Properties of Probability (Proving)

Prove the proposition:
=> P [ ( A ∩ B ' ) U ( A ' ∩ B ) ] = P ( A U B ) - P ( A ∩ B )

please help me I'm having a hard time proving this kind of proposition

∵ A∩B' ≡ A but not B ... and ... B∩A' ≡ B but not A

∴ (A∩B') and (B∩A') are Mutually Exclusive

∴ P[ (A∩B') U (B∩A') ] = P(A∩B') + P(B∩A')

. . . . . . . . . . . . . . . . . . . = [ P(A) - P(A∩B) ] + [ P(B) - P(A∩B) ]

. . . . . . . . . . . . . . . . . . . = [ P(A) + P(B) - P(A∩B) ] - P(A∩B)

. . . . . . . . . . . . . . . . . . . = P(AUB) - P(A∩B) ..................................... Q.E.D.
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P [ ( A ∩ B ' ) U ( A ' ∩ B ) ]
= P(A ∩ B ' ) + P(A ' ∩ B ) - P[( A ∩ B ' ) ∩ ( A ' ∩ B )]
=P(A) - P(A ∩ B) +P(B) - P(A ∩ B) - null
= P(A)+P(B) - 2* P(A ∩ B)
= P(AUB) +P(A ∩ B) - 2* P(A ∩ B)
= P ( A U B ) - P ( A ∩ B )

P [ ( A ∩ B ' ) U ( A ' ∩ B ) ] = P ( A ∩ B ' ) + P ( A ' ∩ B ) [as these 2 events are mutually exclusive]
=P(A) - P(A∩ B ) + P(B) - P(A∩ B ) = P ( A U B ) - P ( A ∩ B ) [as P(X) = P(X∩Y) + P(X∩Y') ]