We define $f \left({x}\right)$ as a '''prime-representing function''' {{iff}}: :$\forall x \in \N: f \left({x}\right) \in \Bbb P$ where: :$\N$ denotes the set of all natural numbers :$\Bbb P$ denotes the set of all prime numbers. Let $p_n$ be the $n$th prime number. From Difference between Consecutive Primes: :$p_{n+1} - p_n < K p_n^{5/8}$ where $K$ is an unknown but fixed positive integer. === Lemma 1 === :$\forall N > K^8 \in \Z: \exists p \in \Bbb P: N^3 < p < \left({N + 1}\right)^3 - 1$ ==== Proof ==== Let $p_n$ be the greatest prime less than $N^3$. {{begin-eqn}} {{eqn | l = N^3       | o = <       | r = p_{n+1}       | c = because $p_n$ is the '''greatest''' prime less than $N^3$ }} {{eqn | o = <       | r = p_n + K {p_n}^{\frac 5 8}       | c = Difference between Consecutive Primes }} {{eqn | o = <       | r = N^3 + K N^{\frac {15} 8}       | c = because $p_n < N^3$ }} {{eqn | o = <       | r = N^3 + N^2       | c = because $N > K^8$ }} {{eqn | o = <       | r = N^3 + 3N^2 + 3N }} {{eqn | r = \left({N + 1}\right)^3 - 1 }} {{end-eqn}} Therefore: :$N^3 < p_{n+1} < \left({N + 1}\right)^3 - 1$ {{qed|lemma}} Let $P_0 > K^8$ be a prime number. By Lemma 1, there exists an infinite sequence of primes: :$P_0, P_1, P_2, \ldots$ such that: :$\forall n \in \N_{>0}: {P_n}^3 < P_{n+1} < \left({P_n + 1}\right)^3 - 1$ Then we define two functions $u, v: \N \to \Bbb P$: :$\forall n \in \N: u \left({n}\right) = {P_n}^{3^{-n} }$ :$\forall n \in \N: v \left({n}\right) = \left({P_n + 1}\right)^{3^{-n} }$ It is trivial that $v \left({n}\right) > u \left({n}\right)$. === Lemma 2 === :$\forall n \in \N_{>0}: u \left({n + 1}\right) > u \left({n}\right)$ ==== Proof ==== {{begin-eqn}} {{eqn | l = u \left({n + 1}\right)       | r = {P_{n+1} }^{3^{-\left({n+1}\right)} } }} {{eqn | o = >       | r = \left({P_n^3}\right)^{3^{-n-1} }       | c = because $P_{n+1} > {P_n}^3$ }} {{eqn | r = {P_n}^{3 \times 3^{-n-1} } }} {{eqn | r = {P_n}^{3^{-n} } }} {{eqn | r = u \left({n}\right) }} {{end-eqn}} {{qed|lemma}} === Lemma 3 === :$\forall n \in \N_{>0}: v \left({n + 1}\right) < v \left({n}\right)$ ==== Proof ==== {{begin-eqn}} {{eqn | l = v \left({n + 1}\right)       | r = \left({P_{n+1} + 1}\right)^{3^{-\left({n+1}\right)} } }} {{eqn | o = <       | r = \left({\left({\left({P_n + 1}\right)^3 - 1}\right) + 1}\right)^{3^{-n-1} }       | c = because $P_{n+1} < \left({P_n + 1}\right)^3 - 1$ }} {{eqn | r = \left({\left({P_n + 1}\right)^3}\right)^{3^{-n-1} } }} {{eqn | r = \left({P_n + 1}\right)^ {3^{-n} } }} {{eqn | r = v \left({n}\right) }} {{end-eqn}} {{qed|lemma}} It follows trivially that $u \left({n}\right)$ is bounded and strictly monotone. Therefore, there exists a number $A$ which is defined as: :$A := \lim_{n \mathop \to \infty} u \left({n}\right)$ From Lemma 2 and Lemma 3, we have: :$u \left({n}\right) < A < v \left({n}\right)$ {{begin-eqn}} {{eqn | l = u \left({n}\right)       | o = <       | r = A       | rr= <       | rrr=\left({n}\right) }} {{eqn | ll= \leadsto       | l = {P_n}^{3^{-n} }       | o = <       | r = A       | rr= <       | rrr=\left({P_n + 1}\right)^{3^{-n} } }} {{eqn | ll= \leadsto       | l = P_n       | o = <       | r = A^{3^n}       | rr= <       | rrr=P_n + 1 }} {{end-eqn}} The result follows. {{qed}} {{Namedfor|William H. Mills|cat = Mills}}