Theorem wfelirr

Description: A well-founded set is not a member of itself. This proof does not require the axiom of regularity, unlike elirr . (Contributed by Mario Carneiro, 2-Jan-2017)

	Ref	Expression
Assertion	wfelirr	$$⊢A\in \bigcup R_1\left[\mathrm{On}\right]\to \neg A\in A$$

Proof

Step	Hyp	Ref	Expression
1		rankon	$$⊢\mathrm{rank}\left(A\right)\in \mathrm{On}$$
2	1	onirri	$$⊢\neg \mathrm{rank}\left(A\right)\in \mathrm{rank}\left(A\right)$$
3		rankelb	$$⊢A\in \bigcup R_1\left[\mathrm{On}\right]\to \left(A\in A\to \mathrm{rank}\left(A\right)\in \mathrm{rank}\left(A\right)\right)$$
4	2 3	mtoi	$$⊢A\in \bigcup R_1\left[\mathrm{On}\right]\to \neg A\in A$$