Predicates and Quanitifiers - Can't understand Question

[Tvtakaveli]

Hi I'm new here but can't get my head around this problem.

We use the predicates O and B, with domain the integers. O(n) is true if n is odd, and
B(n) is true if n if big, which here means that n > 100.
(a) Express ∀m, n ∈ Z|O(m) ∧ B(n) ⇒ B(n − m) in conversational English.
(b) Find a counter-example to this statement.

Now i take it two ways;
Just expressing the imply part of the statement so,
The difference between n and m is big.

Or is it deeper than that like;
For every m, n is an element of Z and so m is odd and n is big. This implies the difference between n and m is big.

Thank you for the help!

 

[Evgeny.Makarov]

(a) Express ∀m, n ∈ Z|O(m) ∧ B(n) ⇒ B(n − m) in conversational English.

I assume that ∧ binds stronger than ⇒ (similar to times and plus, respectively). This is a usual convention. Then the formula should be read literally. The part after | has the form P ⇒ Q. Such formula is read "If P, then Q". Next, the assumption P is O(m) ∧ B(n). This is read "m is odd and n is big". Finally, the conclusion Q of the implication is B(n − m), which is read "n - m is big". Altogether the quantifier-free part is "If m is odd and n is big, then n - m is big". Adding the quantifiers gives the final answer:

For all m and n, if m is odd and n is big, then n - m is big.

I would say that if it is stipulated that the domain is the set of integers, it is not necessary to say "for all integer m and n": this is assumed.

 

[Tvtakaveli]

Hi, thanks for clearing that up, I really appreciate it.

So just to confirm, producing a counter statement would just be substituting values. E. G.

Let m = 9 and n=105. 105-9 =96 which is not big (>100) therefore the statement is irrational.

 

[Evgeny.Makarov]

Let m = 9 and n=105. 105-9 =96 which is not big (>100) therefore the statement is irrational.

Yes. The only remark is that statements can be true or false, and real numbers can be rational or irrational.