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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