Can't figure out this proof.

**tshare1** · March 3rd 2009, 08:06 AM

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!

**Plato** · March 3rd 2009, 08:48 AM

$\begin{array}{rlr} 1 & {\left( {\forall x} \right)\left[ {\left( {A(x) \vee B(x)} \right) \to \left( {M(x) \vee N(x)} \right)} \right]} & {} \\ 2 & {\left( {\exists x} \right)\left[ {A(x) \wedge \neg N(x)} \right]} & {} \\ \hline 3 & {A(t) \wedge \neg N(t)} & 2 \\ \end{array}$
$\begin{array}{rlr} 4 & {\left( {A(t) \vee B(t)} \right) \to \left( {M(t) \vee N(t)} \right)} & \;\;\;\;\;\;\;\;\;\; 1 \\ 5 & {A(t)} & 3 \\ 6 & {A(t) \vee B(t)} & 5 \\ 7 & {M(t) \vee N(t)} & {4,6} \\ 8 & {\neg N(t)} & 3 \\ 9 & {M(t)} & {7,8} \\ {10} & {B(t) \vee M(t)} & 9 \\ \hline \therefore & {\left( {\exists x} \right)\left[ {B(x) \vee M(x)} \right]} & {10} \\ \end{array}$

Now you supply the reasons.

**tshare1** · March 3rd 2009, 09:07 AM

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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