Prove Set Identity: A⋂(B⊕C)=(A⋂B)⊕(A⋂C)

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Homework Statement


Prove that ##A\cap(B\Delta C)=(A\cap B)\Delta(A\cap C)##

Homework Equations

The Attempt at a Solution


[/B]
L.H.S.=##A\cap(B\Delta C)##
=##A\cap[(B - C) \cup (C - B)]##
=##A\cap[(B \cap \bar{C}) \cup (C \cap \bar{B})]##
=##[A\cap (B \cap \bar{C})] \cup [A\cap (C \cap \bar{B})]##
=##[(A\cap B) \cap \bar{C}] \cup [(A\cap C) \cap \bar{B}]##

R.H.S.=##(A \cap B) \Delta (A \cap C)##
=##[(A \cap B) - (A \cap C)] \cup [(A \cap C) - (A \cap B)]##
=##[(A\cap B) \cap \bar{(A\cap C)} ] \cup [(A\cap C) \cap \bar{(A\cap B)}]##
=##[(A\cap B) \cap (\bar{A} \cup \bar{C})] \cup [(A\cap C) \cap (\bar{A} \cup \bar{B})]##

I tried to make L.H.S = R.H.S. But with above results, It's not possible. Can anyone tell me what I've assumed wrong.
 
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Are you required to use that method? The most basic way to prove "X= Y" is to prove both "[itex]X\subseteq Y[/itex]" and "[itex]Y\subseteq X[/itex]"".
And the most basic way to prove "[itex]X\subseteq Y[/itex]" is to start "if x in in X" and use the definitions and properties of X and Y to conclude "therefore x is in Y".

Here we want to prove [itex]A\cap\left(B\Delta C\right)= \left(A\cap B\right)\Delta\left(A\cap C\right)[/itex] so we first prove
[itex]A\cap\left(B\Delta C\right)\subseteq \left(A\cap B\right)\Delta\left(A\cap C\right)[/itex]

To do that:
if [itex]x\in A\cap\left(B\Delta C\right)[/itex] then x is in A and x is in B or C but not both. So look at two cases

1) x is in B but not C. Then x is in [itex]A\cap B[/itex] but not [itex]A\cap C[/itex]. Therefore x is in [itex]\left(A\cap B\right)\Delta\left(A\cap C\right)[/itex].

2) x is in C but not in B. Then x is in [itex]A\cap C[/itex] but not [itex]A\cap B[/itex]. Therefore x is in [itex]\left(A\cap B\right)\Delta\left(A\cap C\right)[/itex].

Now show that [itex]\left(A\cap B\right)\Delta\left(A\cap C\right)\subseteq A\cap\left(B\Delta C\right)[/itex] the same way:
if [itex]x \in \left(A\cap B\right)\Delta\left(A\cap C\right)[/itex] then ...
 
HallsofIvy said:
Are you required to use that method? The most basic way to prove "X= Y" is to prove both "[itex]X\subseteq Y[/itex]" and "[itex]Y\subseteq X[/itex]"".
And the most basic way to prove "[itex]X\subseteq Y[/itex]" is to start "if x in in X" and use the definitions and properties of X and Y to conclude "therefore x is in Y".

Here we want to prove [itex]A\cap\left(B\Delta C\right)= \left(A\cap B\right)\Delta\left(A\cap C\right)[/itex] so we first prove
[itex]A\cap\left(B\Delta C\right)\subseteq \left(A\cap B\right)\Delta\left(A\cap C\right)[/itex]

To do that:
if [itex]x\in A\cap\left(B\Delta C\right)[/itex] then x is in A and x is in B or C but not both. So look at two cases

1) x is in B but not C. Then x is in [itex]A\cap B[/itex] but not [itex]A\cap C[/itex]. Therefore x is in [itex]\left(A\cap B\right)\Delta\left(A\cap C\right)[/itex].

2) x is in C but not in B. Then x is in [itex]A\cap C[/itex] but not [itex]A\cap B[/itex]. Therefore x is in [itex]\left(A\cap B\right)\Delta\left(A\cap C\right)[/itex].

Now show that [itex]\left(A\cap B\right)\Delta\left(A\cap C\right)\subseteq A\cap\left(B\Delta C\right)[/itex] the same way:
if [itex]x \in \left(A\cap B\right)\Delta\left(A\cap C\right)[/itex] then ...

I understand your approach. But I want to prove it through Set identities.