I am self-studying precalculus-level mathematics in perhaps a more formal way than usual, which means that I am reading about logic, sets, proofs, etc.
The text I am looking at contains as an example of a statement that is not a proposition the following:
"This sentence is false."
The text provides the following reasoning:
$\quad$The statement “This sentence is false” is not a proposition because it is neither true nor false. It is an example of a paradox—a situation in which, from premises that look reasonable, one uses apparently acceptable reasoning to derive a conclusion that seems to be contradictory. If the statement “This sentence is false” is true, then by its meaning it must be false. On the other hand, if the given statement is false, then what it claims is false, so it must be true.
I need someone to please elaborate a little more on this explanation. Specifically, I am not quite sure about what it means to "assume" or "suppose" that something is the case. How is this different from asserting the truth of the statement? Second, why is it that if a statement leads to a contradiction, then the statement must not be true? Lastly, this example sometimes doesn't even appear in other texts and it doesn't seem helpful, since it has only confused me. What am I supposed to take from this example and why do some texts omit it?
For Mauro ALLEGRANZA:
After reading your answer, I almost felt satisfied with this explanation, which is very similar to your argument.
Let $p$ be the statement that $p$ is false. As you said, from a "reasonable" conception of truth, we know that (0) $p\rightarrow\lnot p$ and (0') $\lnot p\rightarrow p$. We reason as follows:
(1) $p$ (assumption)
(2) $p\rightarrow\lnot p$ from (0)
(3) $\lnot p$ from (1) and (2) by modus ponens
(4) $p\land\lnot p$ from (1) and (3) by $\land$-introduction
From the assumption that $p$ is true, we derive the contradiction (4) that $p$ is both true and false. Thus, $p$ cannot be true, so it must be false.
But if $p$ is false, we have the following:
(5) $\lnot p$ (assumption)
(6) $\lnot p\rightarrow p$ from (0')
(7) $p$ from (5) and (6) by modus ponens
(8) $p\land\lnot p$ from (5) and (7) by $\land$-introduction
From the assumption $\lnot p$, we are able to derive the contradiction (8) that $p$ is both true and false. Thus, $\lnot p$ cannot be true.
Since assuming $p$ to be true leads to a contradiction and assuming $p$ to be false leads to a contradiction, $p$ cannot have a truth value, and so it is not a proposition.
However, I became confused when using the following argument.
(1') $p\rightarrow\lnot p$ from (0)
(2') $\lnot p\rightarrow p$ from (0')
(3') $p\leftrightarrow\lnot p$ from (1') and (2') by $\leftrightarrow$-introduction.
Now, since for any statement $p$, $p\leftrightarrow\lnot p$ is a contradiction, i.e, always false, (1') or (2') must be false, as [(1')$\land$(2')]$\rightarrow$(3') is a tautology, with $(3')\equiv F$. However, from our conception of truth, we know that (1') and (2') are both true for this particular statement. Yet, for any actual proposition $p$, $(p\rightarrow\lnot p)\land (\lnot p\rightarrow p)$ is a contradiction. What does this mean?