Answer:
See answer below
Step-by-step explanation:
1), Probably, your implication is ∀x∃yP(x,y)⇒∃x∀yP(x,y).
This implication is false. Consider the predicate P(x,y):="x<y" for real numbers x,y. Then, ∀x∃yP(x,y) is true: for all real x, there exists some y greater than x (take y=x+1 for example). However, ∃x∀yP(x,y) is false, as it would imply that there exists some real number x such that x is smaller than all real numbers, which is not true (real numbers do not have a minimum or a lower bound).
A short explanation would be, even if for all elements you can find one that makes a predicate true, you can not find one element that makes the predicate true for all elements.
2) Again, I assume that the predicate is ∃x∀yP(x,y)⇒∀x∃yP(x,y)
This implication is false. Consider the predicate P(x,y):="x is integer and xy=0 " for real numbers x,y. Then ∃x∀yP(x,y) is true, we need to take x=0. However, ∀x∃yP(x,y) is false, if you take x=1/2, 1/2 will never be an integer, no matter the value of y.
A short explanation would be, even if you can find one element that makes a predicate true for all elements, you can not always take an arbitrary element and find some element that makes the predicate true.
pounds of paper;
pounds of cans;
pounds of paper and cans combined in September and October.
pounds of paper;
pounds of cans.