website/content/blog/naive-encodings-transitivity-fol.md

title	date	draft	tags	math	medium_enabled
Naive Encodings of Transitivity within First-Order Logic	2024-12-29T07:28:05-05:00	false	Logic	true	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-example¹.

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 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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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! ↩︎