Let A be a noetherian ring, M a finitely generated A-module, and p and q prime ideals of Supp(M) such that p⊂q. Suppose that dimAq(Mq)=profAq(Mq). Then one has dimAp(Mp)=profAp(Mp) and
dimAq(Mq)=dimAp(Mp)+dim(Aq/pAq).
This follows directly from cor. 1 of prop. 13 of § 1, No. 7.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.