Theorem bj-dfrel2

Description: Alternate definition of relation. Exercise 2 of TakeutiZaring p. 25. (Contributed by NM, 29-Dec-1996) (Proof shortened by BJ, 16-Oct-2026) Do not use opabid2 . (Proof modification is discouraged.)

	Ref	Expression
Assertion	bj-dfrel2	$$⊢\mathrm{Rel}R\leftrightarrow {R^{-1}}^{-1}=R$$

Proof

Step	Hyp	Ref	Expression
1		bj-dfrel4v	$$⊢\mathrm{Rel}R\leftrightarrow \left\{⟨x,y⟩|xRy\right\}=R$$
2		cnvcnv3	$$⊢{R^{-1}}^{-1}=\left\{⟨x,y⟩|xRy\right\}$$
3	2	eqeq1i	$$⊢{R^{-1}}^{-1}=R\leftrightarrow \left\{⟨x,y⟩|xRy\right\}=R$$
4	1 3	bitr4i	$$⊢\mathrm{Rel}R\leftrightarrow {R^{-1}}^{-1}=R$$