I'm reading "Introduction to Logic: And to the Methodology of Deductive Sciences by Tarski", and in this book he first created an axiom system called A, which contains some superfluous axioms (axioms that can be derived from other axioms of the same axiom system), then he got rid of these superfluous axioms and called the new axiom system A¹, then he got rid of all superfluous primitive terms (expressions that seems immediately understandable, like "a set" or the relation "<") and called this new axiom system A².
Then he says: "One can find many axiom systems which are equipollent" specifying: "in this chapter equipollence will ordinarily refer to means of proof" and he gives an example of a new axiom system A³, in which all the axioms are contained in the system A (which means some axioms in system A and A³ are the same) or have been proved on its basis (which means, some theorems that were proved from the axioms of A are now axioms), and since system A and A² are equipollent all the sentences of A³ can also be proved on the bases of A².
Now finally my problem, he says: "Just as in the case of axioms, one can find systems of primitive terms which are equipollent to a given system. This applies, in particular, to the system of the three terms ℝ, <, and + (A³ is a system with only these primitive terms). If, for instance, we replace < by the symbol ≤ we obtain an equipollent system"
One of the axioms of the system A³ states "If x < y, then y≮x" but if I replace < by the symbol ≤ I would get "If x≤y, then y≰x" which is false (when x=y). So how can A³ and this new axiom system be equipollent? What am I missing? What is exactly the definition of equipollent axiom system since he doesn't give any? maybe "two axiom systems A and B are equipollent if all the axioms of A are contained in the system B or have been proved on its basis (or vice versa)"?
Thank you!
The new axioms don't need to be identical. If you introduce another symbol (eg $\leqslant$ instead of $<$), it has another meaning and you have to restate axioms accordingly. eg the antisymmetry would be stated with this axiom : $$ \forall x \ \forall y \ (x \leqslant y \wedge y \leqslant x) \Rightarrow x = y$$