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.

  • $\begingroup$ Short answer: NO. $\endgroup$ Jul 2, 2016 at 8:03
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    $\begingroup$ 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. $\endgroup$ Jul 2, 2016 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. $\endgroup$ Jul 2, 2016 at 8:06
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    $\begingroup$ 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? $\endgroup$ Jul 2, 2016 at 10:29
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    $\begingroup$ If you consider $\top$ an atomic proposition, then it is a tautology*. $\endgroup$ Jul 2, 2016 at 12:17

2 Answers 2


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.

  • $\begingroup$ 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, 2016 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 $\mathfrak M\vDash p$. $\endgroup$ Jul 2, 2016 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". $\endgroup$ Jul 2, 2016 at 14:12

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.

  • $\begingroup$ what exactly is the constant true proposition 1? Why can we choose to include it as element of the languages alphabet or not? $\endgroup$ Dec 17, 2021 at 15:09

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