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 ( 𝐴𝐵 𝐴 𝐵 )

Proof

Step Hyp Ref Expression
1 ssbr ( 𝐴𝐵 → ( 𝑦 𝐴 𝑥𝑦 𝐵 𝑥 ) )
2 1 ssopab2dv ( 𝐴𝐵 → { ⟨ 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 ( 𝐴𝐵 𝐴 𝐵 )