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Thread: Distributive law

  1. October 25th 2009, 08:01 PM #1
    chimp
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    Distributive law

    Prove the distributive law alpha(beta+gamma) = (alpha)(beta) + (alpha)(gamma)
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  2. October 26th 2009, 06:55 AM #2
    novice
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    Quote Originally Posted by chimp View Post
    Prove the distributive law alpha(beta+gamma) = (alpha)(beta) + (alpha)(gamma)
    Prove \alpha(\beta+\gamma)=\alpha \beta + \alpha \gamma

    Let x\in\alpha.
    x\in\alpha\rightarrow x\in \beta\cup \gamma\rightarrow (x\in \beta) \cup (x\in \gamma) \rightarrow (x \in \alpha, x \in\beta)\cup (x \in \alpha, x \in\gamma)
    \because (x\in \beta) \cup (x\in \gamma) \rightarrow (x \in \alpha, x \in\beta)\cup (x \in \alpha, x \in\gamma)
    \therefore \alpha(\beta+\gamma)=\alpha \beta + \alpha \gamma
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