Consistent Sets of Formulas (FOL)

Question

Hello I am hoping to find some direction on solving this try it yourself problem in my textbook.

Let S be an arbitrary set of symbols and let $\Phi = \{v_0 \equiv t | t \in T^S\} \cup \{\exists v_0 \exists v_1 \neg v_0 \equiv v_1\}$.

Note: $T^S$ is the set of all S-terms.

Show that $Con\Phi$ holds and that there is no consistent set in $L^S$ which includes $\Phi$ and contains witnesses.

I think my confusion begins with not understanding exactly what $\Phi$ implies with its 2 formulas. I think it states all terms are equivalent and every term has a negation? If my understanding is correct then to me $\Phi$ seems inconsistent since every S-term is equivalent, and in second formula $\neg v_0 \equiv v_1$ would only be true if the negation of every S-term is itself?

Thank you in advance for helping me wrangle this problem.

Answer 1

So, what's going on here is that all of all $S$-terms are being identified as being the same element, but the second set assets that there are more than one element. I think this might be a point on which you're confused, but the second set just asserts that there exist two distinct elements.

To prove that this is consistent, you should apply the Compactness Theorem. The proof should be quite short, like five sentences.

To prove the second part, assume there is a witness and work towards a contradiction.

In general, the use of this kind of situation is its application to nonstandard models. A nonstandard model of a theory, say of arithmetic, is a model that is probably existent, but doesn't quite follow the usual way we think about the proposition. For example, there are models of arithmetic on the natural numbers that have uncountable many elements.

Technically this is a nonstandard model, but it's a pretty boring one. However, you can "unpack" it by undoing the identification done in the first set of you're careful, and keep the weird element that the second set says exists there still. Then you have something that looks like $T^S$ but also has a bunch of nonstandard elements.

Answer 2

The formula $∃v_0∃v_1 (v_0 \ne v_1)$ is satisfied in any structure with at least two elements.

Irrespective of the number of terms $t_i$ of the language, the set of formulae $\{ v_0 = t_i \mid t_i \in T \}$ can be satisfied by a interpretation with domain $D$ only if all $t_i$ will denote to the same object $c \in D$.

Thus, if we add the requirement that the formulae must have witnesses, we need two terms $t_a, t_b$ such that $t_a \ne t_b$ is satisfied.

This implies that $t_a$ and $t_b$ will denote two different objcets $a$ and $b$ of $D$.

But this conflict with:

$v_0 = t_a$ and $v_0 = t_b$.