What is negation of a tautology?

Question

I was learning concepts of tautology, contradiction, contingent etc. The Tautology page of Wikipedia has following statement:

A formula is satisfiable if it is true under at least one interpretation, and thus a tautology is a formula whose negation is unsatisfiable.

Q1. Is the last part wrong? A statement that is not a tautology can either be a contradiction (which is unsatisfiable) or contingent, but it is not always unsatisfiable. Right?

Q2. I believe that the negation of a satisfiable statement is (obviously) unsatisfiable. Right? (And I believe thats what author of wiki article meant to say, but made a mistake and said that negation of tautology is unsatisifiable.)

Answer 1

Wikipedia hasn't made a mistake on this.

In classical logic, the models that satisfy a formula are precisely those that don't satisfy its negation. Thus a tautology is satisfied in all models and its negation - a contradiction - is satisfied in none, and that's what we mean when we say it's unsatisfiable. A contingent formula is satisfiable, but whether it's satisfied depends on the model.

Answer 2

Defintion: $\varphi$ is satisfiable if there exists an interpretation $\mathfrak{M}\models\varphi$, and unsatisfiable if in every interpretation $\mathfrak{M}$, we have $\mathfrak{M}\not\models \varphi$.

Definition: $\varphi$ is a tautology if for every interpretation $\mathfrak{M}$, we have $\mathfrak{M}\models\varphi$.

Definition: $\varphi$ is a contradiction if for every interpretation $\mathfrak{M}$, we have $\mathfrak{M}\not\models\varphi$, i.e., if $\varphi$ is unsatisfiable.

Lemma: $\mathfrak{M}\models \varphi$ if and only if $\mathfrak{M}\not\models\neg\varphi$.

Observation: $\varphi$ is a contradiction if and only if $\varphi$ is unsatisfiable.

Theorem: $\varphi$ is a tautology if and only if $\neg\varphi$ is not satisfiable.

Proof: ($\Rightarrow$) If $\varphi$ is a tautology then for all $\mathfrak{M}$, we have $\mathfrak{M}\models \varphi$, so by the Lemma, for all $\mathfrak{M}$, we have $\mathfrak{M}\not\models\neg\varphi$, so $\neg\varphi$ is unsatisfiable.

($\Leftarrow$) If $\neg\varphi$ is unsatisfiable, then for all $\mathfrak{M}$ we have $\mathfrak{M}\not\models \neg\varphi$, so by the Lemma, for all $\mathfrak{M}$ we have $\mathfrak{M}\models \varphi$. Therefore $\varphi$ is a tautology. $\square$

A statement which is not a tautology can be either contingent or contradiction,

Yes.

that is unsatisfiable, but it cannot be always unsatifiable.

Either a statement is satisfiable or it is unsatisfiable (which is to say, not satisfiable). There is no such thing as "always unsatisfiable".

Q2. I believe "negation of satisfiable statement is (obviously) unsatisfiable". Right? (And I believe thats what author of wiki article meant to say, but made a mistake and said negation of tautology is unsatisifiable.)

No. The negation of a satisfiable statement can still be satisfiable. "It is cold" is satisfiable, its negation "it is not cold" is also satisfiable, as long as you allow that there are some things which are cold and some (other) things which are not cold.

Answer 3

$p \lor (\lnot p)\equiv \top\;\;$ is a tautology in classical logic.

No matter what truth value we assign to $p$, the statement is true.

It's negation:

\begin{align} \lnot(p \lor (\lnot p))&\equiv (\lnot p \land \lnot(\lnot p))\tag{DeMorgan's}\\ \\ &\equiv (\lnot p \land p)\tag{double negation}\\ \\ &\equiv \bot\end{align}

is a contradiction. It doesn't matter whether $p$ is true or false, the negation of the tautology $p \lor (\lnot p)$ is false.

If $\top$ designates a tautology, then $\lnot \top \equiv \bot$, where $\bot$ designates a contradition.

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With respect to satisfiability, the formula $p \to q$ is contingent: it is satisfiable for all truth-value assignments to p, q, except for the assignment where $p$ is true, and $q$ is false.

The negation of $p \to q$ is given by $\lnot (p \to q) \equiv \lnot(\lnot p \lor q) \equiv (p \land \lnot q).$ It is also contingent because it is satisfiable only when $p$ is true and $q$ is false.

So both $(p\to q)$, and $\lnot(p\to q)$ are satisfiable, but contingent.

Just remember that the negation of a satisfiable formula does not mean the negated formula is unsatisfiable. It is only when a formula is a tautology, satisfiable under every truth value assingment, that the negation of the formula/tautology is a contradiction, and unsatisfiable under any truth value assignment.

Answer 4

A statement that is not a tautology can either be a contradiction (which is unsatisfiable) or contingent, but it is not always unsatisfiable. Right?

Indeed,

1. every statement that is not a tautology is some contingency or contradiction.

However,

2. the negation of every tautology is some contradiction, which is always unsatisfiable.

   [Remember, negation is a logical operation, which means that it has the same effect on a formula (flipping its truth value) regardless of interpretation; so, since a tautology is a formula whose truth-functional form is true regardless of interpretation, its negation's truth-functional form must be false regardless of interpretation, i.e., its negation must be a contradiction.]

I believe that the negation of a satisfiable statement is (obviously) unsatisfiable. Right?

No: $\forall x\,P(x)$ and its negation are both satisfiable.

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Your misunderstanding arises from conflating taking the complement of a set (point #1) and taking the negation of a statement (point #2).

Full explanation here.