Author 
Comment/Response 
jmd

07/21/12 1:53pm
I have a system of linear inequalities in several variables, and know that some of them are consequences of the others. Is the following command the correct one to use?
Reduce[
And[
0 <= x[1, 1],
0 <= x[1, 2],
0 <= x[1, 3],
0 <= x[1, 4],
0 <= x[1, 4]  x[2, 1],
0 <= x[1, 4]  x[2, 2],
0 <= x[1, 4]  x[2, 3],
0 <= x[2, 1] + x[2, 2]  x[2, 4],
0 <= x[2, 1] + x[2, 3]  x[2, 4],
0 <= x[2, 2] + x[2, 3]  x[2, 4],
0 <= x[2, 1] + x[2, 2] + x[2, 3]  x[2, 4],
0 <= x[2, 2]  x[3, 1],
0 <= x[2, 3]  x[3, 1],
0 <= x[2, 2] + x[2, 3]  x[3, 1],
0 <= x[2, 1]  x[3, 2],
0 <= x[2, 3]  x[3, 2],
0 <= x[2, 1] + x[2, 3]  x[3, 2],
0 <= x[2, 4]  x[3, 1]  x[3, 2],
0 <= x[2, 3] + x[2, 4]  x[3, 1]  x[3, 2],
0 <= x[2, 1]  x[3, 3],
0 <= x[2, 2]  x[3, 3],
0 <= x[2, 1] + x[2, 2]  x[3, 3],
0 <= x[2, 4]  x[3, 1]  x[3, 3],
0 <= x[2, 2] + x[2, 4]  x[3, 1]  x[3, 3],
0 <= x[2, 4]  x[3, 2]  x[3, 3],
0 <= x[2, 1] + x[2, 4]  x[3, 2]  x[3, 3],
0 <= 2 x[2, 4]  x[3, 1]  x[3, 2]  x[3, 3],
0 <= x[3, 1] + x[3, 2] + x[3, 3]  2 x[3, 4],
0 <= x[2, 4]  x[3, 4],
0 <= x[3, 1]  x[3, 4],
0 <= x[2, 1] + x[3, 1]  x[3, 4],
0 <= x[3, 2]  x[3, 4],
0 <= x[2, 2] + x[3, 2]  x[3, 4],
0 <= x[3, 3]  x[3, 4],
0 <= x[2, 3] + x[3, 3]  x[3, 4],
0 <= x[2, 1]  x[4, 1],
0 <= x[2, 4]  x[3, 1]  x[4, 1],
0 <= x[3, 2] + x[3, 3]  x[3, 4]  x[4, 1],
0 <= x[4, 1],
0 <= x[2, 2]  x[4, 2],
0 <= x[2, 4]  x[3, 2]  x[4, 2],
0 <= x[3, 1] + x[3, 3]  x[3, 4]  x[4, 2],
0 <= x[3, 3]  x[4, 1]  x[4, 2],
0 <= x[4, 2],
0 <= x[2, 3]  x[4, 3],
0 <= x[2, 4]  x[3, 3]  x[4, 3],
0 <= x[3, 1] + x[3, 2]  x[3, 4]  x[4, 3],
0 <= x[3, 2]  x[4, 1]  x[4, 3],
0 <= x[3, 1]  x[4, 2]  x[4, 3],
0 <= x[3, 4]  x[4, 1]  x[4, 2]  x[4, 3],
0 <= x[4, 3],
0 <= x[4, 4]
],
{x[1, 1], x[1, 2], x[1, 3], x[1, 4], x[2, 1], x[2, 2], x[2, 3],
x[2, 4], x[3, 1], x[3, 2], x[3, 3], x[3, 4], x[4, 1], x[4, 2],
x[4, 3], x[4, 4]},
Integers
]
Out of this computation I want to get a shorter list of linear inequalities.
Thanks in advance to the replier.
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