3
$\begingroup$

Definition: "A tautology is a propositional formula that is true under any truth assignment to each of the atomic propositions in the domain of propositional function."

Let $p$ be a simple (or atomic) proposition (e.g. "9 is a square root of 81").

I understand that a proposition may be either true or false (but not both true and false at the same time). That is, $p$ may be either true or false, exclusively. Under all possible truth assignments, $p$ is not always true. Therefore, from the definition of tautology, $p$ is not a tautology.

However, suppose I proved $p$ to be true. I am tempted to write $p \iff \top$, but this means "$p$ is a tautology". However, the previous paragraph's conclusions was that "$p$ is not a tautology".

What's going on here?

Using the notation in symbolic logic, how does one write that $p$ is indeed true?

Thanks in advance for your help.

$\endgroup$
  • $\begingroup$ Short answer: NO. $\endgroup$ – Mauro ALLEGRANZA Jul 2 '16 at 8:03
  • $\begingroup$ Long answer: the proposiational variables like $p$ in propositional calculus are variables that can be evaluated to true or false: thus an atomic formula like $p$ cannot be a tautology. $\endgroup$ – Mauro ALLEGRANZA Jul 2 '16 at 8:05
  • $\begingroup$ If we move to predicate calculus and we "equate" tautology with valid formula (which is not quite correct...) we have atomic formulae like $\forall x (x=x)$ that are valid. $\endgroup$ – Mauro ALLEGRANZA Jul 2 '16 at 8:06
  • 2
    $\begingroup$ Do you allow 0-ary propositional functions (i.e., propositional functions that bind values to zero variables when evaluated)? Your "9 is a square root of 81" appears to be a 0-ary propositional function, so assigning truth values to its variables yields the same sentence. Or did you intend that your example contained a variable to which a truth value can be assigned? If so, where/what is that variable? $\endgroup$ – Eric Towers Jul 2 '16 at 10:29
  • 2
    $\begingroup$ If you consider $\top$ an atomic proposition, then it is a tautology*. $\endgroup$ – Mauro ALLEGRANZA Jul 2 '16 at 12:17
3
$\begingroup$

However, suppose I proved $p$ to be true. I am tempted to write $p \iff \top$, but this means "$p$ is a tautology".

No it doesn't. $p \Leftrightarrow \top$ is just another formula, one that happens to be true in exactly the same structures as $p$ is.

It also happens that $p \Leftrightarrow \top$ is a tautology exactly if $p$ itself is a tautology. So if you write, "$p\Leftrightarrow\top$ is a tautology", you're also implicitly asserting that $p$ is a tautology.

But simply writing "$p\Leftrightarrow \top$", without further, is either just putting the formula on the table for further study at the metalevel, or implicitly an assertion that "$p\Leftrightarrow \top$" happens to be true under a particular intended interpretation of the symbols -- even though we're often leaving it implicit what the intended interpretation means.

If you want to say that $p$ is a tautology, then you need to say "$p$ is a tautology"; there is no generally understood symbolic way of saying that. You can say, at the metalevel, $$ \vDash p $$ which asserts that $p$ is logically valid, and in some presentations "tautology" is used to mean "logically valid". With the definition you quote, however, that is not the case.

| cite | improve this answer | |
$\endgroup$
  • $\begingroup$ I was using $\Leftrightarrow$ to denote logical equivalence. So, $p \Leftrightarrow \top$ means $p \leftrightarrow \top$ is a tautology. Which, as you've stated implies $p$ is a tautology. Hence the temptation to write $p \Leftrightarrow \top$ to mean $p$ is true. If there is no general way to say $p$ is a tautology, is there a generally understood way to say $p$ is true? Say, $p = 1$? $\endgroup$ – n00b Jul 2 '16 at 13:52
  • $\begingroup$ @n00b: If $\Leftrightarrow$ is a symbol at the metalevel for you, then you won't be justified in writing $p\Leftrightarrow \top$ simply because you have proved $p$ from some particular set of axioms -- logical equivalence is specifically logical equivalence, not equivalence that is forced by non-logical axioms. As for a way to say that "$p$ is true", in order to even say that you need to be speaking about being true in a particular structure $\mathfrak M$, and then you can write $\mathfrak M\vDash p$. $\endgroup$ – hmakholm left over Monica Jul 2 '16 at 14:09
  • $\begingroup$ In theories of arithmetic $(\mathbb N,+,\cdot)$ is usually the intended interpretation, and we can write $\mathbb N\vDash p$ -- but the intended interpretation of the language of set theory does not have any particular symbol (and many dispute that there is even a single intended interpretation), and if you want to speak of truth in some Platonically existing universe of sets, you need to say "$p$ is true" in words. You could write $\mathbf V\vDash p$, but that would usually be interpreted as "true in whichever universe we're working in". $\endgroup$ – hmakholm left over Monica Jul 2 '16 at 14:12
2
$\begingroup$

A simple atomic proposition can qualify as a tautology, if the constant true proposition '1', also denoted 'T', belongs to the vocabulary of the language.

A simple atomic proposition cannot qualify as a tautology if '1' is not part of the vocabulary of the language.

| cite | improve this answer | |
$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.