Predicate Logic Negation Confusion I have a question in my discrete math class that I was having some confusion with.

If:
N(x): x is a non-negative integer 
E(x): x is even 
O(x): x is odd 
P(x): x is prime
Negate each sentence and translate into logical notation.


*

*There exists an even integer. 
Would this be: There is an integer that is not even. $∀x $~$ E(x)$ 

*Every integer is even or odd.
Would this be: There exists an integer that is not even or odd. $∃ x $~$ (E(x)  $V$ O(x))$ 

*All prime integers are non-negative.
Would this be: There exists a prime that is not non-negative. $∃ x $~$ P(x) \Rightarrow N(x)$ 

*The only even prime is 2.
Would this be: The only even prime is not 2. $∀x $~$(P(x) \wedge E(x)) \Rightarrow 2$ 

*Not all integers are odd.
Would this be: All integers are odd. $∀xO(x)$ 
I was wondering if my logic in answering these questions was right.
 A: 

    
*
    
*There exists an even integer.
    
  
Would this be: There is an integer that is not even. $\forall x \neg E(x)$

You managed to get the logic expression correct, but the English should be "All integers are not even", or "No integer is even".



    
*Every integer is even or odd.
    
  
Would this be: There exists an integer that is not even or odd. $∃ x \neg (E(x) \vee O(x))$

Yes, though it is better to say "... neither even nor odd", to clarify that the negation applies to "even or odd", rather than just to "even".



    
*All prime integers are non-negative.
    
  
Would this be: There exists a prime that is not non-negative. $∃ x \neg P(x) \Rightarrow N(x)$

The English is parseable, but the FOL is not.  You want :
$$\underbrace{\exists x }_\text{There is an integer}(\underbrace{P(x)}_\text{that is prime}\;\underbrace{\wedge}_\text{and}\; \underbrace{\neg N(x)}_\text{not non-negative})$$
Remember that the negation of an implication is a conjunction with a negation (of the consequent). $$\;\neg (A\implies B) \;\equiv\; A\wedge \neg B\;$$ 



    
*The only even prime is 2.
    
  
Would this be: The only even prime is not 2. $∀x \neg(P(x) \wedge E(x)) \Rightarrow 2$ 

No. The statement is actually short for "2 is an even prime and there exists no other even prime."  So the negation would be "Either 2 is not an even prime, or there exists an even prime which is not 2".
$$\neg (P(2)\wedge E(2)) \vee \exists x(P(x)\wedge E(x)\wedge (x\neq 2))$$
Also, you can't simply state something implies $2$.  That's nonsensical; $2$ is not a boolean value.  You need to equate something with it (or deny the equality as the case may be).



    
*Not all integers are odd.
    
  
Would this be: All integers are odd. $∀xO(x)$

Yes, it would.
A: (1.) and (2.) are correct.  (3.) should be $\neg \forall x (P(x) \Rightarrow \neg N(x))$.  This is equivalent to both $\exists x \neg(P(x)\Rightarrow N(x))$ and $\exists x (P(x) \land \neg N(x))$.
For (4.) you need the sentence "2 is not the only prime even number," which means either 2 is not a prime even number or there is another prime even number.  It can be formalized as $\neg (P(2)\land E(2) \land \forall x ([P(x) \land E(x) ] \Rightarrow x=2)$.
(5.) is correct.
A: When negating a statement, we interchange "there exists" and "for all," and negate all statements.
1 should be "All integers are not even", in other words, "No integers are even."
2 should be "There exists an integer that is not (even or odd)," so "There exists an integer that is neither even nor odd."
3 is correct.
4 should be "There is a even prime that is not 2."
5 is correct.
A: I found the 1st and 5th to be correct , for the 
2nd ) ∀x ( E(x) <--> ~O(x) ) and its negation would be 
      ∃x ~( E(x) <--> ~O(x) ) and on simplifying we get
      ∃x ( E(x) <--> O(x) )
3rd answer is posted above by graham and
4th) ( P(x) ∧ E(x) ) --> (x=2) so its negation 
          (P(x) ∧ E(x)) ∧ (x≠2)
