Valuation-overring chains compute Krull valuative dimension

Fix $V$ as in the Statement. Every overring $W$ is a valuation ring because it contains $V$, so [F2] supplies its proper nonunit ideal ${\mathfrak{m}}_W$. A product is a unit exactly when both factors are units: if $ab$ has inverse $u$, then $bu$ inverts $a$ and $au$ inverts $b$. Thus ${\mathfrak{m}}_W$ is prime by [F3], and its contraction $\mathfrak{p}={\mathfrak{m}}_W\cap V$ is a proper ideal satisfying the same product criterion. Each $s\in V\setminus \mathfrak{p}$ is invertible in $W$, so $V_{\mathfrak{p}}\subseteq W$. If $w\in W\setminus V$, then $w\mathrm{\ne }0$ and $w^{-1}\in V$ by [F2]; since $w^{-1}$ is invertible in $W$, it lies outside $\mathfrak{p}$, whence $w=1\mathrm{/}(w^{-1})\in V_{\mathfrak{p}}$. This proves $W=V_{\mathfrak{p}}$. Conversely, for a prime $\mathfrak{p}$ of $V$, the fractions in [F3] form an overring and hence a valuation ring. An element $a\in V\setminus \mathfrak{p}$ is invertible in $V_{\mathfrak{p}}$. If a nonzero $a\in \mathfrak{p}$ had inverse $b\mathrm{/}s$ there, then $s=ab\in \mathfrak{p}$, contrary to $s\mathrm{\notin }\mathfrak{p}$; zero is not invertible. Therefore ${\mathfrak{m}}_{V_{\mathfrak{p}}}\cap V=\mathfrak{p}$. The two assignments are inverse. Prime inclusion reverses the denominator sets, and overring inclusion preserves units, so both assignments reverse inclusions; inverse bijections preserve strictness.

By [F1], $V$ has fraction field $K$, so the endpoints under step 1.1 are $V_{(0)}=K$ and $V_{{\mathfrak{m}}_V}=V$. Here $(0)$ is prime because $V$ is a subring of a field, and every prime of $V$ lies in ${\mathfrak{m}}_V$ because a proper ideal contains no unit. Thus any finite strict prime chain can be extended to these endpoints, discarding repetitions. Step 1.1 reverses such chains to strict overring chains from $V$ to $K$, with the same length. Conversely, every such overring chain reverses to a prime chain. By [F4] their length suprema agree, including infinity. Each chain counted by $c(D)$ lies among the overrings of its first ring $V_0$; appending $K$ if necessary shows its length is at most $\dim V_0\le \delta (D)$. Conversely, every prime-chain length in any valuation overring $V$ of $D$ is bounded above by the length of an overring chain from $V$ to $K$, hence by $c(D)$. Taking both suprema gives $\delta (D)=c(D)$ and also proves the assertion with the final ring required to be $K$. These families are sets of subrings of a fixed field and finite sequences of them; they contain a length-zero chain via $K$. Their suprema therefore exist in $\{0,1,\dots ,\mathrm{\infty }\}$. The constructions used a specified ring or chain at each point and no simultaneous selection of witnesses, so no Choice premise is used.