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!
$\displaystyle \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} $
$\displaystyle \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.