Translating Mathematical Statements into Statements Involving Nested Quantifiers

Question

The above question is taken from our textbook

Here, the domain is the set of all integers. And, if x<0 then the predicates evaluate to true.( As per the question ,the predicate must evaluate to True if x,y are +ve integers and x+y>0. ) It means that, the textbook answer is deviating from the requirements mentioned in the question.

I think the more appropriate answer for this question is,

∀x,y ( (x>0) ∧ (y>0) ∧ (x+y>0) )

Here, the predicate evaluates to True only if all the 3 conditions are met.

Please correct me if my answer is wrong.

Answer 1

$$\forall x,y ((x>0) \land (y>0) \land (x+y>0))$$

evidently implies

$$\forall x(x>0)$$

which says all integers are positive, which isn't what you want at all. ("The sum of two positive integers is positive" doesn't imply that all integers are positive!)

The textbook answer is right. You want to capture the thought that IF $x$ and $y$ are positive, so is $x + y$, no matter which integers $x$ and $y$ are. Which is exactly what the given answer does. This thought can't be falsified by looking at negative integers, so the fact that the conditional trivially evaluates as true when $x$ and/or $y$ is negative is exactly what we want (not as you seem to suppose, a bug to be avoided!)

Read any good intro logic book on translating English quantified sentences into their formal equivalents (Paul Teller is good, and his book is freely available at his website http://tellerprimer.ucdavis.edu: I also happen to think that P-t-r Sm-th's Intro to Formal Logic is good on translation!)

Answer 2

What the textbook wrote is correct, which we can write as such:

$$\forall x \forall y (\text{ IF}\;(x >0 \land y>0), \text{ THEN } \,(x + y > 0))$$

If $x\leq 0$ or $y \leq 0$ (or both), the antecedent of the quantified conditional is false, and hence, the quantified conditional is vacuously true.

Note that there is no claim being made in the case that $x\leq 0$ and/or $y \leq 0$. All that matters is that for every two integers GREATER THAN $0$, their sum is also greater than $0$.

Your proposed answer:

$$\forall x,y ((x>0) \land (y>0) \land (x+y>0))$$

is a quantified conjunction: three things must be true of all integers for your proposition to be true:

• Every integer $x$ is positive
• Every integer $y$ is positive
• For every pair of integers x, y, $x+y$ is positive.

For this quantified conjunction, all three conjuncts must be true for the entire conjunction to be true. The problem is, all three conjuncts are false, since there exist negative integers $x, y$, and the sum of two integers can be positive, negative or zero, depending on the integers themselves. That's not at all what is being claimed by the original statement.

Answer 3

Remember that $\; A\to B \;$ is equivalent to $\; \neg A \vee B \;$.

So $\; \forall x\, \forall y\, \big((x>0) \wedge (y> 0) \to (x+y>0)\big) \;$ is equivalent to:

$$\color{navy}{\forall x\, \forall y\, \big((x\leq 0) \vee (y \leq 0) \vee (x+y>0)\big)}$$

Now as at least one of those three atomic predicates will be true for any pair of integers, the statement is valid.

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Where as, as pointed out in the other answers, your statement that $\forall x\, \forall y\, \big((x>0) \wedge (y>0) \wedge (x+y>0)\big)\;$ would only hold if all three atomic predicates were true for all integer pairs; which is to assert that all integers are positive.