# Inequality for XOR

I was reading a book of neural networks in Spanish and I do not understand the inequalities on page 26

I have a table of XOR and inequality restrictions

then:

if there are no contradictions in the previous inequalities, the problem is linearly separable, as observed from the inequalities 2, 3, 4 is impossible

and that its sum is less than zero

my question is how that result of contradiction(inequalities 2,3,4)

Let rewrite the inequalities: \begin{align*} 0 * W_{1,1} + 0 * W_{2,1} + b &< 0 \tag{p_1} \\ 0 * W_{1,1} + 1 * W_{2,1} + b &\geq 0 \tag{p_2} \\ 1 * W_{1,1} + 0 * W_{2,1} + b &\geq 0 \tag{p_3}\\ 1 * W_{1,1} + 1 * W_{2,1} + b &< 0 \tag{p_4} \\ \end{align*}
From ($p_1$), we have $b<0$. From ($p_2$) and ($p_3$), we have $W_{1,1}+b\geq 0$ and $W_{2,1}+b\geq0$ so that $W_{1,1}+W_{2,1}+2b\geq 0$. However, ($p_4$) implies a contradiction $W_{1,1}+W_{2,1}+b<0$, because (as $b<0$) $$W_{1,1}+W_{2,1}+b \geq W_{1,1}+W_{2,1}+2b \geq 0$$
• @x-rw At first ($p_1$) says $b$ is negative. So $b\geq 2b$. OK? Now add ($p_2$) and ($p_3$), then you have $W_{1,1}+W_{2,1}+2b\geq0$. However $b\geq 2b$ so that $W_{1,1}+W_{2,1}+b\geq 0$ too. But it contradict to ($p_4$). – ChoF Apr 15 '18 at 22:22
• @x-rw From ($p_1$) we know $b$ is negative. So $b\geq 2b$. For example, $-1>-2$. – ChoF Apr 16 '18 at 1:10