If G1,G2 are any closed subgroups of Rn, we have
(G1=G2)∗+G1∗∩G2∗and(G1∩G2)∗=G1∗+G2∗.
The real number (u∣x+y) is an integer for all x∈G1 and all y∈G2 if and only if (u∣x) is an integer for all x∈G1 and (u∣y) is an integer for all y∈G2, because (u∣x+y)=(u∣x)+(u∣y); hence
(G1+G2)∗=G1∗∩G2∗
for any two subgroups G1,G2 of Rn. If now we suppose that G1 and G2 are closed, we have (G1∗+G2∗)∗=G1∩G2 by Proposition 6; hence, taking the associated subgroups and applying Proposition 6 again,
(G1∩G2)∗=G1∗+G2∗.
Bourbaki puts the proof directly after the statement, so what is above is the statement and its proof together.