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Math Help - Can't figure out this proof.

  1. #1
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    Can't figure out this proof.

    Premise 1: For all x in D [ (A(x) or B(x)) -> (M(x) or N(x)) ]
    Premise 2: There exists an x in D [ A(x) and ~N(x) ]
    Solution: There exists an x in D [ B(x) or M(x) ]

    Any help is appreciated. Thanks!
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  2. #2
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    \begin{array}{rlr}<br />
   1 & {\left( {\forall x} \right)\left[ {\left( {A(x) \vee B(x)} \right) \to \left( {M(x) \vee N(x)} \right)} \right]} & {}  \\<br />
   2 & {\left( {\exists x} \right)\left[ {A(x) \wedge \neg N(x)} \right]} & {}  \\<br />
\hline<br />
   3 & {A(t) \wedge \neg N(t)} & 2  \\<br /> <br />
 \end{array}
    \begin{array}{rlr}<br />
   4 & {\left( {A(t) \vee B(t)} \right) \to \left( {M(t) \vee N(t)} \right)} & \;\;\;\;\;\;\;\;\;\; 1  \\<br />
   5 & {A(t)} & 3  \\<br />
   6 & {A(t) \vee B(t)} & 5  \\<br />
   7 & {M(t) \vee N(t)} & {4,6}  \\<br />
   8 & {\neg N(t)} & 3  \\<br />
   9 & {M(t)} & {7,8}  \\<br />
   {10} & {B(t) \vee M(t)} & 9  \\<br />
\hline   \therefore  & {\left( {\exists x} \right)\left[ {B(x) \vee M(x)} \right]} & {10}  \\ \end{array}

    Now you supply the reasons.
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  3. #3
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    Thanks for your help. I understand all the reasons except the last step how did you get to B(t) or M(t)?
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