Theorem relexprnd

Description: The range of an exponentiation of a relation a subset of the relation's field. (Contributed by Drahflow, 12-Nov-2015) (Revised by RP, 30-May-2020) (Revised by AV, 12-Jul-2024)

	Ref	Expression
Hypothesis	relexprnd.1	$$⊢\phi \to N\in {\mathbb{N}}_0$$
Assertion	relexprnd	$$⊢\phi \to \mathrm{ran}\left(R{\uparrow }_{r}N\right)\subseteq \bigcup \bigcup R$$

Proof

Step	Hyp	Ref	Expression
1		relexprnd.1	$$⊢\phi \to N\in {\mathbb{N}}_0$$
2		relexprn	$$⊢\left(N\in {\mathbb{N}}_0\wedge R\in \mathrm{V}\right)\to \mathrm{ran}\left(R{\uparrow }_{r}N\right)\subseteq \bigcup \bigcup R$$
3	1 2	sylan	$$⊢\left(\phi \wedge R\in \mathrm{V}\right)\to \mathrm{ran}\left(R{\uparrow }_{r}N\right)\subseteq \bigcup \bigcup R$$
4	3	ex	$$⊢\phi \to \left(R\in \mathrm{V}\to \mathrm{ran}\left(R{\uparrow }_{r}N\right)\subseteq \bigcup \bigcup R\right)$$
5		reldmrelexp	$$⊢\mathrm{Rel}\mathrm{dom}{\uparrow }_r$$
6	5	ovprc1	$$⊢\neg R\in \mathrm{V}\to R{\uparrow }_{r}N=\varnothing$$
7	6	rneqd	$$⊢\neg R\in \mathrm{V}\to \mathrm{ran}\left(R{\uparrow }_{r}N\right)=\mathrm{ran}\varnothing$$
8		rn0	$$⊢\mathrm{ran}\varnothing =\varnothing$$
9	7 8	eqtrdi	$$⊢\neg R\in \mathrm{V}\to \mathrm{ran}\left(R{\uparrow }_{r}N\right)=\varnothing$$
10		0ss	$$⊢\varnothing \subseteq \bigcup \bigcup R$$
11	9 10	eqsstrdi	$$⊢\neg R\in \mathrm{V}\to \mathrm{ran}\left(R{\uparrow }_{r}N\right)\subseteq \bigcup \bigcup R$$
12	4 11	pm2.61d1	$$⊢\phi \to \mathrm{ran}\left(R{\uparrow }_{r}N\right)\subseteq \bigcup \bigcup R$$