Translating "If a function $f$ has an inverse, then that inverse also has an inverse"

Question

Suppose that we are in an interpretation $\mathcal{I}$ with the domain being the collection of all functions. Define the predicate $Pxy:= "y\circ x=i_{\text{Dom}(x)} \enspace \wedge \enspace x\circ y=i_{\text{Dom}(y)}"$.

• If a function $f$ has an inverse, then that inverse also has an inverse. $\quad (\ast)$

Which of these predicate-logic translations captures (or more accurately captures) the above statement?

1. $\forall f \,\Big( (\exists g \:Pfg) \longrightarrow \exists h \:Pgh\Big)$

2. $\forall f\;\exists g\, \Big( Pfg \longrightarrow \exists h \:Pgh\Big)$

Thoughts:

Writing out what it means for 1. to be true in $\mathcal{I}$:

\begin{gather} \mathcal{I} \vDash_{v} \forall f [ (\exists g Pfg) \longrightarrow \exists h Pgh] \quad \text{for every $\mathcal{I}-$ assignment $v$}\\ \text{iff} \\ \enspace \left( \left(\mathcal{I} \not\vDash_{v \frac{f'}{f}\frac{g'}{g}} Pfg \enspace \text{for every $g'$ in $D$}\right) \text{or}\left( \mathcal{I} \vDash_{v\frac{f'}{f}\frac{h'}{h} } Pgh \enspace \text{ for some $h$ in $D$} \right) \right) \text{ for every $f'$ in $D$} \quad \text{for every $\mathcal{I}-$ assignment $v$} \tag{1} \\ \end{gather}

and similarly for 2. to be true in $\mathcal{I}$:

\begin{gather} \mathcal{I} \vDash_v \forall f\exists g [Pfg \longrightarrow \exists h Pgh] \quad \text{for every $\mathcal{I}-$ assignment $v$} \\ \text{iff} \\ \left( \left( \left(\mathcal{I} \not\vDash_{v\frac{g'}{g}} Pfg \right) \text{or} \left( \mathcal{I} \vDash_{v\frac{g'}{g}\frac{h'}{h} } Pgh \enspace \text{for some $h'$ in $D$} \right) \right) \text{for some $g'$ in $D$} \right) \text{for every $f'$ in $D$} \quad \text{for every $\mathcal{I}-$ assignment $v$} \tag{2} \\ \end{gather}

It is easy to see that $(2)$ is true, thus 2. is true in $\mathcal{I}$. However 2. allows for the possibility that for every arbitrary $f_0$ in $D$ that does not have an inverse we can choose a $g_0$ (of course any $g_0$ in $D$ would work since $f_0$ does not have an inverse) such that $\mathcal{I} \not\vDash_{v\frac{f_0}{f}\frac{g_0}{g}} Pfg$ for every $\mathcal{I}-$assignment $v$ so that $(2)$ is true. You may ask, so what? This case is still consistent with our meaning with $(\ast)$ since this is the case where the antecedent is false, i.e. $f$ has no inverse. But when we look at 1. it excludes this possibility because the existential quantifier becomes part of the implication part of the expression, therefore $(1)$ cannot necessarily be true for an $f$ that does not have an inverse by choosing some $g_0$ but rather becomes true because $\mathcal{I} \not\vDash_{v \frac{f'}{f}\frac{g'}{g}} Pfg \enspace \text{for every $g'$ in $D$}$, which is more accurate in some sense since this is a consequence of a function not having an inverse. However 1. has a free variable $g$ in $\exists h Pgh$ so $(1)$ is not necessarily true for every $\mathcal{I}-$assignment $v$. But suppose we are allowed to assume that we could somehow assign the same $g'$ in $D$ such that $\mathcal{I} \vDash_{v\frac{f'}{f}\frac{g'}{g}} Pfg$ for some $g'$ in $D$ to the $g$ that occurs free in 1. so that the free variable $g'$ is in effect in the "scope" of the existential quantifier in 1., then under this assumption $(1)$ is true. So it seems to me that 1. captures $(\ast)$ better than 2., but the free variable is causing trouble because I can't make 1. true under this interpretation.

Having said all this, my questions then are:

I. Is there a way to re-write 1. formally so that somehow the $g$ that you choose is the same $g$ in the consequent of the implication without moving the existential quantifier outside the closed bracket (or else that becomes the same as 2.)? Or is there no way of doing this because of the way the scope of a quantifier is defined? Is this just the limitation of what can be expressed within the limits of the "grammar" of wffs?

II. And again as asked above, which of the two translations (1. or 2.) captures (or more accurately captures) what we mean by $(\ast)$?

P.S. This post is inspired by What is the difference between $\forall x [ (\exists y Pxy) \longrightarrow \phi ]$ and $\forall x \exists y [Pxy \longrightarrow \phi]$?

Answer 1

• If a function $f$ has an inverse, then that inverse also has an inverse. $\quad (\ast)$

1. First, notice that you intend for the above statement not to be interpreted as “If each function $f$...” but as

   • If an arbitrary function $f$ has an inverse, then that inverse also has an inverse.

2. Next, observe that, because you want the 1st and 2nd of the three copies of ‘inverse’ to have the same referent, there is an implicit universal quantification:

   • For each function $\boldsymbol {g,}$ if $g$ is the inverse of $f,$ then $g$ also has an inverse.

3. More explicitly:

   • For each function $g,$ if $g$ is the inverse of $f,$ then for some function $\boldsymbol {h,}$ $h$ is the inverse of $g.$

4. Finally, invoking universal generalisation, the required formalisation is $$∀f\;∀g\,\Big(Pfg → ∃h\:Pgh\Big)\tag{*}$$ or, equivalently, $$∀f\;∀g\,∃h\,\Big(Pfg → Pgh\Big).\tag{*}$$ This can be read simply as

   • Every function that is the inverse of some function itself has an inverse. $\quad (\ast)$

1. $\forall f \,\Big( (\exists g \:Pfg) \to \exists h \:Pgh\Big)$

2. $\forall f\;\exists g\, \Big( Pfg \to \exists h \:Pgh\Big)$

Converting these sentences to prenex form:

1. $\forall f \:\forall j \:\exists h \,\Big( Pfj \to Pgh\Big)$

   this formula contains a free variable, so can't be logically equivalent to $(*);$

2. $\forall f\:\exists g\: \exists h \,\Big( Pfg \to Pgh\Big)$

   this sentence is true in a universe containing exactly two functions $p$ and $q$ such that $Ppq$ and $\lnot Ppp$ and $\lnot\exists h\, Pqh;$ however, this universe falsifies $(*);$ thus, this sentence too is not logically equivalent to $(*).$