eqvrelqseqdisj3

Description: Implication of eqvreldisj3 , lemma for the Member Partition Equivalence Theorem mpet3 . (Contributed by Peter Mazsa, 27-Oct-2020) (Revised by Peter Mazsa, 24-Sep-2021)

		Ref	Expression
	Assertion	eqvrelqseqdisj3	$⊢ (EqvRel R \land B / R = A) \to Disj ({E^{-1} ↾}_{A})$

Proof

Step	Hyp	Ref	Expression
1		eqvreldisj3	$⊢ EqvRel R \to Disj ({E^{-1} ↾}_{(B / R)})$
2	1	adantr	$⊢ (EqvRel R \land B / R = A) \to Disj ({E^{-1} ↾}_{(B / R)})$
3		reseq2	$⊢ B / R = A \to {E^{-1} ↾}_{(B / R)} = {E^{-1} ↾}_{A}$
4	3	disjeqd	$⊢ B / R = A \to (Disj ({E^{-1} ↾}_{(B / R)}) \leftrightarrow Disj ({E^{-1} ↾}_{A}))$
5	4	adantl	$⊢ (EqvRel R \land B / R = A) \to (Disj ({E^{-1} ↾}_{(B / R)}) \leftrightarrow Disj ({E^{-1} ↾}_{A}))$
6	2 5	mpbid	$⊢ (EqvRel R \land B / R = A) \to Disj ({E^{-1} ↾}_{A})$

Metamath Proof Explorer

Theorem eqvrelqseqdisj3

Proof