Let A be a filtered ring and (Jn)n∈Z its filtration. Assume that one has J0=A and p.1A∈J1. Let m and n be integers such that m⩾1 and n⩾0, and let a0,...,an,b0,...,bn be elements of A.
a) If one has ai≡bimodJm for 0⩽i⩽n, then one has
Φi(a0,...,ai)≡Φi(b0,...,bi)modJm+ifor 0⩽i⩽n.
b) Suppose that, for every integer k⩾1, and every x∈A, the relation p.x∈Jk+1 entails x∈Jk. If one has Φi(a0,...,ai)≡Φi(b0,...,bi)modJm+i for 0⩽i⩽n, then one has ai≡bimodJm for 0⩽i⩽n.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.
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