Definition of Predicate Calculus.

This question is from Introduction to Mathematical Logic by Elliot Mendelson.

in page 69-70 , a first order theory $K$ is defined to have 5 logical axioms and a (possibly empty) set of proper axioms.A first. order theory in which there are no proper axioms is called a first-order predicate calculus.Now , if a theory $K$ has no proper axioms , then it will be a predicate calculus and it has 5 specific axioms.So there seems to be only 1 predicate calculus.

But then at exercise 2.63(c) of the book ,there is this exercise:

Prove that no predicate calculus is a scapegoat theory.

This seems to say that there can be multiple kinds of predicate calculus , but I cant really understand how.Am I missing something?

We have to start from the definition of first-order language [page 57]:

A [emphasis mine] first-order langugage $\mathscr L$ is [...]

(f) a non-empty set of predicate letters.

Thus, we have e.g. the first-order language of set theory, with the binary predicate symbol $\in$ [in Mendelson's formalism $A_1^2$] and a constant symbol $\emptyset$, as well as the first-order language of arithmetic, with the function symbol $s$ (the successor), the binary predicate symbol $\lt$ and a constant symbol $0$.

If there are no proper axioms, the corresponding theory will be called a [emphasis mine] first-order predicate calculus.

Thus

are there multiple predicate calculus?

Yes; there will be the predicate calculus for the first-order language of set theory, with non-logical symbols: $\in, \emptyset$, but without proper axioms involving them, as well as the predicate calculus for the first-order language of arithmetic, with non-logical symbols: $s, \lt, 0$, but without proper axioms involving them.

The non-logical symbols will occur in instances of logical axioms, like e.g.: $\forall x (x \in \emptyset) \to (\emptyset \in \emptyset)$.

In this predicate calculus we will prove all valid formulas using the proper symbols of the language, like the formula above as well as $\emptyset = \emptyset$, in the corresponding version of predicate calculus with equality [page 94].

Of course, we cannot prove with them "proper" theorems of the corresponding theories, like e.g. $\lnot \exists x (x \in \emptyset)$, because they are not valid.