If υ(i,x) were total, then d(j) would be defined, leading to a contradiction via diagonalization

Having just seen that there is a universal partial computable function \(\upsilon(i,x)\), a natural question is whether this function is also computable (i.e., total). A negative answer is provided immediately by observing that by using \(\upsilon(i,x)\) we may define another modified diagonal function \(d(x) = \upsilon(x,x) + 1\) which is partial computable (since \(\upsilon(i,x)\) is). This in turn implies that \(d(x) \simeq \phi_j(x)\) for some \(j\). But now note that if \(\upsilon(i,x)\) we