994)

Ref Expression
Assertion relss ⊢ A ⊆ B → Rel ⁡ B → Rel ⁡ A

Proof

Step Hyp Ref Expression
1 sstr2 ⊢ A ⊆ B → B ⊆ V × V → A ⊆ V × V
2 df-rel ⊢ Rel ⁡ B ↔ B ⊆ V × V
3 df-rel ⊢ Rel ⁡ A ↔ A ⊆ V × V
4 1 2 3 3imtr4g ⊢ A ⊆ B → Rel ⁡ B → Rel ⁡ A