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\mu(f)(x_1, \ldots, x_k) = z \stackrel{\mathrm{def}}{\iff}\ f(i, x_1, \ldots, x_k)&>0 \quad \text{for}\quad i=0, \ldots, z-1 \quad\text{and}\\
f(z, x_1, \ldots, x_k)&=0\quad
\end{align}</math><br />Intuitively, minimisation seeks—beginning the search from 0 and proceeding upwards—the smallest argument that causes the function to return zero; if there is no such argument, or if one encounters an argument for which {{mvar|f}} is not defined, then the search never terminates, and <math> \mu(f)</math> is not defined for the argument <math>(x_1, \ldots, x_k).</math> <br> (''Note'': While some textbooks use the μ-operator as defined here,<ref name="Enderton.1972">Enderton, H. B., A Mathematical Introduction to Logic, Academic Press, 1972</ref><ref name="Boolos.Burgess.Jeffrey.2007">Boolos, G. S., Burgess, J. P., Jeffrey, R. C., Computability and Logic, Cambridge University Press, 2007</ref> others like<ref name="Jones.1997">Jones, N. D., Computability and Complexity: From a Programming Perspective, The MIT Press, Cambridge, Massachusetts, London, England, 1997</ref><ref>Kfoury, A. J., R. N. Moll, and M. A. Arbib, A Programming Approach to Computability, 2nd ed., Springer-Verlag, Berlin, Heidelberg, New York, 1982</ref> demand that the μ-operator is applied to ''total'' functions <math>f</math> only. Although this restricts the μ-operator as compared to the definition given here, the class of μ-recursive functions remains the same, which follows from Kleene's Normal Form Theorem (see [[#Normal form theorem|below]]).<ref name="Enderton.1972"/><ref name="Boolos.Burgess.Jeffrey.2007"/> The only difference is, that it becomes undecidable whether
The '''strong equality''' operator <math>\simeq</math> can be used to compare partial μ-recursive functions. This is defined for all partial functions ''f'' and ''g'' so that
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