mo4f

Description: At-most-one quantifier expressed using implicit substitution. Note that the disjoint variable condition on y , ph can be replaced by the nonfreeness hypothesis |- F/ y ph with essentially the same proof. (Contributed by NM, 10-Apr-2004) Remove dependency on ax-13 . (Revised by Wolf Lammen, 19-Jan-2023)

	Ref	Expression
Hypotheses	mo4f.1	$⊢ Ⅎ x ψ$
	mo4f.2	$⊢ x = y \to (φ \leftrightarrow ψ)$
Assertion	mo4f	$⊢ \exists^{*} x φ \leftrightarrow \forall x \forall y ((φ \land ψ) \to x = y)$

Proof

Step	Hyp	Ref	Expression
1		mo4f.1	$⊢ Ⅎ x ψ$
2		mo4f.2	$⊢ x = y \to (φ \leftrightarrow ψ)$
3		nfv	$⊢ Ⅎ y φ$
4	3	mo3	$⊢ \exists^{*} x φ \leftrightarrow \forall x \forall y ((φ \land [y/ x] φ) \to x = y)$
5	1 2	sbiev	$⊢ [y/ x] φ \leftrightarrow ψ$
6	5	anbi2i	$⊢ (φ \land [y/ x] φ) \leftrightarrow (φ \land ψ)$
7	6	imbi1i	$⊢ ((φ \land [y/ x] φ) \to x = y) \leftrightarrow ((φ \land ψ) \to x = y)$
8	7	2albii	$⊢ \forall x \forall y ((φ \land [y/ x] φ) \to x = y) \leftrightarrow \forall x \forall y ((φ \land ψ) \to x = y)$
9	4 8	bitri	$⊢ \exists^{*} x φ \leftrightarrow \forall x \forall y ((φ \land ψ) \to x = y)$

Metamath Proof Explorer

Theorem mo4f

Proof