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| title | math |
|---|---|
| Finding Counter-Models through Truth Functional Expansions | 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:
$$
\forall x P(x) \iff P(c_1) \wedge \dots \wedge P(c_n)
$$
$$
\exists x P(x) \iff P(c_1) \vee \dots \vee P(c_n)
To find a counter-example, you'll need to find a model in which the formula does not hold. A model consists of a universe of discourse and whether or not each possible predicate holds. For example, let's take a binary predicate F. For a domain of discourse of size two, the following predicates may hold: F(c_1, c_1), F(c_1, c_2), F(c_2, c_1), F(c_2, c_2). For any given model, each of those instantiations can hold or not, amounting to 2^4 = 16 possible models for a domain of discourse of size 2 with one binary predicate.
Now let's find a counter-example for an invalid formula. Consider \exists x F(x, x) \implies \forall x F(x, x).
To perform a truth functional expansion, we need to first convert to prenex normal form. This works out to be \forall x \forall y (F(x, x) \implies F(y, y))
{{<details "Details">}} $$ \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*} $$ {{}}
Let \mathcal{D} = \\{ c_1 \\}. The truth functional expansion is F(c_1, c_1) \implies F(c_1, c_1). This is a tuatlogy so the formula is valid for domain of size 1.
Now let \mathcal{D} = \\{ c_1, c_2 \\}. After removing the tautologies, the truth functional expansion works out to be (F(c_1, c_1) \implies F(c_2, c_2)) \wedge (F(c_2, c_2) \implies F(c_1, c_1))
{{<details "Details">}}
$$
\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))
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.