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The clique decision problem is a decidable language: on a valid finite graph and an integer $k$, reject if $k$ exceeds the number of vertices, and otherwise examine all $k$-element subsets of vertices, accepting if one is a clique. Each candidate can be checked by examining all its pairs of distinct vertices. This finite search gives at least one Turing machine deciding the problem, without any claim of polynomial time.

By well-ordering of the natural numbers, there is a least state count $n_0$ among Turing machines deciding the clique decision problem. For every $n\ge n_0$, add $n-n_0$ unreachable states to a machine with $n_0$ states. Its behaviour on every input is unchanged, and it has exactly $n$ states. For $n<n_0$ there is no such machine, by minimality. ThusA Turing machine with the fixed constant $n_0$ in its description decides this threshold language by validating the input and comparing its value with $n_0$. The argument proves the existence of a decider; it does not compute the smallest possible clique decision problem decider. There is no need to search through machine descriptions and attempt to certify their behaviour on every input.