Metamath Proof Explorer


Theorem lpadlem3

Description: Lemma for lpadlen1 . (Contributed by Thierry Arnoux, 7-Aug-2023)

Ref Expression
Hypotheses lpadlen.1 ( 𝜑𝐿 ∈ ℕ0 )
lpadlen.2 ( 𝜑𝑊 ∈ Word 𝑆 )
lpadlen.3 ( 𝜑𝐶𝑆 )
lpadlen1.1 ( 𝜑𝐿 ≤ ( ♯ ‘ 𝑊 ) )
Assertion lpadlem3 ( 𝜑 → ( ( 0 ..^ ( 𝐿 − ( ♯ ‘ 𝑊 ) ) ) × { 𝐶 } ) = ∅ )

Proof

Step Hyp Ref Expression
1 lpadlen.1 ( 𝜑𝐿 ∈ ℕ0 )
2 lpadlen.2 ( 𝜑𝑊 ∈ Word 𝑆 )
3 lpadlen.3 ( 𝜑𝐶𝑆 )
4 lpadlen1.1 ( 𝜑𝐿 ≤ ( ♯ ‘ 𝑊 ) )
5 lencl ( 𝑊 ∈ Word 𝑆 → ( ♯ ‘ 𝑊 ) ∈ ℕ0 )
6 2 5 syl ( 𝜑 → ( ♯ ‘ 𝑊 ) ∈ ℕ0 )
7 6 nn0zd ( 𝜑 → ( ♯ ‘ 𝑊 ) ∈ ℤ )
8 1 nn0zd ( 𝜑𝐿 ∈ ℤ )
9 fzo0n ( ( ( ♯ ‘ 𝑊 ) ∈ ℤ ∧ 𝐿 ∈ ℤ ) → ( 𝐿 ≤ ( ♯ ‘ 𝑊 ) ↔ ( 0 ..^ ( 𝐿 − ( ♯ ‘ 𝑊 ) ) ) = ∅ ) )
10 9 biimpa ( ( ( ( ♯ ‘ 𝑊 ) ∈ ℤ ∧ 𝐿 ∈ ℤ ) ∧ 𝐿 ≤ ( ♯ ‘ 𝑊 ) ) → ( 0 ..^ ( 𝐿 − ( ♯ ‘ 𝑊 ) ) ) = ∅ )
11 7 8 4 10 syl21anc ( 𝜑 → ( 0 ..^ ( 𝐿 − ( ♯ ‘ 𝑊 ) ) ) = ∅ )
12 11 xpeq1d ( 𝜑 → ( ( 0 ..^ ( 𝐿 − ( ♯ ‘ 𝑊 ) ) ) × { 𝐶 } ) = ( ∅ × { 𝐶 } ) )
13 0xp ( ∅ × { 𝐶 } ) = ∅
14 12 13 eqtrdi ( 𝜑 → ( ( 0 ..^ ( 𝐿 − ( ♯ ‘ 𝑊 ) ) ) × { 𝐶 } ) = ∅ )