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Thread: Solve Sets

  1. #1
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    Solve Sets

    Solve Sets-q3.jpgCan some one please help me to solve these question using SETS (de Morgan's law)
    Solve Sets-ned-ant-_5.jpg
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  2. #2
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    Re: Solve Sets

    this is pretty straightforward if you just apply the definitions.

    The general way to show two sets are equal is to show they are both subsets of one another, i.e. $A \subset B \wedge B \subset A \Rightarrow A=B$

    I'll do one direction and leave the other to you.

    let $(x,y) \in (A \times B) \cap (C \times D)$

    then by definition $(x,y) \in (A \times B) \wedge (x,y) \in (C \times D)$

    then again by definition this implies

    $x \in A \wedge x \in C,~~y \in B \wedge y \in D$

    thus $x \in A \cap C, ~~y \in B \cap D$ and thus $(x,y) \in (A \cap C) \times (B \cap D)$

    thus $ (A \times B) \cap (C \times D) \subset (A \cap C) \times (B \cap D)$

    now you do the other direction.
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  3. #3
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    Re: Solve Sets

    Quote Originally Posted by osamamansoor2016 View Post
    Can some one please help me to solve these question using SETS (de Morgan's law)
    Am I wrong, is the first question not, Prove that A\times (B\setminus C)=(A\times B)\setminus (A\times C)\Large\color{red}??

     \begin{align*}(a,d)\in (A\times (B\setminus C)&\iff (a\in A) \wedge [d\in (B\setminus C)]\\&\iff (a\in A) \wedge[d\in B\wedge d\notin C]\\&\iff [(a\in A)\wedge (d\in B)] \wedge[d\in A\wedge d\notin C] \\&\iff (a,d)\in [A\times B]\setminus [A\times C]\end{align*}
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