By A. Bengtsson

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12) H(M ; x) ≻ qy H(M ; x) ≻ ⊲ if M ; x ∈ H . 14) But then H is precisely a universal machine programmed so that it halts in the accepting configuration qy whenever the machine M halts on input x. Theorem H is not recursive. Proof Suppose contrary to the proposition that there exist a Turing machine H that decides H. 11), that we have H(M ; x) ≻ qy H(M ; x) ≻ qn if M ; x ∈ H . 15) Now since both the input data and the program are encoded using the same alphabet, it is possible to write programs for the universal machine that takes programs for other Turing machines as input.

But in this case, the anti-diagonal function is intuitively computable. This is often phrased as saying that we can diagonalize out of any computational model for total functions. This intuitive computation of the anti-diagonal function relies on examining the list of functions and computing its values based on this list, so it could also be seen as a meta-computation. The question of whether it is possible to diagonalize out of the model or not when partial functions are allowed, depends on whether there is any general procedure to determine if a function is defined for a certain number or not.

1 The circuit model and non-computable functions We will now prove that there are circuits to compute any function f : {0, 1}k → {0, 1}l . The proof uses induction over the number of input bits. Theorem Every function f : {0, 1}k → {0, 1}l has a circuit that computes it. Proof First note that it suffices to prove the assertion for functions f : {0, 1}k → {0, 1} as the l-bit output case is easily put together from l 1-bit output functions. For k = 0 there is nothing to prove. For k = 1 there are four possible functions: 1.