Metamath Proof Explorer


Theorem clsk1indlem2

Description: The ansatz closure function ( r e. ~P 3o |-> if ( r = { (/) } , { (/) , 1o } , r ) ) has the K2 property of expanding. (Contributed by RP, 6-Jul-2021)

Ref Expression
Hypothesis clsk1indlem.k ⊢ K = r ∈ 𝒫 3 𝑜 ⟼ if r = ∅ ∅ 1 𝑜 r
Assertion clsk1indlem2 ⊢ ∀ s ∈ 𝒫 3 𝑜 s ⊆ K ⁡ s

Proof

Step Hyp Ref Expression
1 clsk1indlem.k ⊢ K = r ∈ 𝒫 3 𝑜 ⟼ if r = ∅ ∅ 1 𝑜 r
2 id ⊢ s = ∅ → s = ∅
3 snsspr1 ⊢ ∅ ⊆ ∅ 1 𝑜
4 2 3 eqsstrdi ⊢ s = ∅ → s ⊆ ∅ 1 𝑜
5 4 ancli ⊢ s = ∅ → s = ∅ ∧ s ⊆ ∅ 1 𝑜
6 5 con3i ⊢ ¬ s = ∅ ∧ s ⊆ ∅ 1 𝑜 → ¬ s = ∅
7 ssid ⊢ s ⊆ s
8 6 7 jctir ⊢ ¬ s = ∅ ∧ s ⊆ ∅ 1 𝑜 → ¬ s = ∅ ∧ s ⊆ s
9 8 orri ⊢ s = ∅ ∧ s ⊆ ∅ 1 𝑜 ∨ ¬ s = ∅ ∧ s ⊆ s
10 9 a1i ⊢ s ∈ 𝒫 3 𝑜 → s = ∅ ∧ s ⊆ ∅ 1 𝑜 ∨ ¬ s = ∅ ∧ s ⊆ s
11 sseq2 ⊢ if s = ∅ ∅ 1 𝑜 s = ∅ 1 𝑜 → s ⊆ if s = ∅ ∅ 1 𝑜 s ↔ s ⊆ ∅ 1 𝑜
12 sseq2 ⊢ if s = ∅ ∅ 1 𝑜 s = s → s ⊆ if s = ∅ ∅ 1 𝑜 s ↔ s ⊆ s
13 11 12 elimif ⊢ s ⊆ if s = ∅ ∅ 1 𝑜 s ↔ s = ∅ ∧ s ⊆ ∅ 1 𝑜 ∨ ¬ s = ∅ ∧ s ⊆ s
14 10 13 sylibr ⊢ s ∈ 𝒫 3 𝑜 → s ⊆ if s = ∅ ∅ 1 𝑜 s
15 eqeq1 ⊢ r = s → r = ∅ ↔ s = ∅
16 id ⊢ r = s → r = s
17 15 16 ifbieq2d ⊢ r = s → if r = ∅ ∅ 1 𝑜 r = if s = ∅ ∅ 1 𝑜 s
18 prex ⊢ ∅ 1 𝑜 ∈ V
19 vex ⊢ s ∈ V
20 18 19 ifex ⊢ if s = ∅ ∅ 1 𝑜 s ∈ V
21 17 1 20 fvmpt ⊢ s ∈ 𝒫 3 𝑜 → K ⁡ s = if s = ∅ ∅ 1 𝑜 s
22 14 21 sseqtrrd ⊢ s ∈ 𝒫 3 𝑜 → s ⊆ K ⁡ s
23 22 rgen ⊢ ∀ s ∈ 𝒫 3 𝑜 s ⊆ K ⁡ s