Metamath Proof Explorer


Theorem clwwlknonex2lem1

Description: Lemma 1 for clwwlknonex2 : Transformation of a special half-open integer range into a union of a smaller half-open integer range and an unordered pair. This Lemma would not hold for N = 2 , i.e., ( #W ) = 0 , because ( 0 ..^ ( ( ( #W ) + 2 ) - 1 ) ) = ( 0 ..^ ( ( 0 + 2 ) - 1 ) ) = ( 0 ..^ 1 ) = { 0 } =/= { -u 1 , 0 } = ( (/) u. { -u 1 , 0 } ) = ( ( 0 ..^ ( 0 - 1 ) ) u. { ( 0 - 1 ) , 0 } ) = ( ( 0 ..^ ( ( #W ) - 1 ) ) u. { ( ( #W ) - 1 ) , ( #W ) } ) . (Contributed by AV, 22-Sep-2018) (Revised by AV, 26-Jan-2022)

Ref Expression
Assertion clwwlknonex2lem1 ⊢ N ∈ ℤ ≥ 3 ∧ W = N − 2 → 0 ..^ W + 2 - 1 = 0 ..^ W − 1 ∪ W − 1 W

Proof

Step Hyp Ref Expression
1 eluzelcn ⊢ N ∈ ℤ ≥ 3 → N ∈ ℂ
2 2cnd ⊢ N ∈ ℤ ≥ 3 → 2 ∈ ℂ
3 1 2 subcld ⊢ N ∈ ℤ ≥ 3 → N − 2 ∈ ℂ
4 3 adantr ⊢ N ∈ ℤ ≥ 3 ∧ W = N − 2 → N − 2 ∈ ℂ
5 eleq1 ⊢ W = N − 2 → W ∈ ℂ ↔ N − 2 ∈ ℂ
6 5 adantl ⊢ N ∈ ℤ ≥ 3 ∧ W = N − 2 → W ∈ ℂ ↔ N − 2 ∈ ℂ
7 4 6 mpbird ⊢ N ∈ ℤ ≥ 3 ∧ W = N − 2 → W ∈ ℂ
8 2cnd ⊢ N ∈ ℤ ≥ 3 ∧ W = N − 2 → 2 ∈ ℂ
9 1cnd ⊢ N ∈ ℤ ≥ 3 ∧ W = N − 2 → 1 ∈ ℂ
10 7 8 9 addsubd ⊢ N ∈ ℤ ≥ 3 ∧ W = N − 2 → W + 2 - 1 = W - 1 + 2
11 10 oveq2d ⊢ N ∈ ℤ ≥ 3 ∧ W = N − 2 → 0 ..^ W + 2 - 1 = 0 ..^ W - 1 + 2
12 oveq1 ⊢ W = N − 2 → W − 1 = N - 2 - 1
13 12 adantl ⊢ N ∈ ℤ ≥ 3 ∧ W = N − 2 → W − 1 = N - 2 - 1
14 uznn0sub ⊢ N ∈ ℤ ≥ 3 → N − 3 ∈ ℕ 0
15 1cnd ⊢ N ∈ ℤ ≥ 3 → 1 ∈ ℂ
16 1 2 15 subsub4d ⊢ N ∈ ℤ ≥ 3 → N - 2 - 1 = N − 2 + 1
17 2p1e3 ⊢ 2 + 1 = 3
18 17 oveq2i ⊢ N − 2 + 1 = N − 3
19 16 18 eqtrdi ⊢ N ∈ ℤ ≥ 3 → N - 2 - 1 = N − 3
20 nn0uz ⊢ ℕ 0 = ℤ ≥ 0
21 20 eqcomi ⊢ ℤ ≥ 0 = ℕ 0
22 21 a1i ⊢ N ∈ ℤ ≥ 3 → ℤ ≥ 0 = ℕ 0
23 14 19 22 3eltr4d ⊢ N ∈ ℤ ≥ 3 → N - 2 - 1 ∈ ℤ ≥ 0
24 23 adantr ⊢ N ∈ ℤ ≥ 3 ∧ W = N − 2 → N - 2 - 1 ∈ ℤ ≥ 0
25 13 24 eqeltrd ⊢ N ∈ ℤ ≥ 3 ∧ W = N − 2 → W − 1 ∈ ℤ ≥ 0
26 fzosplitpr ⊢ W − 1 ∈ ℤ ≥ 0 → 0 ..^ W - 1 + 2 = 0 ..^ W − 1 ∪ W − 1 W - 1 + 1
27 25 26 syl ⊢ N ∈ ℤ ≥ 3 ∧ W = N − 2 → 0 ..^ W - 1 + 2 = 0 ..^ W − 1 ∪ W − 1 W - 1 + 1
28 7 9 npcand ⊢ N ∈ ℤ ≥ 3 ∧ W = N − 2 → W - 1 + 1 = W
29 28 preq2d ⊢ N ∈ ℤ ≥ 3 ∧ W = N − 2 → W − 1 W - 1 + 1 = W − 1 W
30 29 uneq2d ⊢ N ∈ ℤ ≥ 3 ∧ W = N − 2 → 0 ..^ W − 1 ∪ W − 1 W - 1 + 1 = 0 ..^ W − 1 ∪ W − 1 W
31 11 27 30 3eqtrd ⊢ N ∈ ℤ ≥ 3 ∧ W = N − 2 → 0 ..^ W + 2 - 1 = 0 ..^ W − 1 ∪ W − 1 W