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(2) (a)FALSE: because 1+1=2 is true but 2+3=4 is false. (b)TRUE: because false implies false is true and 1+1=3 is false and monkey can is false (c)TRUE:p->q is true if either ~p is true or q is true or both .Since ~p=monkey can't fly is true and therefore p->q is true (d)TRUE: p : 1+1=3 and q : 2+2=4 . Since ~p is true and q is also true ,therefore if p then q is true 3\begin{aligned}(p \rightarrow q) \wedge(p \rightarrow r) & \equiv(\neg p \vee q) \wedge(\neg p \vee r) \\& \equiv \neg p \vee(q \wedge r) \\& \equiv p \rightarrow(q \wedge r)\end{aligned} ...