Theorem wrdexg

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	$$⊢S\in V\to \mathrm{Word}S\in \mathrm{V}$$

Proof

Step	Hyp	Ref	Expression
1		wrdval	$$⊢S\in V\to \mathrm{Word}S=\bigcup _{l\in {\mathbb{N}}_0}\left(S^{\left(0..^l\right)}\right)$$
2		nn0ex	$$⊢{\mathbb{N}}_0\in \mathrm{V}$$
3		ovexd	$$⊢\left(S\in V\wedge l\in {\mathbb{N}}_0\right)\to S^{\left(0..^l\right)}\in \mathrm{V}$$
4	3	ralrimiva	$$⊢S\in V\to \forall l\in {\mathbb{N}}_0{S}^{\left(0..^l\right)}\in \mathrm{V}$$
5		iunexg	$$⊢\left({\mathbb{N}}_0\in \mathrm{V}\wedge \forall l\in {\mathbb{N}}_0{S}^{\left(0..^l\right)}\in \mathrm{V}\right)\to \bigcup _{l\in {\mathbb{N}}_0}\left(S^{\left(0..^l\right)}\right)\in \mathrm{V}$$
6	2 4 5	sylancr	$$⊢S\in V\to \bigcup _{l\in {\mathbb{N}}_0}\left(S^{\left(0..^l\right)}\right)\in \mathrm{V}$$
7	1 6	eqeltrd	$$⊢S\in V\to \mathrm{Word}S\in \mathrm{V}$$