Metamath Proof Explorer


Theorem scottidm

Description: The Scott operation is idempotent. (Contributed by BTernaryTau, 23-Aug-2026)

Ref Expression
Assertion scottidm Scott Scott 𝐴 = Scott 𝐴

Proof

Step Hyp Ref Expression
1 scottss ⊢ Scott Scott 𝐴 ⊆ Scott 𝐴
2 elscott ⊢ ( 𝑥 ∈ Scott 𝐴 ↔ ( 𝑥 ∈ 𝐴 ∧ ∀ 𝑦 ∈ 𝐴 ( rank ‘ 𝑥 ) ⊆ ( rank ‘ 𝑦 ) ) )
3 scottss ⊢ Scott 𝐴 ⊆ 𝐴
4 ssralv ⊢ ( Scott 𝐴 ⊆ 𝐴 → ( ∀ 𝑦 ∈ 𝐴 ( rank ‘ 𝑥 ) ⊆ ( rank ‘ 𝑦 ) → ∀ 𝑦 ∈ Scott 𝐴 ( rank ‘ 𝑥 ) ⊆ ( rank ‘ 𝑦 ) ) )
5 3 4 ax-mp ⊢ ( ∀ 𝑦 ∈ 𝐴 ( rank ‘ 𝑥 ) ⊆ ( rank ‘ 𝑦 ) → ∀ 𝑦 ∈ Scott 𝐴 ( rank ‘ 𝑥 ) ⊆ ( rank ‘ 𝑦 ) )
6 2 5 simplbiim ⊢ ( 𝑥 ∈ Scott 𝐴 → ∀ 𝑦 ∈ Scott 𝐴 ( rank ‘ 𝑥 ) ⊆ ( rank ‘ 𝑦 ) )
7 6 ancli ⊢ ( 𝑥 ∈ Scott 𝐴 → ( 𝑥 ∈ Scott 𝐴 ∧ ∀ 𝑦 ∈ Scott 𝐴 ( rank ‘ 𝑥 ) ⊆ ( rank ‘ 𝑦 ) ) )
8 elscott ⊢ ( 𝑥 ∈ Scott Scott 𝐴 ↔ ( 𝑥 ∈ Scott 𝐴 ∧ ∀ 𝑦 ∈ Scott 𝐴 ( rank ‘ 𝑥 ) ⊆ ( rank ‘ 𝑦 ) ) )
9 7 8 sylibr ⊢ ( 𝑥 ∈ Scott 𝐴 → 𝑥 ∈ Scott Scott 𝐴 )
10 9 ssriv ⊢ Scott 𝐴 ⊆ Scott Scott 𝐴
11 1 10 eqssi ⊢ Scott Scott 𝐴 = Scott 𝐴