# Thread: Logical expressions

1. ## Logical expressions

Let $R(x)$ be an open sentence over a domain $S$.

$\forall S, R(x)$ is a false statement and that the set $T$ of counterexamples is a proper subset of $S$.

If I let $R(x)$ be true and $\sim R(x)$ be false, is

$\forall x ( x \in T \subset S \Rightarrow \sim R(x))$ a correct expression?

2. Originally Posted by novice
Let $R(x)$ be an open sentence over a domain $S$.

$\forall S, R(x)$ is a false statement and that the set $T$ of counterexamples is a proper subset of $S$.

If I let $R(x)$ be true and $\sim R(x)$ be false, is

$\forall x ( x \in T \subset S \Rightarrow \sim R(x))$ a correct expression?
I have absolutely no idea what any of that says.
Is that something you have written yourself?
If not can you post the exact question as it appears wherever you found it?

3. Originally Posted by Plato
I have absolutely no idea what any of that says.
Is that something you have written yourself?
If not can you post the exact question as it appears wherever you found it?
Exercise 5.34, from Mathematical Proofs by Chartrand, page 125.

Let R(x) be an open sentence over a domain $S$. Suppose that $\forall x \in S, R(x)$ 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 $\forall x \in W, R(x)$ is true.

My attempt:

Let $R(x)$ be true and $R(x)$ be false. Then

$\forall x (x \in T \subsetneq S \Rightarrow \sim R(x))$

Since $T \subsetneq S$, it follows that $S-T \not=\emptyset$.

The negation of the statment is

$\exists x (x \in S-t \Rightarrow R(x))$ or equivalently

$\exists x (x \in W \Rightarrow R(x))$, where $W = S-T \not= \emptyset$

I don't know how to get
$\forall x \in W, R(x)$ is true.

4. I think that is self-contradictory problem.
How can an open sentence be false $\left( {\forall x \in S} \right)$?
And then true on some nonempty subset of $S$.

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.

5. Originally Posted by Plato
I think that is self-contradictory problem.
How can an open sentence be false $\left( {\forall x \in S} \right)$?
And then true on some nonempty subset of $S$.

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.
Plato,

I absolutely agree with you, but being a novice I thought it better to have you look at it. There were two proofs that I found laughable, but over all it's not a bad book.

Thanks for you time.

6. is a correct expression?
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, $x\in T\subset S$ is not considered a well-formed formula, though it makes sense to view it as an abbreviation for $x\in T\land T\subset S$. In fact, I don't think " ${}\subset S$" part is needed in the formula at all.

Concerning semantics, this formula is true according to the description of $R$, $T$ and $S$.

The negation of the statment is

$\exists x (x \in S-t \Rightarrow R(x))$
No, if you are talking about the negation of $\forall x\,(x\in T\Rightarrow\sim R(x))$, it is $\exists x\,(x\in T\land R(x))$.

The formula $\forall x\,(x\in W\Rightarrow R(x))$ does not follow from $\forall x\,(x\in T\Rightarrow\sim R(x))$, or its negation, alone. The latter formula says that elements of $T$ make $R$ false, but it does not say anything about elements outside $T$.

On the other hand, $\forall x\,(x\in T\Leftrightarrow\sim R(x))$ does imply $\forall x\,(x\in W\Rightarrow R(x))$ (in fact, the implication to the left is enough).

7. Originally Posted by emakarov
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, $x\in T\subset S$ is not considered a well-formed formula, though it makes sense to view it as an abbreviation for $x\in T\land T\subset S$. In fact, I don't think " ${}\subset S$" part is needed in the formula at all.

Concerning semantics, this formula is true according to the description of $R$, $T$ and $S$.

No, if you are talking about the negation of $\forall x\,(x\in T\Rightarrow\sim R(x))$, it is $\exists x\,(x\in T\land R(x))$.

The formula $\forall x\,(x\in W\Rightarrow R(x))$ does not follow from $\forall x\,(x\in T\Rightarrow\sim R(x))$, or its negation, alone. The latter formula says that elements of $T$ make $R$ false, but it does not say anything about elements outside $T$.

On the other hand, $\forall x\,(x\in T\Leftrightarrow\sim R(x))$ does imply $\forall x\,(x\in W\Rightarrow R(x))$ (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.