Metamath Proof Explorer

Theorem sbel2x

Description: Elimination of double substitution. Usage of this theorem is discouraged because it depends on ax-13 . (Contributed by NM, 5-Aug-1993) (Proof shortened by Wolf Lammen, 29-Sep-2018) (New usage is discouraged.)

		Ref	Expression
	Assertion	sbel2x	$⊢ φ \leftrightarrow \exists x \exists y ((x = z \land y = w) \land [y/ w] [x/ z] φ)$

Proof

Step	Hyp	Ref	Expression
1		nfv	$⊢ Ⅎ y φ$
2		nfv	$⊢ Ⅎ x φ$
3	1 2	2sb5rf	$⊢ φ \leftrightarrow \exists y \exists x ((y = w \land x = z) \land [y/ w] [x/ z] φ)$
4		ancom	$⊢ (y = w \land x = z) \leftrightarrow (x = z \land y = w)$
5	4	anbi1i	$⊢ ((y = w \land x = z) \land [y/ w] [x/ z] φ) \leftrightarrow ((x = z \land y = w) \land [y/ w] [x/ z] φ)$
6	5	2exbii	$⊢ \exists y \exists x ((y = w \land x = z) \land [y/ w] [x/ z] φ) \leftrightarrow \exists y \exists x ((x = z \land y = w) \land [y/ w] [x/ z] φ)$
7		excom	$⊢ \exists y \exists x ((x = z \land y = w) \land [y/ w] [x/ z] φ) \leftrightarrow \exists x \exists y ((x = z \land y = w) \land [y/ w] [x/ z] φ)$
8	3 6 7	3bitri	$⊢ φ \leftrightarrow \exists x \exists y ((x = z \land y = w) \land [y/ w] [x/ z] φ)$