Metamath Proof Explorer


Theorem scottsn

Description: Applying Scott's trick to a singleton leaves it unchanged. (Contributed by BTernaryTau, 3-Jul-2026)

Ref Expression
Assertion scottsn ⊢ Scott A = A

Proof

Step Hyp Ref Expression
1 df-scott ⊢ Scott A = x ∈ A | ∀ y ∈ A rank ⁡ x ⊆ rank ⁡ y
2 velsn ⊢ x ∈ A ↔ x = A
3 velsn ⊢ y ∈ A ↔ y = A
4 eqtr3 ⊢ x = A ∧ y = A → x = y
5 2 3 4 syl2anb ⊢ x ∈ A ∧ y ∈ A → x = y
6 fveq2 ⊢ x = y → rank ⁡ x = rank ⁡ y
7 6 eqimssd ⊢ x = y → rank ⁡ x ⊆ rank ⁡ y
8 5 7 syl ⊢ x ∈ A ∧ y ∈ A → rank ⁡ x ⊆ rank ⁡ y
9 8 ralrimiva ⊢ x ∈ A → ∀ y ∈ A rank ⁡ x ⊆ rank ⁡ y
10 9 rabeqc ⊢ x ∈ A | ∀ y ∈ A rank ⁡ x ⊆ rank ⁡ y = A
11 1 10 eqtri ⊢ Scott A = A