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Metamath Proof Explorer

Theorem clwlkclwwlklem2a2

Description: Lemma 2 for clwlkclwwlklem2a . (Contributed by Alexander van der Vekens, 21-Jun-2018)

Ref Expression
Hypothesis clwlkclwwlklem2.f ⊢ 𝐹 = ( 𝑥 ∈ ( 0 ..^ ( ( ♯ ‘ 𝑃 ) − 1 ) ) ↦ if ( 𝑥 < ( ( ♯ ‘ 𝑃 ) − 2 ) , ( ◡ 𝐸 ‘ { ( 𝑃 ‘ 𝑥 ) , ( 𝑃 ‘ ( 𝑥 + 1 ) ) } ) , ( ◡ 𝐸 ‘ { ( 𝑃 ‘ 𝑥 ) , ( 𝑃 ‘ 0 ) } ) ) )
Assertion clwlkclwwlklem2a2 ( ( 𝑃 ∈ Word 𝑉 ∧ 2 ≤ ( ♯ ‘ 𝑃 ) ) → ( ♯ ‘ 𝐹 ) = ( ( ♯ ‘ 𝑃 ) − 1 ) )

Proof

Step Hyp Ref Expression
1 clwlkclwwlklem2.f ⊢ 𝐹 = ( 𝑥 ∈ ( 0 ..^ ( ( ♯ ‘ 𝑃 ) − 1 ) ) ↦ if ( 𝑥 < ( ( ♯ ‘ 𝑃 ) − 2 ) , ( ◡ 𝐸 ‘ { ( 𝑃 ‘ 𝑥 ) , ( 𝑃 ‘ ( 𝑥 + 1 ) ) } ) , ( ◡ 𝐸 ‘ { ( 𝑃 ‘ 𝑥 ) , ( 𝑃 ‘ 0 ) } ) ) )
2 lencl ⊢ ( 𝑃 ∈ Word 𝑉 → ( ♯ ‘ 𝑃 ) ∈ ℕ0 )
3 nn0z ⊢ ( ( ♯ ‘ 𝑃 ) ∈ ℕ0 → ( ♯ ‘ 𝑃 ) ∈ ℤ )
4 3 adantr ⊢ ( ( ( ♯ ‘ 𝑃 ) ∈ ℕ0 ∧ 2 ≤ ( ♯ ‘ 𝑃 ) ) → ( ♯ ‘ 𝑃 ) ∈ ℤ )
5 0red ⊢ ( ( ( ♯ ‘ 𝑃 ) ∈ ℕ0 ∧ 2 ≤ ( ♯ ‘ 𝑃 ) ) → 0 ∈ ℝ )
6 2re ⊢ 2 ∈ ℝ
7 6 a1i ⊢ ( ( ( ♯ ‘ 𝑃 ) ∈ ℕ0 ∧ 2 ≤ ( ♯ ‘ 𝑃 ) ) → 2 ∈ ℝ )
8 nn0re ⊢ ( ( ♯ ‘ 𝑃 ) ∈ ℕ0 → ( ♯ ‘ 𝑃 ) ∈ ℝ )
9 8 adantr ⊢ ( ( ( ♯ ‘ 𝑃 ) ∈ ℕ0 ∧ 2 ≤ ( ♯ ‘ 𝑃 ) ) → ( ♯ ‘ 𝑃 ) ∈ ℝ )
10 2pos ⊢ 0 < 2
11 10 a1i ⊢ ( ( ( ♯ ‘ 𝑃 ) ∈ ℕ0 ∧ 2 ≤ ( ♯ ‘ 𝑃 ) ) → 0 < 2 )
12 simpr ⊢ ( ( ( ♯ ‘ 𝑃 ) ∈ ℕ0 ∧ 2 ≤ ( ♯ ‘ 𝑃 ) ) → 2 ≤ ( ♯ ‘ 𝑃 ) )
13 5 7 9 11 12 ltletrd ⊢ ( ( ( ♯ ‘ 𝑃 ) ∈ ℕ0 ∧ 2 ≤ ( ♯ ‘ 𝑃 ) ) → 0 < ( ♯ ‘ 𝑃 ) )
14 elnnz ⊢ ( ( ♯ ‘ 𝑃 ) ∈ ℕ ↔ ( ( ♯ ‘ 𝑃 ) ∈ ℤ ∧ 0 < ( ♯ ‘ 𝑃 ) ) )
15 4 13 14 sylanbrc ⊢ ( ( ( ♯ ‘ 𝑃 ) ∈ ℕ0 ∧ 2 ≤ ( ♯ ‘ 𝑃 ) ) → ( ♯ ‘ 𝑃 ) ∈ ℕ )
16 2 15 sylan ⊢ ( ( 𝑃 ∈ Word 𝑉 ∧ 2 ≤ ( ♯ ‘ 𝑃 ) ) → ( ♯ ‘ 𝑃 ) ∈ ℕ )
17 nnm1nn0 ⊢ ( ( ♯ ‘ 𝑃 ) ∈ ℕ → ( ( ♯ ‘ 𝑃 ) − 1 ) ∈ ℕ0 )
18 16 17 syl ⊢ ( ( 𝑃 ∈ Word 𝑉 ∧ 2 ≤ ( ♯ ‘ 𝑃 ) ) → ( ( ♯ ‘ 𝑃 ) − 1 ) ∈ ℕ0 )
19 fvex ⊢ ( ◡ 𝐸 ‘ { ( 𝑃 ‘ 𝑥 ) , ( 𝑃 ‘ ( 𝑥 + 1 ) ) } ) ∈ V
20 fvex ⊢ ( ◡ 𝐸 ‘ { ( 𝑃 ‘ 𝑥 ) , ( 𝑃 ‘ 0 ) } ) ∈ V
21 19 20 ifex ⊢ if ( 𝑥 < ( ( ♯ ‘ 𝑃 ) − 2 ) , ( ◡ 𝐸 ‘ { ( 𝑃 ‘ 𝑥 ) , ( 𝑃 ‘ ( 𝑥 + 1 ) ) } ) , ( ◡ 𝐸 ‘ { ( 𝑃 ‘ 𝑥 ) , ( 𝑃 ‘ 0 ) } ) ) ∈ V
22 21 1 fnmpti ⊢ 𝐹 Fn ( 0 ..^ ( ( ♯ ‘ 𝑃 ) − 1 ) )
23 ffzo0hash ⊢ ( ( ( ( ♯ ‘ 𝑃 ) − 1 ) ∈ ℕ0 ∧ 𝐹 Fn ( 0 ..^ ( ( ♯ ‘ 𝑃 ) − 1 ) ) ) → ( ♯ ‘ 𝐹 ) = ( ( ♯ ‘ 𝑃 ) − 1 ) )
24 18 22 23 sylancl ⊢ ( ( 𝑃 ∈ Word 𝑉 ∧ 2 ≤ ( ♯ ‘ 𝑃 ) ) → ( ♯ ‘ 𝐹 ) = ( ( ♯ ‘ 𝑃 ) − 1 ) )