XOR of three values?

Question

What is the simplest way to do a three-way exclusive OR? In other words, I have three values, and I want a statement that evaluates to true IFF only *one* of the three values is true. So far, this is what I've come up with: ((a ^ b) && (a ^ c) && !(b && c)) || ((b ^ a) && (b ^ c) && !(a && c)) || ((c ^ a) && (c ^ b) && !(a && b)) Is there something simpler to do the same thing? --- Here's the proof that the above accomplishes the task: a = true; b = true; c = true ((a ^ b) && (a ^ c) && !(b && c)) || ((b ^ a) && (b ^ c) && !(a && c)) || ((c ^ a) && (c ^ b) && !(a && b)) => false a = true; b = true; c = false ((a ^ b) && (a ^ c) && !(b && c)) || ((b ^ a) && (b ^ c) && !(a && c)) || ((c ^ a) && (c ^ b) && !(a && b)) => false a = true; b = false; c = true ((a ^ b) && (a ^ c) && !(b && c)) || ((b ^ a) && (b ^ c) && !(a && c)) || ((c ^ a) && (c ^ b) && !(a && b)) => false a = true; b = false; c = false ((a ^ b) && (a ^ c) && !(b && c)) || ((b ^ a) && (b ^ c) && !(a && c)) || ((c ^ a) && (c ^ b) && !(a && b)) => true a = false; b = true; c = true ((a ^ b) && (a ^ c) && !(b && c)) || ((b ^ a) && (b ^ c) && !(a && c)) || ((c ^ a) && (c ^ b) && !(a && b)) => false a = false; b = true; c = false ((a ^ b) && (a ^ c) && !(b && c)) || ((b ^ a) && (b ^ c) && !(a && c)) || ((c ^ a) && (c ^ b) && !(a && b)) => true a = false; b = false; c = true ((a ^ b) && (a ^ c) && !(b && c)) || ((b ^ a) && (b ^ c) && !(a && c)) || ((c ^ a) && (c ^ b) && !(a && b)) => true a = false; b = false; c = false ((a ^ b) && (a ^ c) && !(b && c)) || ((b ^ a) && (b ^ c) && !(a && c)) || ((c ^ a) && (c ^ b) && !(a && b)) => false

Answer 1

(a∧~(b∨c))∨(b∧~(a∨c))∨(c∧~(a∨b))