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Thread: How do you distribute AND and OR?

  1. #1
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    How do you distribute AND and OR?

    Hi all,
    I'm working through a book on set theory and I haven't gotten very far yet. I'm sort of struck by the looseness of the proofs for the associative, distributive, and commutative laws provided by the book and I think this stems from the fact that I'm not sure how to relate / distribute the ANDs and ORs of written logic.

    Example:
    If x is an element of (A intersect B) union C, then x is an element of A AND x is an element of B, OR x is an element of C.

    Why does the proof jump from this kind of formulation to: (A U C) intersect (B U C)?
    Why is the OR back distributed over the AND?

    Any help clarifying my confusion would be much appreciated.
    Thanks,
    Ultros
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    Quote Originally Posted by Ultros88 View Post
    Hi all,
    I'm working through a book on set theory and I haven't gotten very far yet. I'm sort of struck by the looseness of the proofs for the associative, distributive, and commutative laws provided by the book and I think this stems from the fact that I'm not sure how to relate / distribute the ANDs and ORs of written logic.

    Example:
    If x is an element of (A intersect B) union C, then x is an element of A AND x is an element of B, OR x is an element of C.

    Why does the proof jump from this kind of formulation to: (A U C) intersect (B U C)?
    Why is the OR back distributed over the AND?

    Any help clarifying my confusion would be much appreciated.
    Thanks,
    Ultros
    Using $\displaystyle \wedge$ for AND and $\displaystyle \vee$ for OR, we have the following properties:

    $\displaystyle (A\wedge B)\vee C\Leftrightarrow (A\vee C)\wedge(B\vee C)$

    $\displaystyle (A\vee B)\wedge C\Leftrightarrow (A\wedge C)\vee (B\wedge C)$

    $\displaystyle \emph{Proof: }$ We use a truth table (note that $\displaystyle \text{T}$ means "true" and $\displaystyle \text{F}$ "false"):

    $\displaystyle \begin{tabular}{c|c|c|c|c}
    $A$ & $B$ & $C$ & $A\wedge B$ & $(A\wedge B)\vee C$\\\hline
    T & T & T & T & T\\
    T & T & F & T & T\\
    T & F & T & F & T\\
    T & F & F & F & F\\
    F & T & T & F & T\\
    F & T & F & F & F\\
    F & F & T & F & T\\
    F & F & F & F & F
    \end{tabular}$

    $\displaystyle \begin{tabular}{c|c|c|c|c|c}
    $A$ & $B$ & $C$ & $A\vee C$ & $B\vee C$ & $(A\vee C)\wedge (B\vee C)$\\\hline
    T & T & T & T & T & T\\
    T & T & F & T & T & T\\
    T & F & T & T & T & T\\
    T & F & F & T & F & F\\
    F & T & T & T & T & T\\
    F & T & F & F & T & F\\
    F & F & T & T & T & T\\
    F & F & F & F & F & F
    \end{tabular}$

    Since the final columns of the two truth tables are the same, we can say that the two statements are equivalent.

    Using a similar method, you can show that $\displaystyle (A\vee B)\wedge C\Leftrightarrow (A\wedge C)\vee (B\wedge C)$ $\displaystyle \square$


    Thus, we can show that $\displaystyle x\in(A\cap B)\cup C\Leftrightarrow x\in(A\cup C)\cap(B\cup C)$

    $\displaystyle \emph{Proof: }$

    $\displaystyle x\in(A\cap B)\cup C\Leftrightarrow (x\in A\cap B)\vee(x\in C)$

    $\displaystyle \Leftrightarrow (x\in A\wedge x\in B)\vee(x\in C)$

    $\displaystyle \Leftrightarrow (x\in A\vee x\in C)\wedge(x\in B\vee x\in C)$ (from the result above)

    $\displaystyle \Leftrightarrow (x\in A\cup C)\wedge(x\in B\cup C)$

    $\displaystyle \Leftrightarrow x\in(A\cup C)\cap(B\cup C)\quad\square$
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