4-(suite): axiom A (or equivalently axiom TG) have powerfull consequences. (i) It is easy to see that A1 and A2 prove the power-set axiom, by separation, because P(x is included inside the set y; (ii)from A1 and iterations of A2, we also can derive the axiom of infinity. In fact all sets P(n,x), with P(0)=x, P(1,x)=P(x) and P(n+1, x)=P(P(n,x)) must be member elements of the set y and so are forming a set z by separation. But, by iteration of the Cantor's theorem, no injection from P(n+m,x) into P(n,x) is possible for m>0; so that all sets P(n,x) are distinct sets. We then have an injection from z into the class of finite ordinals, and by replacement this class must be a set, so that we prove the axiom of infinity. (iii) It is also possible, but this is much more difficult and uses essentially the regularity axiom, to prove thecaxiom of choice from axiom A. The three preceding proofs can be built so that we can prove that the theory "ZFC + Axiom A" (or equivalently the theory "ZFC + "There exists a proper class of inaccessible cardinals"") is equivalent to the theory having as axioms: Replacement, Extensionality, Pair, Union and Regularity [and the axiom asserting the existence of some set , if this is not considered as a logical truth]. So, it seems that the price to be paid for deleting power-set, infinity and choice is to add the axiom of the Pair, but Tarski asserts that the Pair axiom can be derived from axiom A and replacement. QUESTION 4 "Is it possible to dispense with the addition of the axiom of the Pair and to preserve the equivalence with "ZFC + axiom A ?"