Description: The set of words over a set is a set. (Contributed by Mario Carneiro, 26-Feb-2016) (Proof shortened by JJ, 18-Nov-2022)
Ref | Expression | ||
---|---|---|---|
Assertion | wrdexg | ⊢ ( 𝑆 ∈ 𝑉 → Word 𝑆 ∈ V ) |
Step | Hyp | Ref | Expression |
---|---|---|---|
1 | wrdval | ⊢ ( 𝑆 ∈ 𝑉 → Word 𝑆 = ∪ 𝑙 ∈ ℕ0 ( 𝑆 ↑m ( 0 ..^ 𝑙 ) ) ) | |
2 | nn0ex | ⊢ ℕ0 ∈ V | |
3 | ovexd | ⊢ ( ( 𝑆 ∈ 𝑉 ∧ 𝑙 ∈ ℕ0 ) → ( 𝑆 ↑m ( 0 ..^ 𝑙 ) ) ∈ V ) | |
4 | 3 | ralrimiva | ⊢ ( 𝑆 ∈ 𝑉 → ∀ 𝑙 ∈ ℕ0 ( 𝑆 ↑m ( 0 ..^ 𝑙 ) ) ∈ V ) |
5 | iunexg | ⊢ ( ( ℕ0 ∈ V ∧ ∀ 𝑙 ∈ ℕ0 ( 𝑆 ↑m ( 0 ..^ 𝑙 ) ) ∈ V ) → ∪ 𝑙 ∈ ℕ0 ( 𝑆 ↑m ( 0 ..^ 𝑙 ) ) ∈ V ) | |
6 | 2 4 5 | sylancr | ⊢ ( 𝑆 ∈ 𝑉 → ∪ 𝑙 ∈ ℕ0 ( 𝑆 ↑m ( 0 ..^ 𝑙 ) ) ∈ V ) |
7 | 1 6 | eqeltrd | ⊢ ( 𝑆 ∈ 𝑉 → Word 𝑆 ∈ V ) |