Theorem coi1res

Description: Precomposing with the identity is the same as taking the relation part of a class. (Contributed by BJ, 16-Aug-2026)

	Ref	Expression
Assertion	coi1res	$$⊢A\circ \mathrm{I}={A\upharpoonright }_{\mathrm{V}}$$

Proof

Step	Hyp	Ref	Expression
1		vex	$$⊢z\in \mathrm{V}$$
2	1	ideq	$$⊢x\mathrm{I}z\leftrightarrow x=z$$
3		equcom	$$⊢x=z\leftrightarrow z=x$$
4	2 3	bitri	$$⊢x\mathrm{I}z\leftrightarrow z=x$$
5	4	anbi1i	$$⊢\left(x\mathrm{I}z\wedge zAy\right)\leftrightarrow \left(z=x\wedge zAy\right)$$
6	5	exbii	$$⊢\exists z\left(x\mathrm{I}z\wedge zAy\right)\leftrightarrow \exists z\left(z=x\wedge zAy\right)$$
7		breq1	$$⊢z=x\to \left(zAy\leftrightarrow xAy\right)$$
8	7	equsexvw	$$⊢\exists z\left(z=x\wedge zAy\right)\leftrightarrow xAy$$
9	6 8	bitri	$$⊢\exists z\left(x\mathrm{I}z\wedge zAy\right)\leftrightarrow xAy$$
10	9	opabbii	$$⊢\left\{⟨x,y⟩|\exists z\left(x\mathrm{I}z\wedge zAy\right)\right\}=\left\{⟨x,y⟩|xAy\right\}$$
11		df-co	$$⊢A\circ \mathrm{I}=\left\{⟨x,y⟩|\exists z\left(x\mathrm{I}z\wedge zAy\right)\right\}$$
12		relpart	$$⊢{A\upharpoonright }_{\mathrm{V}}=\left\{⟨x,y⟩|xAy\right\}$$
13	10 11 12	3eqtr4i	$$⊢A\circ \mathrm{I}={A\upharpoonright }_{\mathrm{V}}$$