i need help on this question
$\displaystyle \forall x\forall y[ O(x) +O(y) = E(x+y)]$
Now the negation of :
..........$\displaystyle \exists m[ m\in Z\wedge O(2m)]$ is.............
............$\displaystyle \forall m[\neg( m\in Z\wedge O(2m)]$......................
Now drop $\displaystyle \forall,m$ and we are left with:
............~$\displaystyle (m\in Z\wedge O(2m))$.................................................. .....................................1
.But (1) is equivalent to:
.......................$\displaystyle [m\in Z\rightarrow\neg O(2m)$.................................................. .......................................2
And introducing back the for all quantifier we get:
..................$\displaystyle \forall m[ m\in z\rightarrow\neg O(2m)]$.................................................. ........................................3
FOR all integers 2m is not odd............................................... ...........
The negation is correct and well detailled, no blank to fill in!
For a), $\displaystyle O$ and $\displaystyle E$ are predicates, so $\displaystyle O(x)+O(y)=E(x+y)$ has no sense. But the answer is not so far from the formula proposed, only two symbols to change (and maybe a pair of parenthesis to add).