# Is there any way to simplify the following boolean expression?

I was trying to manipulate with litarals and minterms of this booleans expression but it really did not lead to anything that could simplify the expression further.. Not sure if I am doing it wrong or it indeed cannot be simplified any further.

$$x_1'x_2'x_3 + x_1'x_2x_3'+x_1x_2'x_3' + x_1x_2x_3$$

well, I can do $$x_1(x_2'x_3' + x_2x_3) + x_1'(x_2'x_3 + x_2x_3')$$ I was thinking if expressions in parenthesis can be simplified and I was looking for appropriate identities.. but it seems they cannot be simplified, there is no tautology. Thus, I stuck here and I believe this expression cannot be simplified really. Right?

Thanks

This is the parity function for three inputs (it's true iff the number of true inputs is odd). In sum of product form, it indeed can't be simplified further. We can alternatively express it more concisely via the exclusive-or operation: $$x_1'x_2'x_3 + x_1'x_2x_3'+x_1x_2'x_3' + x_1x_2x_3 = x_1 \oplus x_2 \oplus x_3$$