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I'm pretty clear with definition of substitutability for primary cases, within which the target variable to be replaced is totally free or completely bound, all over the formula.

For example, assume we're gonna find $\phi_{t}^{x}$ (Let $S$ be a unary function):

1- $x$ is bound and not free in $\phi$: $\phi:\equiv \forall x(x=y \rightarrow Sx = Sy)$ where $t=S0$

So, $\phi_{t}^{x} = (S0=y \rightarrow SS0 = Sy)$. But in view of the substitutability test, we can say $t$ is substitutable in $\phi$ instead of $x$.

2- $x$ is free in $\phi$ and $y$ does occur in $t$: $\phi:\equiv \forall y(x=y \rightarrow Sx = Sy)$ where $t=Sy$

So, $\phi_{t}^{x} = \forall y(Sy=y \rightarrow SSy = Sy)$. In view of the substitutability test, $t$ is NOT substitutable in $\phi$ instead of $x$.

But I'm not completely clear about "partially-free occurrences" of $x$:

3- $x$ is partially-free (or partially-bound): $\phi:\equiv x=y \rightarrow (\forall x)(Sx = Sy)$ where $t=Sy$

In this case, one is authorized to replace first $x$ with $t$ (because its a free occurrence), but not the second one (due to its bound property). Therefore, it can't pass the substitutability test, in overall case. Mathematically:

$\phi_{t}^{x} = Sy=y \rightarrow (SSy = Sy)$

So, can I conclude results below, in summary?

Substitutability is

1- defined

2- undefined

3- undefined

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    In the third case, the result must be : $\phi_t^x := Sy=y \to (\forall x)(Sx=Sy)$. – 2017-01-15
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    Thanks... But what about the overall result of substitutability case?... So, the substitutability is defined for the third case, by your trick, right? – 2017-01-15
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    The details are : $t$ is *substitutable for* $x$ in $(α →β)$ iff it is substitutable for $x$ in both $α$ and $β$. And $t$ is *substitutable for* $x$ in $∀y α$ iff either : (a) $x$ does not occur free in $∀y α$, or (b) $y$ does not occur in $t$ and $t$ is substitutable for $x$ in $α$. Thus, calculate... – 2017-01-15
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    We have that : $Sy$ is subst for $x$ in $x=y$ and it is subst in $(∀x)(Sx=Sy)$ because $x$ is **not** free in it. – 2017-01-15
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    We have to recall that substitution is always defined : $(∀y α)^x_t = ∀y α$ if $x = y$ and $=∀y (α^x_t)$ if $x \ne y$. – 2017-01-15
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    Sorry for my confusion! By the trick, there will be only one $x$ which is substitutable, then overall x=y→(∀z)(Sz=Sy) will be substitutable. But if we do not use the trick, $t$ will not be substitutable, because it couldn't be replaced for second $x$. But If I've understood truly, first and second $x$ are independent. So, the first argument is true and overall $(\alpha \rightarrow \beta)$ is substitutable. Do you confirm it? – 2017-01-15
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    NO; $Sy$ is overall substitutable for $x$ because there are **no** $Qy$ quantifiers that can "capture" the $y$. The second occurrence of $x$ is not *free* and thus must be "skipped" because we cannot substitute a bound var. – 2017-01-15
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    The subst test must be performed only on the free occurrences of the $x$. – 2017-01-15
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    Much appreciated for your contribution. – 2017-01-15

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