# How to construct and evaluate a boolean expression using symbols?

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Jay Vaidya on 15 Nov 2020
Answered: Walter Roberson on 15 Nov 2020
If I have a boolean expression that is
y = (x1barx2)(x3barx4bar)(x3barx5)
I am currently using a simple 1 and 0 matrix to evaluate this sometimes manually and sometimes in a for a loop. However, it is difficult to do the same for larger expressions. Is there a way that I can evaluate these expressions using symbols like the ones we have poly2sym()? In other words, using and or logical operators and the values of the variables, can I find the answer of any boolean expression without using a for a loop as my expressions should less time to evaluate compared to the other part of the code.

Walter Roberson on 15 Nov 2020
[X1, X2, X3, X4, X5] = ndgrid([false,true]);
X1bar = ~X1;
X3bar = ~X3;
X4bar = ~X4;
X1barvX2 = X1bar | X2;
X3barvX4bar = X3bar | X4bar;
X3barvX5 = X3bar | X5;
y = (X1barvX2) & (X3barvX4bar) & (X3barvX5);
y_table = [X1(y), X2(y), X3(y), X4(y), X5(y)];
disp(y_table)
0 0 0 0 0 0 1 0 0 0 1 1 0 0 0 0 0 0 1 0 0 1 0 1 0 1 1 0 1 0 0 0 0 0 1 0 1 0 0 1 1 1 0 0 1 0 0 1 0 1 0 1 1 0 1 1 1 1 0 1 0 0 0 1 1 0 1 0 1 1 1 1 0 1 1

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