S. =. Proof, N. S. , C. S. , P. S. , H. et al., Let a in H S . Let V be the valuation on S Let V ? be the valuation on S defined by a |= ?(p) iff (S, V ? ), a |= ?(p) Since (S, V ), a |= ?(p), (S, V ? ), a |= ?(p) Therefore, by ( * ) Suppose [C] is definable with [N ] and [P ] in the class of all normal frames. Hence, there exists a formula ?(p) in [N ] and [P ] such that ( * ) for all normal frames Let a in H S . Let V be the valuation on S defined by: a contradiction. (3) Suppose [P ] is definable with, Moreover, for all formulas ?(p) in the class of all normal frames. Hence, there exists a formula ?(p) in [N ] and [C] such that ( * ) for all standard frames Let a in H S . By Sym(C), Sym(P ) and Inf 1 (C, P ), there exists b in H S such that C S (a, b) and not P S (a, b). Let V be the valuation on S defined by V (p) = [a] C S \{b}. Obviously, (S, V ), a |= [P ]p. Thus, by ( * ), (S, V ), a |= ?(p). By Sym(P ) and Inf 1 (P, ?), there exists c in H S such that P SS, V ? ), a |= [P ]p. Moreover, for all formulas ?(p) in [N ] and [C], (S, V ), a |= ?(p) iff (S, V ? ), a |= ?(p)

V. (. Since and |. , Therefore Variants Other primitives may be defined as well. In this section, we consider the predicate symbol < of precedence between positive hyperreals and the function symbol + of addition between positive hyperreals

. Proof, <. Suppose, N. Is-definable-with, C. , P. et al., Obviously, the second player wins all Ehrenfeucht games over (S P H , [a, b]) and (S P H , [b, a]) with respect to N , C, P and ?. Thus, by [12, theorem 2.2.8], for all formulas ?(x, y) b: a contradiction. ? What about the axiomatization/completeness or the decidability/complexity of the first-order theory based on the predicates N , C, P and <? As for the modal logic option, the obvious road consists in adding a connective [<] interpreted in S P H in the following way Adding addition Let us add a function symbol + of arity 2 to our first-order language, What about the axiomatization/completeness or the decidability/complexity of the modal logic based on the connectives The formulas are now given by the rule: ? ? ::= N (s, t) | C(s, t) | P (s, t) | s ? t | ? | ¬? | (? ? ?) | ?x.? where s and t range over the set of terms defined by the rule ? s ::= x |

+. Let and S. Interpreted-in, it appears that if we restrict the language to the predicate N or if we restrict the language to the predicates N and C then the function symbol + can be eliminated To see this, it suffices to observe that the following sentences hold in S P H : ? ?x, ? ?x.?y.?z.(C(x, y + z) ? (C(x, y) ? ¯ N (x, z)) ? (C(x, z) ? ¯ N (x, y)))

H. S. Proof, S. P. , C. S. , ?. , ?. P. et al., Since Obviously, the second player wins all Ehrenfeucht games over(x, y, z) [a ?1 , b ?1 , c ?1 ] Since ( * ), C S P H (a ?1 + S P H b ?1 , c ?1 ): a contradiction Suppose there exists a formula ?(x, y, z) in N , C, P and ? such that ( * ) for all a, b, c in H Since ( * ), S |= ?(x, y, z) [a, b, c]. Obviously, the second player wins all Ehrenfeucht games over ): a contradiction. (3) Suppose there exists a formula ?(x, y, z) in N , C, P and ? such that ( * ) for all a, b, Thus, by [12, theorem 2.2.8], for all formulas ? H S P H be such that a + S P H b = c and a 2 + S P H b 2 = c 2 . Since ( * ), S |= ?(x Obviously, the second player wins all Ehrenfeucht games over (S P H Since ( * ), a 2 + S P H b 2 = c 2 : a contradiction. ? What about the axiomatization/completeness or the decidability/complexity of the first-order theory based on the predicates N , C and P and the function +? As for the modal logic option, the obvious road consists in adding a connective ? interpreted in S P H in the following way: ? (S P H , V ), a |= ? ? ? iff there exist b

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