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 ( 𝐴𝐵 → ( Rel 𝐵 → Rel 𝐴 ) )

Proof

Step Hyp Ref Expression
1 sstr2 ( 𝐴𝐵 → ( 𝐵 ⊆ ( V × V ) → 𝐴 ⊆ ( V × V ) ) )
2 df-rel ( Rel 𝐵𝐵 ⊆ ( V × V ) )
3 df-rel ( Rel 𝐴𝐴 ⊆ ( V × V ) )
4 1 2 3 3imtr4g ( 𝐴𝐵 → ( Rel 𝐵 → Rel 𝐴 ) )