Metamath Proof Explorer


Theorem brcnvssrid

Description: Any set is a converse subset of itself. (Contributed by Peter Mazsa, 9-Jun-2021)

Ref Expression
Assertion brcnvssrid ( 𝐴 ∈ 𝑉 → 𝐴 ◡ S 𝐴 )

Proof

Step Hyp Ref Expression
1 ssid ⊢ 𝐴 ⊆ 𝐴
2 brcnvssr ⊢ ( 𝐴 ∈ 𝑉 → ( 𝐴 ◡ S 𝐴 ↔ 𝐴 ⊆ 𝐴 ) )
3 1 2 mpbiri ⊢ ( 𝐴 ∈ 𝑉 → 𝐴 ◡ S 𝐴 )