Theorem 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	$$⊢Ⅎx\psi$$
	mo4f.2	$$⊢x=y\to \left(\phi \leftrightarrow \psi \right)$$
Assertion	mo4f	$$⊢{\exists }^{*}x\phi \leftrightarrow \forall x\forall y\left(\left(\phi \wedge \psi \right)\to x=y\right)$$

Proof

Step	Hyp	Ref	Expression
1		mo4f.1	$$⊢Ⅎx\psi$$
2		mo4f.2	$$⊢x=y\to \left(\phi \leftrightarrow \psi \right)$$
3		nfv	$$⊢Ⅎy\phi$$
4	3	mo3	$$⊢{\exists }^{*}x\phi \leftrightarrow \forall x\forall y\left(\left(\phi \wedge \left[y/x\right]\phi \right)\to x=y\right)$$
5	1 2	sbiev	$$⊢\left[y/x\right]\phi \leftrightarrow \psi$$
6	5	anbi2i	$$⊢\left(\phi \wedge \left[y/x\right]\phi \right)\leftrightarrow \left(\phi \wedge \psi \right)$$
7	6	imbi1i	$$⊢\left(\left(\phi \wedge \left[y/x\right]\phi \right)\to x=y\right)\leftrightarrow \left(\left(\phi \wedge \psi \right)\to x=y\right)$$
8	7	2albii	$$⊢\forall x\forall y\left(\left(\phi \wedge \left[y/x\right]\phi \right)\to x=y\right)\leftrightarrow \forall x\forall y\left(\left(\phi \wedge \psi \right)\to x=y\right)$$
9	4 8	bitri	$$⊢{\exists }^{*}x\phi \leftrightarrow \forall x\forall y\left(\left(\phi \wedge \psi \right)\to x=y\right)$$