Let E be a left vector space over K. A mapping p of E in R+={0,+∞}, is called a semi-norm on E if it satisfies the following axioms : (SN_I) If x∈E and λ∈K then p(λx)=∣λ∣p(x). (SN_II) If x∈E and y∈E then p(x+y)≤p(x)+p(y).
Since p(x)≤p(y)+p(x−y) and p(y)≤p(x)+p(y−x), from p(y−x)=p(x−y), we deduce
∣p(x)−p(y)∣≤p(x−y).
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.