Let be an open sentence over a domain .
is a false statement and that the set of counterexamples is a proper subset of .
If I let be true and be false, is
a correct expression?
Exercise 5.34, from Mathematical Proofs by Chartrand, page 125.
Let R(x) be an open sentence over a domain . Suppose that is a false statement and that the set T of counterexamples is a proper subset. Show that there exists a nonempty subset W of S such that is true.
My attempt:
Let be true and be false. Then
Since , it follows that .
The negation of the statment is
or equivalently
, where
I don't know how to get is true.
I think that is self-contradictory problem.
How can an open sentence be false ?
And then true on some nonempty subset of .
I don’t think I have ever seen that textbook. There may be some way it is using the expression “the set T of counterexamples is a proper subset ” .
I find that idea a bit odd.
Sorry, this depends on the definition of a correct expression. I assume you are talking about syntactically well-formed expressions, regardless of whether they are true or false. This again depends on the definition of an expression, or a formula.
Usually, is not considered a well-formed formula, though it makes sense to view it as an abbreviation for . In fact, I don't think " " part is needed in the formula at all.
Concerning semantics, this formula is true according to the description of , and .
No, if you are talking about the negation of , it is .The negation of the statment is
The formula does not follow from , or its negation, alone. The latter formula says that elements of make false, but it does not say anything about elements outside .
On the other hand, does imply (in fact, the implication to the left is enough).
Thank you for pointing at the well formed formula. I noticed that my math book has violated wff left and right. I read a book on logic over a Christmas holiday out of curiosity. I finished reading the entire book in two weeks. Ha ha, now you know what quality education I obtained in logic.