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+---
+title: "Naive Encodings of Transitivity within First-Order Logic"
+date: 2024-12-29T07:28:05-05:00
+draft: false
+tags: ["Logic"]
+math: true
+medium_enabled: false
+---
+
+> The transitive closure of a binary relation cannot, in general, be expressed in first-order logic (FO)
+
+In Ray Reiter's book "Knowledge in Action: Logical Foundations for Specifying and Implementing Dynamical Systems", he goes over a typical naive encoding of transitivity in first-order logic and goes over a counter-example[^1].
+
+[^1]: Unfortunately there was an errata in Example 2.1.1. The set `T` should be described as `{(b, a), (b, b)}` instead of the typo `{(a, b), (b, b)}`. Thanks James for noticing this!
+
+Let $G$ be a binary relation that represents whether or not there is a direct edge between two nodes.
+
+We want to axiomatize the transitive closure into the binary relation $T$. This means we need to have a formula that holds when we have a transitive closure, and does not hold when we do not.
+
+A naive axiomatization is the following:
+$$
+T(x, y) \iff G(x, y) \vee \exists z(G(x, z) \wedge T(z, y))
+$$
+Ray's counter-example is the following. Consider the following valuation of $G$:
+
+```
+G: {(b, b)}
+```
+
+{{< unsafe >}}
+
+
+
+
+
+{{< /unsafe >}}
+
+Now consider the following valuation of $T$:
+
+```
+T: {(b, a), (b, b)}
+```
+
+We falsely state that `b` is connected to `a` via transitive closure. If our axiomatization is sound, then it should be falsified.
+$$
+\begin{align*}
+T(b, a) &\iff G(b, a) \vee \exists z (G(b, z) \wedge T(z, a)) \\\\
+&\impliedby \bot \vee (G(b, b) \wedge T(b, a)) \\\\
+&\impliedby \bot \vee (\top \wedge \top) \\\\
+&\impliedby \top
+\end{align*}
+$$
+However, the formula is satisfied! Therefore, this cannot be used to axiomatize transitive closure.
+
+That's the end of the original counter-example in the book. However, I thought it would be fun to extend the exercise.
+
+The issue in the last example, is that we had a cycle in which we were able to define $T(b, a)$ in terms of $T(b, a)$. What if we add a constraint so that isn't the case?
+$$
+T(x, y) \iff G(x, y) \vee \exists z(z \ne x \wedge G(x, z) \wedge T(z, y))
+$$
+Does this new formula axiomatize transitive closure? The quote at the beginning begs to differ, so let's find a counter-example!
+
+Consider the following valuations for $G$ and $T$:
+
+```
+G: {(b, a), (a, b)}
+T: {(b, c), (a, c)}
+```
+
+{{< unsafe >}}
+
+
+
+
+
+{{< /unsafe >}}
+
+As before, this model should not have transitive closure. Let's evaluate our modified formula.
+$$
+\begin{align*}
+T(b, c) &\iff G(b, c) \vee \exists z (z \ne b \wedge G(b, z) \wedge T(z,c)) \\\\
+&\impliedby \bot \vee (a \ne b \wedge G(b, a) \wedge T(a, c)) \\\\
+&\impliedby \bot \vee (\top \wedge \top \wedge \top) \\\\
+&\impliedby \top
+\end{align*}
+$$
+
+$$
+\begin{align*}
+T(a, c) &\iff G(a, c) \vee \exists z (G(a, z) \wedge T(z, c)) \\\\
+&\impliedby G(a, c) \vee (b \ne a \wedge G(a, b) \wedge T(b, c)) \\\\
+&\impliedby \bot \vee (\top \wedge \top \wedge \top) \\\\
+&\impliedby \top
+\end{align*}
+$$
+
+Here we can see that the above valuations depending on each other.
+
+Wikipedia has a great [article](https://en.wikipedia.org/wiki/Transitive_closure) on transitive closure including a section on its use within logic and computational complexity.
+
+> (First-order Transitive-Closure Logic) FO(TC) is strictly more expressive than FO.
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