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 ⊆ ◡ 𝐵 )

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 = { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝑦 𝐴 𝑥 }
4 df-cnv ⊢ ◡ 𝐵 = { ⟨ 𝑥 , 𝑦 ⟩ ∣ 𝑦 𝐵 𝑥 }
5 2 3 4 3sstr4g ⊢ ( 𝐴 ⊆ 𝐵 → ◡ 𝐴 ⊆ ◡ 𝐵 )