Theorem eldisjeq

Description: Equality theorem for disjoint elementhood. (Contributed by Peter Mazsa, 23-Sep-2021)

	Ref	Expression
Assertion	eldisjeq	$$⊢A=B\to \left(ElDisjA\leftrightarrow ElDisjB\right)$$

Proof

Step	Hyp	Ref	Expression
1		reseq2	$$⊢A=B\to {{\mathrm{E}}^{-1}\upharpoonright }_A={{\mathrm{E}}^{-1}\upharpoonright }_B$$
2	1	disjeqd	$$⊢A=B\to \left(Disj\left({{\mathrm{E}}^{-1}\upharpoonright }_A\right)\leftrightarrow Disj\left({{\mathrm{E}}^{-1}\upharpoonright }_B\right)\right)$$
3		df-eldisj	$$⊢ElDisjA\leftrightarrow Disj\left({{\mathrm{E}}^{-1}\upharpoonright }_A\right)$$
4		df-eldisj	$$⊢ElDisjB\leftrightarrow Disj\left({{\mathrm{E}}^{-1}\upharpoonright }_B\right)$$
5	2 3 4	3bitr4g	$$⊢A=B\to \left(ElDisjA\leftrightarrow ElDisjB\right)$$