Metamath Proof Explorer


Theorem cnvssb

Description: Subclass theorem for converse. (Contributed by RP, 22-Oct-2020)

Ref Expression
Assertion cnvssb ⊢ Rel ⁡ A → A ⊆ B ↔ A -1 ⊆ B -1

Proof

Step Hyp Ref Expression
1 cnvss ⊢ A ⊆ B → A -1 ⊆ B -1
2 cnvss ⊢ A -1 ⊆ B -1 → A -1 -1 ⊆ B -1 -1
3 dfrel2 ⊢ Rel ⁡ A ↔ A -1 -1 = A
4 3 biimpi ⊢ Rel ⁡ A → A -1 -1 = A
5 4 eqcomd ⊢ Rel ⁡ A → A = A -1 -1
6 5 adantr ⊢ Rel ⁡ A ∧ A -1 -1 ⊆ B -1 -1 → A = A -1 -1
7 id ⊢ A -1 -1 ⊆ B -1 -1 → A -1 -1 ⊆ B -1 -1
8 cnvcnvss ⊢ B -1 -1 ⊆ B
9 7 8 sstrdi ⊢ A -1 -1 ⊆ B -1 -1 → A -1 -1 ⊆ B
10 9 adantl ⊢ Rel ⁡ A ∧ A -1 -1 ⊆ B -1 -1 → A -1 -1 ⊆ B
11 6 10 eqsstrd ⊢ Rel ⁡ A ∧ A -1 -1 ⊆ B -1 -1 → A ⊆ B
12 11 ex ⊢ Rel ⁡ A → A -1 -1 ⊆ B -1 -1 → A ⊆ B
13 2 12 syl5 ⊢ Rel ⁡ A → A -1 ⊆ B -1 → A ⊆ B
14 1 13 impbid2 ⊢ Rel ⁡ A → A ⊆ B ↔ A -1 ⊆ B -1