Answer :
Answer:
Step-by-step explanation:
a) Recall the quantifiers [tex]\forall, \exists[/tex].
Then, we can translate the proposition as follows
[tex]\forall m \in M \exists x \in D \exists y \in D S(x,m)\land S(y,m)[/tex]
b) Recall that an expression of the form [tex]\exists x P(x)[/tex] its negation is of the form [tex]\forall x \neg P(x)[/tex] which means that it is not true that for all elements the proposition P holds. Equivalently, we have that the negation of an expression of the form [tex]\forall x P(x)[/tex] is [tex]\exists x \neg P(x)[/tex] which means that there is at least one x such that P doesn't hold. Using this, we get the following
[tex]\neg(\forall m \in M \exists x \in D \exists y \in D S(x,m)\land S(y,m))= \exists m \in M \neg (\exists x \in D \exists y \in D S(x,m)\land S(y,m))= \exists m \in M \forall x \in D \neg (\exists y \in D S(x,m)\land S(y,m))= \exists m \in M \forall x \in D \forall y \in D \neg(S(x,m)\land S(y,m))[/tex]
By De Morgan's law, we have that [tex]\neg (A \land B) = \neg A \lor \neg B[/tex]
So, the final statement is
[tex]\exists m \in M \forall x \in D \forall y \in D \neg S(x,m) \lor \neg S(y,m)[/tex]
c)
This statement means: There is a movie that for every pair of students, at least one of the students hasn't seen the movie yet.