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Differential Equations 18 Online
OpenStudy (anonymous):

Please help, trying to get the knack for negating down early in the quarter so I don't have problems later....The domain of the following predicates is the set of all plants. P(x) = "x is poisonous." Q(x) = "Jeff has eaten x." Translate the following statement into predicate logic. Jeff has never eaten a poisonous plant.

OpenStudy (perl):

There does not exist a plant such that the plant is poisonous and Jeff has eaten it this is equivalent to there does not exist x such that P(x) and Q(x) using E = exist ~ (E x s.t. P(x) & Q(x) )

OpenStudy (perl):

`~ (E x s.t. P(x) & Q(x) )` is equivalent to , if you distribute the negation sign `For all x ~( P(x) & Q(x) )` `For all x (~P(x) or ~ Q(x))` since (~p or ~q) = p -> ~q `For all x ( if P(x) then ~Q(x))` In other words, to bring it back to english: for all plants , if the plant is poisonous then Jeff did not eat it. this is also equivalent by similar reasoning for all plants, if Jeff ate it then it is not poisonous

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