Theorem wlkiswwlks2lem3

Description: Lemma 3 for wlkiswwlks2 . (Contributed by Alexander van der Vekens, 20-Jul-2018)

	Ref	Expression
Hypothesis	wlkiswwlks2lem.f	⊢ 𝐹 = ( 𝑥 ∈ ( 0 ..^ ( ( ♯ ‘ 𝑃 ) − 1 ) ) ↦ ( ◡ 𝐸 ‘ { ( 𝑃 ‘ 𝑥 ) , ( 𝑃 ‘ ( 𝑥 + 1 ) ) } ) )
Assertion	wlkiswwlks2lem3	⊢ ( ( 𝑃 ∈ Word 𝑉 ∧ 1 ≤ ( ♯ ‘ 𝑃 ) ) → 𝑃 : ( 0 ... ( ♯ ‘ 𝐹 ) ) ⟶ 𝑉 )

Proof

Step	Hyp	Ref	Expression
1		wlkiswwlks2lem.f	⊢ 𝐹 = ( 𝑥 ∈ ( 0 ..^ ( ( ♯ ‘ 𝑃 ) − 1 ) ) ↦ ( ◡ 𝐸 ‘ { ( 𝑃 ‘ 𝑥 ) , ( 𝑃 ‘ ( 𝑥 + 1 ) ) } ) )
2	1	wlkiswwlks2lem1	⊢ ( ( 𝑃 ∈ Word 𝑉 ∧ 1 ≤ ( ♯ ‘ 𝑃 ) ) → ( ♯ ‘ 𝐹 ) = ( ( ♯ ‘ 𝑃 ) − 1 ) )
3		wrdf	⊢ ( 𝑃 ∈ Word 𝑉 → 𝑃 : ( 0 ..^ ( ♯ ‘ 𝑃 ) ) ⟶ 𝑉 )
4		lencl	⊢ ( 𝑃 ∈ Word 𝑉 → ( ♯ ‘ 𝑃 ) ∈ ℕ0 )
5		nn0z	⊢ ( ( ♯ ‘ 𝑃 ) ∈ ℕ0 → ( ♯ ‘ 𝑃 ) ∈ ℤ )
6		fzoval	⊢ ( ( ♯ ‘ 𝑃 ) ∈ ℤ → ( 0 ..^ ( ♯ ‘ 𝑃 ) ) = ( 0 ... ( ( ♯ ‘ 𝑃 ) − 1 ) ) )
7	5 6	syl	⊢ ( ( ♯ ‘ 𝑃 ) ∈ ℕ0 → ( 0 ..^ ( ♯ ‘ 𝑃 ) ) = ( 0 ... ( ( ♯ ‘ 𝑃 ) − 1 ) ) )
8		oveq2	⊢ ( ( ( ♯ ‘ 𝑃 ) − 1 ) = ( ♯ ‘ 𝐹 ) → ( 0 ... ( ( ♯ ‘ 𝑃 ) − 1 ) ) = ( 0 ... ( ♯ ‘ 𝐹 ) ) )
9	8	eqcoms	⊢ ( ( ♯ ‘ 𝐹 ) = ( ( ♯ ‘ 𝑃 ) − 1 ) → ( 0 ... ( ( ♯ ‘ 𝑃 ) − 1 ) ) = ( 0 ... ( ♯ ‘ 𝐹 ) ) )
10	7 9	sylan9eq	⊢ ( ( ( ♯ ‘ 𝑃 ) ∈ ℕ0 ∧ ( ♯ ‘ 𝐹 ) = ( ( ♯ ‘ 𝑃 ) − 1 ) ) → ( 0 ..^ ( ♯ ‘ 𝑃 ) ) = ( 0 ... ( ♯ ‘ 𝐹 ) ) )
11	10	feq2d	⊢ ( ( ( ♯ ‘ 𝑃 ) ∈ ℕ0 ∧ ( ♯ ‘ 𝐹 ) = ( ( ♯ ‘ 𝑃 ) − 1 ) ) → ( 𝑃 : ( 0 ..^ ( ♯ ‘ 𝑃 ) ) ⟶ 𝑉 ↔ 𝑃 : ( 0 ... ( ♯ ‘ 𝐹 ) ) ⟶ 𝑉 ) )
12	11	biimpcd	⊢ ( 𝑃 : ( 0 ..^ ( ♯ ‘ 𝑃 ) ) ⟶ 𝑉 → ( ( ( ♯ ‘ 𝑃 ) ∈ ℕ0 ∧ ( ♯ ‘ 𝐹 ) = ( ( ♯ ‘ 𝑃 ) − 1 ) ) → 𝑃 : ( 0 ... ( ♯ ‘ 𝐹 ) ) ⟶ 𝑉 ) )
13	12	expd	⊢ ( 𝑃 : ( 0 ..^ ( ♯ ‘ 𝑃 ) ) ⟶ 𝑉 → ( ( ♯ ‘ 𝑃 ) ∈ ℕ0 → ( ( ♯ ‘ 𝐹 ) = ( ( ♯ ‘ 𝑃 ) − 1 ) → 𝑃 : ( 0 ... ( ♯ ‘ 𝐹 ) ) ⟶ 𝑉 ) ) )
14	3 4 13	sylc	⊢ ( 𝑃 ∈ Word 𝑉 → ( ( ♯ ‘ 𝐹 ) = ( ( ♯ ‘ 𝑃 ) − 1 ) → 𝑃 : ( 0 ... ( ♯ ‘ 𝐹 ) ) ⟶ 𝑉 ) )
15	14	adantr	⊢ ( ( 𝑃 ∈ Word 𝑉 ∧ 1 ≤ ( ♯ ‘ 𝑃 ) ) → ( ( ♯ ‘ 𝐹 ) = ( ( ♯ ‘ 𝑃 ) − 1 ) → 𝑃 : ( 0 ... ( ♯ ‘ 𝐹 ) ) ⟶ 𝑉 ) )
16	2 15	mpd	⊢ ( ( 𝑃 ∈ Word 𝑉 ∧ 1 ≤ ( ♯ ‘ 𝑃 ) ) → 𝑃 : ( 0 ... ( ♯ ‘ 𝐹 ) ) ⟶ 𝑉 )