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 ⊆ ◡ 𝐵 ) )

Proof

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