Theorem eqvreldmqs2

Description: Two ways to express comember equivalence relation on its domain quotient. (Contributed by Peter Mazsa, 30-Dec-2024)

	Ref	Expression
Assertion	eqvreldmqs2	$$⊢\left(EqvRel\wr \left({{\mathrm{E}}^{-1}\upharpoonright }_A\right)\wedge \mathrm{dom}\wr \left({{\mathrm{E}}^{-1}\upharpoonright }_A\right)/\wr \left({{\mathrm{E}}^{-1}\upharpoonright }_A\right)=A\right)\leftrightarrow \left(EqvRel\sim A\wedge \bigcup A/\sim A=A\right)$$

Proof

Step	Hyp	Ref	Expression
1		df-coels	$$⊢\sim A=\wr \left({{\mathrm{E}}^{-1}\upharpoonright }_A\right)$$
2	1	eqvreleqi	$$⊢EqvRel\sim A\leftrightarrow EqvRel\wr \left({{\mathrm{E}}^{-1}\upharpoonright }_A\right)$$
3	2	bicomi	$$⊢EqvRel\wr \left({{\mathrm{E}}^{-1}\upharpoonright }_A\right)\leftrightarrow EqvRel\sim A$$
4		dmqs1cosscnvepreseq	$$⊢\mathrm{dom}\wr \left({{\mathrm{E}}^{-1}\upharpoonright }_A\right)/\wr \left({{\mathrm{E}}^{-1}\upharpoonright }_A\right)=A\leftrightarrow \bigcup A/\sim A=A$$
5	3 4	anbi12i	$$⊢\left(EqvRel\wr \left({{\mathrm{E}}^{-1}\upharpoonright }_A\right)\wedge \mathrm{dom}\wr \left({{\mathrm{E}}^{-1}\upharpoonright }_A\right)/\wr \left({{\mathrm{E}}^{-1}\upharpoonright }_A\right)=A\right)\leftrightarrow \left(EqvRel\sim A\wedge \bigcup A/\sim A=A\right)$$