Consider the statement: for every integer such that is odd, is a multiple of . Which of the following is correct?
AIt is false, as shows.
BIt is true, vacuously: is always even, so there is no such .
CIt is false, as shows.
DIt is true, and every such happens to be a multiple of .
EIt is neither true nor false, because no such exists.
FIt is false, as shows.
GIt is true, but only because is prime.
HIt cannot be decided without knowing which integers are allowed.