  Thus if \(f(i,x_{1},\ldots ,x_{k}) = 1\), and for all \(j \lt i, f(j,x_{1},\ldots ,x_{k})\) is defined, but not equal to 1, then \(\mu [f](x_{1}, \ldots ,x_{k}) = i\). Otherwise \(\mu[f](x_{1}, \ldots ,x_{k})\) is undefined.