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Let $A$ and $B$ be two events of a sample space $Ω$.

Prove that $A \nabla B=(A \cup B)-(A \cap B)$

By writing the formulas, how I can prove this in an easy way?

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    If you prove it using Venn diagrams, you will se that the proof is trivial. See https://en.wikipedia.org/wiki/Symmetric_difference – 2017-01-22
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    I'll try it using the formulas. But thanks for the info. – 2017-01-22

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I assume that $A \nabla B$ is the 'symmetric difference' or 'exclusive-or (XOR)' and that it is defined as $A \nabla B \equiv AB^c \cup A^cB,$ where intersection is written as multiplication for brevity and ${}^c$ indicates a complement.

Then $$P(A \nabla B) = P(AB^c) + P(A^cB) = [P(AB^c) + P(AB) + P(A^cB)] - P(AB)\\ = P(A \cup B) - P(AB),$$ as required. In the above the first equality holds because the events are disjoint, the second by arithmetic, and the third because the three quantities inside square brackets add to the probability of the union.

Because you have not shown us what you have tried, I have no idea what results about sums of probabilities you have already proved. So I am not sure this is an acceptable answer at this point in your course. Please check on that, and try to make any necessary modifications. (As @Crostul has commented, this relationship is easily seen in a Venn Diagram, but you have asked for a proof using formulas.)

A related equation involving the symmetric difference is that $$P(A \nabla B) = P(A) + P(B) - 2P(AB).$$

Notes: (a) In case anyone else wants to show an alternate method, $\nabla$ is rendered in MathJax as $\nabla$; I had to look it up. (b) Some texts use $A \Delta B$ instead of $A \nabla B.$

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    Can you explain the second equality and how do you reduce it to reach the third? Thanks. – 2017-01-22
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    3rd expression obtained by adding and subtracting $P(AB).$ 4th uses $[P(AB^c) + P(AB) + P(A^cB)] = P(A \cup B).$ Does that help? – 2017-01-22
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    So you add $P(AB)$ to the expression, you reduce it and then you remove $P(AB)$ and that way you keep the equality? Another thing: It always $X-Y=P(X-Y)=P(X)-P(Y)$? – 2017-01-22