Metamath Proof Explorer


Theorem qsid

Description: A set is equal to its quotient set modulo the converse membership relation. (Note: the converse membership relation is not an equivalence relation.) (Contributed by NM, 13-Aug-1995) (Revised by Mario Carneiro, 9-Jul-2014)

Ref Expression
Assertion qsid ( 𝐴 / ◡ E ) = 𝐴

Proof

Step Hyp Ref Expression
1 vex ⊢ 𝑥 ∈ V
2 1 ecid ⊢ [ 𝑥 ] ◡ E = 𝑥
3 2 eqeq2i ⊢ ( 𝑦 = [ 𝑥 ] ◡ E ↔ 𝑦 = 𝑥 )
4 equcom ⊢ ( 𝑦 = 𝑥 ↔ 𝑥 = 𝑦 )
5 3 4 bitri ⊢ ( 𝑦 = [ 𝑥 ] ◡ E ↔ 𝑥 = 𝑦 )
6 5 rexbii ⊢ ( ∃ 𝑥 ∈ 𝐴 𝑦 = [ 𝑥 ] ◡ E ↔ ∃ 𝑥 ∈ 𝐴 𝑥 = 𝑦 )
7 vex ⊢ 𝑦 ∈ V
8 7 elqs ⊢ ( 𝑦 ∈ ( 𝐴 / ◡ E ) ↔ ∃ 𝑥 ∈ 𝐴 𝑦 = [ 𝑥 ] ◡ E )
9 risset ⊢ ( 𝑦 ∈ 𝐴 ↔ ∃ 𝑥 ∈ 𝐴 𝑥 = 𝑦 )
10 6 8 9 3bitr4i ⊢ ( 𝑦 ∈ ( 𝐴 / ◡ E ) ↔ 𝑦 ∈ 𝐴 )
11 10 eqriv ⊢ ( 𝐴 / ◡ E ) = 𝐴