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General recursive function
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==Examples== Examples not involving the minimization operator can be found at [[Primitive recursive function#Examples]]. The following examples are intended just to demonstrate the use of the minimization operator; they could also be defined without it, albeit in a more complicated way, since they are all primitive recursive. {{unordered list |The [[integer square root]] of {{mvar|x}} can be defined as the least {{mvar|z}} such that <math>(z+1)^2 > x</math>. Using the minimization operator, a general recursive definition is <math>\operatorname{Isqrt} = \mu(\operatorname{Not} \circ \operatorname{Gt} \circ (\operatorname{Mul} \circ (S \circ P_1^2,S \circ P_1^2), P_2^2))</math>, where {{math|Not}}, {{math|Gt}}, and {{math|Mul}} are [[logical negation]], greater-than, and multiplication,<ref>defined in [[Primitive recursive function#Junctors]], [[Primitive recursive function#Equality predicate]], and [[Primitive recursive function#Multiplication]]</ref> respectively. In fact, <math>(\operatorname{Not} \circ \operatorname{Gt} \circ (\operatorname{Mul} \circ (S \circ P_1^2,S \circ P_1^2), P_2^2)) \; (z,x) = (\lnot S(z)*S(z) > x)</math> is {{val|0}} if, and only if, <math>S(z)*S(z) > x</math> holds. Hence <math>\operatorname{Isqrt}(x)</math> is the least {{mvar|z}} such that <math>S(z)*S(z) > x</math> holds. The negation [[junctor]] {{math|Not}} is needed since {{math|Gt}} encodes truth by {{val|1}}, while {{mvar|μ}} seeks for {{val|0}}. }} The following examples define general recursive functions that are not primitive recursive; hence they cannot avoid using the minimization operator. {{unordered list |{{example needed|date=November 2021}} }}
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