Theorem eqvreldmqs

Description: Two ways to express membership equivalence relation on its domain quotient. (Contributed by Peter Mazsa, 26-Sep-2021) (Revised by Peter Mazsa, 17-Jul-2023)

	Ref	Expression
Assertion	eqvreldmqs	$$⊢\left(EqvRel\wr \left({{\mathrm{E}}^{-1}\upharpoonright }_A\right)\wedge \mathrm{dom}\wr \left({{\mathrm{E}}^{-1}\upharpoonright }_A\right)/\wr \left({{\mathrm{E}}^{-1}\upharpoonright }_A\right)=A\right)\leftrightarrow \left(CoElEqvRelA\wedge \bigcup A/\sim A=A\right)$$

Proof

Step	Hyp	Ref	Expression
1		df-coeleqvrel	$$⊢CoElEqvRelA\leftrightarrow EqvRel\wr \left({{\mathrm{E}}^{-1}\upharpoonright }_A\right)$$
2	1	bicomi	$$⊢EqvRel\wr \left({{\mathrm{E}}^{-1}\upharpoonright }_A\right)\leftrightarrow CoElEqvRelA$$
3		dmqs1cosscnvepreseq	$$⊢\mathrm{dom}\wr \left({{\mathrm{E}}^{-1}\upharpoonright }_A\right)/\wr \left({{\mathrm{E}}^{-1}\upharpoonright }_A\right)=A\leftrightarrow \bigcup A/\sim A=A$$
4	2 3	anbi12i	$$⊢\left(EqvRel\wr \left({{\mathrm{E}}^{-1}\upharpoonright }_A\right)\wedge \mathrm{dom}\wr \left({{\mathrm{E}}^{-1}\upharpoonright }_A\right)/\wr \left({{\mathrm{E}}^{-1}\upharpoonright }_A\right)=A\right)\leftrightarrow \left(CoElEqvRelA\wedge \bigcup A/\sim A=A\right)$$