Let $\gamma=\exists!x\alpha$ . Suppose that variable $y$ is not a free variable of the wff $\alpha$ , and let $\beta$ be $\alpha$ where we substituted $y$ for all free occurrences of $x$ (note, that $x$ is not a free variable of $\beta$ ). Consider $\delta=\exists x(\alpha\wedge\forall y(\beta\rightarrow x=y))$ . We show that $\gamma$ and $\delta$ are logically equivalent. For a structure $\mathfrak{A}$ and $s:V\rightarrow|\mathfrak{A}|$ , $\vDash_{\mathfrak{A}}\gamma[s]$ iff there is unique $a\in|\mathfrak{A}|$ such that $\vDash_{\mathfrak{A}}\alpha[s(x|a)]$ iff there is $a\in|\mathfrak{A}|$ such that $\vDash_{\mathfrak{A}}\alpha[s(x|a)]$ and for any other $b\in|\mathfrak{A}|$ , $\not\vDash_{\mathfrak{A}}\alpha[s(x|b)]$ iff there is $a\in|\mathfrak{A}|$ such that $\vDash_{\mathfrak{A}}\alpha[s(x|a)]$ and for every $b\in|\mathfrak{A}|$ , $\not\vDash_{\mathfrak{A}}\alpha[s(x|b)]$ or $b=a$ iff there is $a\in|\mathfrak{A}|$ such that $\vDash_{\mathfrak{A}}\alpha[s(x|a)]$ and for every $b\in|\mathfrak{A}|$ , $\vDash_{\mathfrak{A}}\neg\alpha[s(x|b)]$ or $\vDash_{\mathfrak{A}}x=y[s(x|a)(y|b)]$ iff there is $a\in|\mathfrak{A}|$ such that $\vDash_{\mathfrak{A}}\alpha[s(x|a)]$ and for every $b\in|\mathfrak{A}|$ , $\vDash_{\mathfrak{A}}\neg\beta[s(y|b)]$ or $\vDash_{\mathfrak{A}}x=y[s(x|a)(y|b)]$ iff there is $a\in|\mathfrak{A}|$ such that $\vDash_{\mathfrak{A}}\alpha[s(x|a)]$ and for every $b\in|\mathfrak{A}|$ , $\vDash_{\mathfrak{A}}\neg\beta[s(x|a)s(y|b)]$ ($x$ is not a free variable of $\beta$ ) or $\vDash_{\mathfrak{A}}x=y[s(x|a)(y|b)]$ iff there is $a\in|\mathfrak{A}|$ such that $\vDash_{\mathfrak{A}}\alpha[s(x|a)]$ and for every $b\in|\mathfrak{A}|$ , $\vDash_{\mathfrak{A}}\neg\beta\vee x=y[s(x|a)s(y|b)]$ iff there is $a\in|\mathfrak{A}|$ such that $\vDash_{\mathfrak{A}}\alpha[s(x|a)]$ and $\vDash_{\mathfrak{A}}\forall y(\neg\beta\vee x=y)[s(x|a)]$ iff there is $a\in|\mathfrak{A}|$ such that $\vDash_{\mathfrak{A}}\alpha\wedge\forall y(\neg\beta\vee x=y)[s(x|a)]$ iff $\vDash_{\mathfrak{A}}\exists x(\alpha\wedge\forall y(\neg\beta\vee x=y))[s]$ iff $\vDash_{\mathfrak{A}}\exists x(\alpha\wedge\forall y(\beta\rightarrow x=y))[s]$ iff $\vDash_{\mathfrak{A}}\delta[s]$ .