Metamath Proof Explorer


Theorem elzs12

Description: Membership in the dyadic fractions. (Contributed by Scott Fenton, 7-Aug-2025)

Ref Expression
Assertion elzs12 ( 𝐴 ∈ ℤs[1/2] ↔ ∃ 𝑥 ∈ ℤs𝑦 ∈ ℕ0s 𝐴 = ( 𝑥 /su ( 2ss 𝑦 ) ) )

Proof

Step Hyp Ref Expression
1 elex ( 𝐴 ∈ ℤs[1/2] → 𝐴 ∈ V )
2 id ( 𝐴 = ( 𝑥 /su ( 2ss 𝑦 ) ) → 𝐴 = ( 𝑥 /su ( 2ss 𝑦 ) ) )
3 ovex ( 𝑥 /su ( 2ss 𝑦 ) ) ∈ V
4 2 3 eqeltrdi ( 𝐴 = ( 𝑥 /su ( 2ss 𝑦 ) ) → 𝐴 ∈ V )
5 4 a1i ( ( 𝑥 ∈ ℤs𝑦 ∈ ℕ0s ) → ( 𝐴 = ( 𝑥 /su ( 2ss 𝑦 ) ) → 𝐴 ∈ V ) )
6 5 rexlimivv ( ∃ 𝑥 ∈ ℤs𝑦 ∈ ℕ0s 𝐴 = ( 𝑥 /su ( 2ss 𝑦 ) ) → 𝐴 ∈ V )
7 eqeq1 ( 𝑧 = 𝐴 → ( 𝑧 = ( 𝑥 /su ( 2ss 𝑦 ) ) ↔ 𝐴 = ( 𝑥 /su ( 2ss 𝑦 ) ) ) )
8 7 2rexbidv ( 𝑧 = 𝐴 → ( ∃ 𝑥 ∈ ℤs𝑦 ∈ ℕ0s 𝑧 = ( 𝑥 /su ( 2ss 𝑦 ) ) ↔ ∃ 𝑥 ∈ ℤs𝑦 ∈ ℕ0s 𝐴 = ( 𝑥 /su ( 2ss 𝑦 ) ) ) )
9 df-zs12 s[1/2] = { 𝑧 ∣ ∃ 𝑥 ∈ ℤs𝑦 ∈ ℕ0s 𝑧 = ( 𝑥 /su ( 2ss 𝑦 ) ) }
10 8 9 elab2g ( 𝐴 ∈ V → ( 𝐴 ∈ ℤs[1/2] ↔ ∃ 𝑥 ∈ ℤs𝑦 ∈ ℕ0s 𝐴 = ( 𝑥 /su ( 2ss 𝑦 ) ) ) )
11 1 6 10 pm5.21nii ( 𝐴 ∈ ℤs[1/2] ↔ ∃ 𝑥 ∈ ℤs𝑦 ∈ ℕ0s 𝐴 = ( 𝑥 /su ( 2ss 𝑦 ) ) )