# Thread: logic sets predicates calculus

1. ## logic sets predicates calculus

i need help on this question

2. Originally Posted by srk619
i need help on this question
$\forall x\forall y[ O(x) +O(y) = E(x+y)]$

Now the negation of :

.......... $\exists m[ m\in Z\wedge O(2m)]$ is.............

............ $\forall m[\neg( m\in Z\wedge O(2m)]$......................

Now drop $\forall,m$ and we are left with:

............~ $(m\in Z\wedge O(2m))$.................................................. .....................................1

.But (1) is equivalent to:

....................... $[m\in Z\rightarrow\neg O(2m)$.................................................. .......................................2

And introducing back the for all quantifier we get:

.................. $\forall m[ m\in z\rightarrow\neg O(2m)]$.................................................. ........................................3

FOR all integers 2m is not odd............................................... ...........

3. can anyone help to fill in the blanks

4. The negation is correct and well detailled, no blank to fill in!

For a), $O$ and $E$ are predicates, so $O(x)+O(y)=E(x+y)$ 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).