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I have problem determining truth value of statements involving three quantifiers like this one.

$\forall x\ \exists y$ such that $\forall z, x+y = z,$ assuming all variables are real numbers.

I normally start these types of problems by trial and error, checking what happens if I fix one variable and vary the other. But since I have three here, I tried picking say $x = 4$ and $z = 3$ and see if I can find one y so $x+y = z$. Is this correct?

If so, is the statement then equivalent to the following? $\forall x$ and $\forall z,\exists y$ such that , $\ x+y = z$

I appreciate pointers or ways to tackle this problem. Thanks.

user74954
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The "for all x" quantifier (assuming a domain of real numbers) means that x is a real number (but no other properties are assumed). The following "there exists y" quantifier means there exists at least one real number y and that this y may depend on x. The "for all z" quantifier means that z is a real number but no other properties are assumed.

The second statement is not equivalent to the first, because the "there exists y" quantifier means that there exists at least one real number y and that this y may depend on both x and z. So the second statement is true (because y can depend on x and z, and be chosen as y = z - x) but the first statement is false because the choice of y cannot depend on z.

Joseph Myers
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  • Its saying for every x there exists a y such that for all z x+y=z. Its false since 1+3=4 but 1+3 /=5. – user60887 Apr 29 '13 at 04:40
  • Thanks all. @JosephMyers Let me see if I understand, if this is a game, I challenge someone to give me arbitrary x, say 1 (to use user608877's value) and I fix that number to find one real y value that will solve 1+y=z for all z, correct? So I cannot fix x and z values like I did in my original post. – user74954 Apr 29 '13 at 11:56
  • @user794954 your first statement is false as I said, and your example shows why it is wrong. Your second statement is true and is not equivalent to the first. – Joseph Myers Apr 30 '13 at 02:05