Truth Functional Expansions

content/research/atp/_index.md

@@ -6,3 +6,4 @@ description: Notes about Automated Theorem Proving
 More links coming soonish:
 - [Definitional CNF](definitional-cnf)
 - [Davis Putnam](davis-putnam)
+- [Finding Counter-Models through Truth Functional Expansions](truth-functional-expansion)

content/research/atp/truth-functional-expansion.md

@@ -0,0 +1,47 @@
+---
+title: Finding Counter-Models through Truth Functional Expansions
+math: true
+---
+
+The best way to disprove a statement is to find a counter-example.
+For first order formulas, the quantifiers are over some *universe of discourse* $\mathcal{D}$.
+
+For example, $\forall x, P(x)$ says that for every $x$ within the universe of discourse, $x$ has property $P$. Let's say that our universe of discourse consists of two elements $c_1$ and $c_2$. Then the statement $\forall x, P(x)$ is equivalent to $P(c_1) \wedge P(c_2)$. That is, both elements in the universe of discourse hold property $P$. This equivalence is called the *truth-functional expansion*.
+
+Let $\mathcal{D} = \\{ c_1, \dots, c_n \\}$ be the universe of discourse. Then the following equivalences hold:
+$$
+    \forall x P(x) \iff P(c_1) \wedge \dots \wedge P(c_n)
+$$
+$$
+    \exists x P(x) \iff P(c_1) \vee \dots \vee P(c_n)
+$$
+
+To find a counter-example, you'll need to find a model in which the formula does not hold. A model consists of a universe of discourse and whether or not each possible predicate holds. For example, let's take a binary predicate $F$. For a domain of discourse of size two, the following predicates may hold: $F(c_1, c_1), F(c_1, c_2), F(c_2, c_1), F(c_2, c_2)$. For any given model, each of those instantiations can hold or not, amounting to $2^4 = 16$ possible models for a domain of discourse of size $2$ with one binary predicate.
+
+Now let's find a counter-example for an invalid formula. Consider $\exists x F(x, x) \implies \forall x F(x, x)$.
+To perform a truth functional expansion, we need to first convert to [prenex normal form](https://en.wikipedia.org/wiki/Prenex_normal_form). This works out to be $\forall x \forall y (F(x, x) \implies F(y, y))$
+
+{{<details "Details">}}
+$$
+\begin{align*}
+\exists x F(x, x) \implies \forall x F(x, x) &\iff \forall x (F(x, x) \implies \forall y F(y, y)) \\\\
+&\iff \forall x \forall y (F(x, x) \implies F(y, y))
+\end{align*}
+$$
+{{</details>}}
+
+Let $\mathcal{D} = \\{ c_1 \\}$. The truth functional expansion is $F(c_1, c_1) \implies F(c_1, c_1)$. This is a tuatlogy so the formula is valid for domain of size 1.
+
+Now let $\mathcal{D} = \\{ c_1, c_2 \\}$. After removing the tautologies, the truth functional expansion works out to be $(F(c_1, c_1) \implies F(c_2, c_2)) \wedge (F(c_2, c_2) \implies F(c_1, c_1))$
+
+{{<details "Details">}}
+$$
+\forall x \forall y (F(x, x) \implies F(y, y)) \\\\
+(\forall y (F(c_1, c_1) \implies F(y, y))) \wedge (\forall y (F(c_2, c_2) \implies F(y, y))) \\\\
+((F(c_1, c_1) \implies F(c_1, c_1)) \wedge (F(c_1, c_1) \implies F(c_2, c_2))) \wedge ((F(c_2, c_2) \implies F(c_1, c_1)) \wedge (F(c_2, c_2) \implies F(c_2, c_2))) \\\\
+(F(c_1, c_1) \implies F(c_2, c_2)) \wedge (F(c_2, c_2) \implies F(c_1, c_1))
+$$
+</details>
+{{</details>}}
+
+Consider the model $\\{ F(c_1, c_1), \neg F(c_1, c_2), \neg F(c_2, c_1), \neg F(c_2, c_2) \\}$. This causes the left hand side of the conjunction to fail. Hence the formula is invalid and we have found a counterexample.