Outside the strong regime, the MAC intersection can be too small

Under independent uniform inputs, $I_2(X_1;Y_1\mathrm{\mid }{X}_2)=1$ while $I_2(X_1;Y_2\mathrm{\mid }{X}_2)=0$. Thus the first strong-interference inequality fails, so the hypothesis of [F2] is absent.

Each receiver observes its own input noiselessly. A binary code of rate below one can therefore be sent independently by each user. Since each output alphabet has size two, no rate can exceed one bit per use. Taking closure gives the true capacity square.

For every finite time-sharing variable $Q$ and every conditional product input law, each output is binary, so $I_2(X_1,X_2;Y_i\mathrm{\mid }Q)\le H_2(Y_i\mathrm{\mid }Q)\le 1$. Consequently every rate pair in the entire union and its closure asserted by the formula in [F2] satisfies $R_1+R_2\le 1$ bit per use. This excludes $(1,1)$, which step 1.2 places in the true capacity square. Hence the formula without the strong-interference hypothesis is false.