By Abraham Robinson

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**Sample text**

1 . 2 3 . establish the connection between Q and S and P. Interpreting E ( x , y) here, as everywhere in the sequel, as the relation of equality, while S(x, y, z ) stands for ' z is the product of x and y', we see that axioms 4 . 1 . 1 . - 4 . 1 . 4 . , 4 . 1 . 7 . , 4 . 1 . 8 . , 4 . 1 . , 4 . 1 . 1 2 . , 4 . 1 . 1 6 . , together constitute the axiomatic system of a group. We shall denote this set by A,. Adding 4 . 1 . 1 4 . to A,, we obtain the axiomatic system A, of a commutative group. Again axioms 4 .

X\ or Y. ), the validity of 3 . 7 . 3 . can again be deduced from the validity of 3 . 7 . 2 . Now X, may be identical with one of the remaining XL, in that case Y can be deduced from the set { X i , X i , . . X;-l, Xi+l . . Xi} above, and the degree of that set is certainly smaller than d. If this is not the case, then we distinguish the following possibilities. Either p ( k ) = 1, and in that case X: belongs to Kn-l and therefore to some K:-'), so that the degree of { X i , X i , . , X:, . .

T(A,,)is the same for the two valuations - of S, and of Qi - for all A,, E SJ. This condition is satisfied by the valuation of S,,as defined above. We may then show similarly as in 3 . 4 . 1 . above, that either the definition T ( A J = 1, together with the given valuation of S,, agrees with at least one admissible valuation of every Qi, u < y, or the same applies to T(A,) = 0, (or both). I n t h e first case we define T ( A J = 1, in the alternative case T ( A J = 0. However, in order to prove that this procedure of induction can be carried on indefinitely, we still have to show that for every p, the valuation of Spfound in this way does in fact agree with at least one admissible valuation of every Ql as explained above.