Predicate calculus is not scapegoat theory ?!

Question

A theory $T$ is a scapegoat theory if for every formula $A$ with only one free variable there exist a closed term s such that $T$ proves: $(\exists x(\neg A(x)))\Rightarrow \neg A(s)$

Why is any predicate calculus not a scapegoat theory?

Please prove it.

I think this property is expected for any theory, since it says that we can subsitute a term to a free variable. So why can’t we see this property in any predicate calculus? By completeness theorem we know theorems are exactly logically valid.

Do you think "not scapegoateness property" is logically valid?

Answer 1

Here are three examples. I wrote this out at length because I want to be able to link to it for future questions about the "scapegoat property".

Not every theory is a scapegoat theory

First, here is an example showing that general theories are not scapegoat theories. Consider a theory in the language of pure equality. The only relation symbol is "$=$", and there are no constant symbols or function symbols. This theory proves $(\exists x)[x = x]$. But there are no closed terms at all, so the theory cannot be a scapegoat theory.

Every theory can be extended to a scapegoat theory in a larger language

Now here is an example of how to make any theory into a scapegoat theory. Given a theory $T$, for each formula $A(x)$ with one free variable we add a new constant symbol $c_A$ and a new axiom $$ (\exists x) A(x) \to A(c_A) $$ This may cause new formulas with one free variable to appear, so we repeat the process of adding new constants until we obtain a theory in which every formula $A(x)$ with one free variable has an associated constant $c_A$. This new theory will have the scapegoat property, and will be conservative over the original theory: if a formula in the language of the original theory is provable in the extended theory, that formula is also provable in the original theory.

This is a special case of a more general process called Skolemization.

Peano arithmetic does not have the scapegoat property

This example will be a more nontrivial theory without the scapegoat property. For this example, some familiarity with the incompleteness theorems is needed.

Let $T$ be first-order Peano arithmetic, PA. Unlike in the first example, PA has many closed terms, such as $1+1+1$. In particular it has a term for every natural number.

Let $A(x)$ be the statement "$x$ codes a proof of $0=1$ from the axioms of first-order Peano arithmetic". The key points of the incompleteness theorem are:

• $A(x)$ is expressible as a formula of PA with only bounded quantifiers
• $(\exists x)A(x)$ is consistent with PA, but not provable or disprovable

Assume for a contradiction that PA has the scapegoat property. Pick a nonstandard model of PA in which $(\exists x)A(x)$ holds. According to the scapegoat property, we can choose a term $s$ such that PA proves $(\exists x)A(x) \to A(s)$. Thus $A(s)$ holds in our nonstandard model.

Because $A(x)$ has only bounded quantifiers, if $A(s)$ holds in any model of PA then it holds in every model of PA. But this means $A(s)$ must be provable from PA, so $(\exists x)A(x)$ is also provable. This contradicts the second bullet above. Thus PA does not have the scapegoat property for the formula $A(x)$.

Answer 2

In a previous post I've pointed at the following exercise in Elliott Mendelson, Introduction to mathematical logic (4ed - 1997), page 94 : Ex.2.63(c) : "prove that no predicate calculus is a scapegoat theory".

For the definition of predicate calculus see page 70 : for a language which contains denumerable many individual variables, individual constants (possibly none), function letters (possibly none) and predicate letters :

a theory without non-logical (or proper) axioms.

Thus, a predicate calculus is a "degenerate" firts-order theory in any signature which has no proper axioms.

Following Mendelson's definition of scapegoat theory [page 84], for a predicate calculus we have that :

for any formula $\mathcal B$ with only one free variable $x$, there is a closed term $t$ such that :

(A) --- $\vdash \exists x \lnot \mathcal B(x) \rightarrow \mathcal \lnot B(t)$.

The formula above is equivalent to :

(B) --- $\vdash \mathcal B(t)\rightarrow \forall x \mathcal B(x)$.

But this formula is not valid.

Consider the language with constant $c$ and unary predicate letter $P$.

Consider the following instance of (B) :

$P(c) \rightarrow \forall x P(x)$;

we can find interpretations which falsify it : choose a domain with more than one object and assign to $P$ a non-empty proper subset of the domain.

Thus, if (B) is not valid, neither (A) is. By soundness, (A) is not provable.