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How can one prove the following identity:

$$B\times(C\cap A) = B\times(C∪A) \setminus B\times(C\bigtriangleup A)$$ I tried it so many times but there's nothing.

There are three photos of one of my attempt. The solution progress may not be absolutely clear - sorry for this. I can rewrite, if you need.

Thanks a lot for any help!

Okey. Thanks to @User1006 I prove the identity.

  1. By this post I prove that $B×(C∪A)-B×(C△A)=B×((C∪A)-(C△A))$.

    $(x,y) ∈ B × (C ∪ A) - B × (C △ A) ⟺ (x,y) ∈ B × (C ∪ A) ∧ (x,y) ∉ B × (C △ A) ⟺ (x ∈ B ∧ y ∈ (C ∪ A)) ∧ (x ∉ B ∨ y ∉ (C △ A)) ⟺ (x ∈ B ∧ y ∈ (C ∪ A) ∧ x ∉ B) ∨ (x ∈ B ∧ y ∈ (C ∪ A) ∧ y ∉ (C △ A)) ⟺ 0 ∨ (x ∈ B ∧ y ∈ (C ∪ A) ∧ y ∉ (C △ A)) ⟺ x ∈ B ∧ y ∈ (C ∪ A) ∧ y ∉ (C △ A) ⟺ x ∈ B ∧ y ∈ (C∪A)-(C△A) ⟺ (x,y) ∈ B×((C∪A)-(C△A))$

  2. Now return to both parts of the identity. We can see that:
    1) For the right part of the identity $(x,y) ∈ B×(C∪A)-B×(C△A)⟺ (x,y) ∈ B×((C∪A)-(C△A))⟺ x ∈ B ∧ y ∈ (C∪A)-(C△A)$
    2) For the left part of the identity $(x,y) ∈ B×(C∩A)⟺ x ∈ B ∧ y ∈ (C∩A)$

  3. So our task now is to prove that $C∩A=C∪A−C△A$
    The approval is in User1006's answer below.

  • 0
    What is $\oplus$? Symmetric difference? – 2017-01-25
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    @Arthur Yes, exactly. – 2017-01-25
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    As of now you've gotten 4 downvotes. It is _most likely_ because you haven't expanded anything on "I tried it so many times but there's nothing." If you edit your question and _tell_ us one or two things you tried, and exactly what made it not work, preferably in as much detail as possible, this question would've been much better recieved. – 2017-01-25
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    Are you able to demonstrate the single-component identity $C \cap A = (C \cup A) \setminus (C \oplus A)$? – 2017-01-25
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    All images are here : http://dropmefiles.com/sbobj – 2017-01-25

1 Answers 1

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First prove that $$ C\cap A=C\cup A -C\bigtriangleup A $$ (We use the standard symbol $\bigtriangleup $ for Symmetric difference.). We use the fact that $$ C\bigtriangleup A = (C-A) \cup(A-C)=(C\cap A^c) \cup(A\cap C^c) $$ The proof is as follows: \begin{align} C\cup A -C\bigtriangleup A &= C\cup A -((C\cap A^c) \cup(A\cap C^c)) \\ &=(C\cup A) \cap ((C\cap A^c) \cup(A\cap C^c))^c \\ &=(C\cup A) \cap ((C\cap A^c)^c \cap(A\cap C^c)^c) \\ &=(C\cup A) \cap ((C^c\cup A) \cap(A^c\cup C)) \\ &=(C\cup A) \cap ((C^c\cap A^c) \cup(C^c\cap C)\cup(A\cap A^c)\cup(A\cap C)) \\ &=(C\cup A) \cap ((C^c\cap A^c) \cup\varnothing\cup\varnothing\cup(A\cap C)) \\ &=(C\cup A) \cap ((C^c\cap A^c) \cup(A\cap C)) \\ &=((C\cup A) \cap (C^c\cap A^c)) \cup((C\cup A) \cap (A\cap C)) \\ &=((C \cap (C^c\cap A^c))\cup( A \cap (C^c\cap A^c)))\cup((C\cup A) \cap (A\cap C)) \\ &=(((C \cap C^c)\cap A^c)\cup( (A \cap A^c)\cap C^c))\cup((C\cup A) \cap (A\cap C)) \\ &=((\varnothing\cap A^c)\cup(\varnothing\cap C^c))\cup((C \cap (A\cap C)\cup(A \cap (A\cap C))) \\ &=((\varnothing\cap\varnothing)\cup((A \cap (C\cap C)\cup(C \cap (A\cap A))) \\ &=\varnothing\cup(A \cap C)\cup(C \cap A) \\ &=(A \cap C)\cup(C \cap A) \\ &=(C \cap A)\cup(C \cap A) \\ &=C \cap A \end{align} Then by this post Prove that $A \times (B -C) = (A \times B) - (A \times C)$, it follows $$ B\times(C\cap A) = B\times(C∪A) \setminus B\times(C\bigtriangleup A) $$