Metamath Proof Explorer


Theorem cnvss

Description: Subset theorem for converse. (Contributed by NM, 22-Mar-1998) (Proof shortened by Kyle Wyonch, 27-Apr-2021)

Ref Expression
Assertion cnvss ⊢ A ⊆ B → A -1 ⊆ B -1

Proof

Step Hyp Ref Expression
1 ssbr ⊢ A ⊆ B → y A x → y B x
2 1 ssopab2dv ⊢ A ⊆ B → x y | y A x ⊆ x y | y B x
3 df-cnv ⊢ A -1 = x y | y A x
4 df-cnv ⊢ B -1 = x y | y B x
5 2 3 4 3sstr4g ⊢ A ⊆ B → A -1 ⊆ B -1