**Proof.**
Let suppose that $X$ is catenary. According to Definition 5.11.4, for every pair of irreducible closed subsets $Y \subset Y'$ we have $\text{codim}(Y,Y') < \infty $. Let $Y \subset Y' \subset Y''$ be a triple of irreducible closed subsets of $X$. Let

\[ Y = Y_0 \subset Y_1 \subset ... \subset Y_{e_1} = Y' \]

be a maximal chain of irreducible closed subsets between $Y$ and $Y'$ where $e_1 = \text{codim}(Y,Y')$. Let also

\[ Y' = Y_{e_1} \subset Y_{e_1 + 1}\subset ... \subset Y_{e_1 + e_2} = Y'' \]

be a maximal chain of irreducible closed subsets between $Y'$ and $Y''$ where $e_2 = \text{codim}(Y',Y'')$. As the two chains are maximal, the concatenation

\[ Y = Y_0\subset Y_1 \subset ... \subset Y_{e_1} = Y' = Y_{e_1} \subset Y_{e_1+1}\subset ... \subset Y_{e_1+e_2}=Y'' \]

is maximal too (between $Y$ and $Y''$) and its length equals to $e_1 + e_2$. As $X$ is catenary, each maximal chain has the same length equals to the codimension. Thus the point (2) that $\text{codim}(Y,Y'') = e_1 + e_2 = \text{codim}(Y,Y') + \text{codim}(Y',Y'')$ is verified.

For the reciprocal, we show by induction that : if $Y = Y_1 \subset ... \subset Y_ n = Y'$, then $ \text{codim}(Y,Y') = \text{codim}(Y_1,Y_2) + ... + \text{codim}(Y_{n-1},Y_ n)$. Therefore, it forces maximal chains to have the same length.
$\square$

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