Tautologies and Contradictions

Consider the sentence "it will rain tomorrow or it will not rain tomorrow." Without knowing anything about tomorrow's weather, you can already declare this sentence true. Why does this happen? What does it reveal about the machinery of logical reasoning itself? The answer is that some compound propositions are built from their pieces in a special way. No assignment of truth values to those pieces can ever make the whole statement false. Or, at the opposite extreme, no assignment can ever make it true. This phenomenon matters because it separates two kinds of statements. Some statements have truth that depends on facts about the world. Others have truth guaranteed (or forbidden) by logical form alone. A tautology is true purely by the shape of its logical connectives. A contradiction is false for the same structural reason. Mathematicians exploit this constantly. Every valid rule of inference — modus ponens, proof by cases, reductio ad absurdum — is secretly a tautology in disguise. Every proof by contradiction relies on deriving a contradiction from a false assumption. The big picture is this: tautologies and contradictions are the two "degenerate" ends of a spectrum. This spectrum ranks propositions by how many truth assignments satisfy them. A contingent proposition is true under some assignments and false under others. A tautology is true under all of them. A contradiction is true under none. Recognizing which category a compound statement falls into is an essential first skill. It must come before one can compare two statements for logical equivalence. It is also needed to justify a rule of inference as always safe to use. This article gives the precise truth-table definition of tautology and contradiction. It proves the basic closure and duality theorems relating the two notions. It works through fully expanded truth-table verifications. It also shows where the concept quietly powers the rest of mathematical logic — from Logical Equivalence to Proof by Contradiction to the deduction theorem in propositional calculus.

Key takeaways
  • A tautology is a compound proposition that is true under every possible truth assignment to its component propositions.
  • A contradiction is a compound proposition that is false under every possible truth assignment.
  • Tautologies and contradictions are negations of each other: ¬P\neg P is a contradiction iff PP is a tautology.
  • Verifying tautology/contradiction status is decidable by exhaustive truth table: nn variables need 2n2^n rows.
  • Tautologies formalize valid argument forms; a rule of inference is valid exactly when its associated conditional is a tautology.
  • Key idea: truth or falsity that is guaranteed by logical form alone, independent of the truth values of the atomic pieces, is what makes an argument logically valid rather than merely factually correct.

Where this lives today

Tautology-checking is not a museum piece; it underlies the practical engine of automated reasoning. Determining whether a propositional formula is a tautology is the complementary problem to Boolean satisfiability (SAT) — PP is a tautology exactly when ¬P\neg P is unsatisfiable — and SAT solvers built on this duality now check hardware circuit designs, verify compiler optimizations, and search combinatorial spaces in industrial software, all by testing enormous formulas for tautology or contradiction far too large to check by hand-drawn truth tables. In mathematical logic itself, tautologies are the semantic backbone of the deduction theorem: a formula QQ is derivable from premises P1,…,PnP_1,\dots,P_n in a proof system exactly when (P1∧⋯∧Pn)→Q(P_1 \wedge \cdots \wedge P_n) \rightarrow Q is a tautology (or, in first-order settings, valid). Every application of Proof by Contradiction rests on the contradiction half of this article: assuming ¬Q\neg Q and deriving a proposition of the form R∧¬RR \wedge \neg R — a contradiction — licenses the conclusion QQ, precisely because R∧¬RR \wedge \neg R can never be true under any assignment. The notion of Logical Equivalence between two propositions PP and QQ is itself defined as the biconditional P↔QP \leftrightarrow Q being a tautology, making tautology the parent concept from which equivalence is derived. Outside pure logic, cryptographic protocol verification uses tautology checking to certify that security properties hold under every possible adversarial input, and circuit designers use it to prove two logic-gate networks compute identical Boolean functions. In statistics and decision theory, recognizing a "tautological" hypothesis (one unfalsifiable by any data) is a basic sanity check that a claim carries no empirical content — mirroring exactly the mathematical fact that a tautology's truth value never depends on its atomic inputs.

Ask the tutor: Why does the deduction theorem let us treat "premises imply conclusion" as a single tautology, and how does that connect to writing a formal proof?

DefinitionTautology, Contradiction, and Contingency

Let PP be a compound proposition built from atomic propositions p1,p2,…,pnp_1, p_2, \dots, p_n using logical connectives (¬,∧,∨,→,↔\neg, \wedge, \vee, \rightarrow, \leftrightarrow). A truth assignment to p1,…,pnp_1,\dots,p_n is a function v:{p1,…,pn}→{T,F}v : \{p_1,\dots,p_n\} \to \{T, F\} assigning a truth value to each atomic proposition. Each truth assignment extends uniquely to a truth value v(P)∈{T,F}v(P) \in \{T,F\} for the compound proposition PP, via the usual recursive rules for the connectives (this extension is exactly what a truth table computes row by row).

  • PP is a tautology if v(P)=Tv(P) = T for every truth assignment vv to its atomic propositions. Notation: ⊨P\models P.
  • PP is a contradiction (or unsatisfiable) if v(P)=Fv(P) = F for every truth assignment vv.
  • PP is a contingency (or contingent) if it is neither a tautology nor a contradiction — that is, v(P)=Tv(P)=T for some assignments and v(P)=Fv(P) = F for others.
  • PP is satisfiable if v(P)=Tv(P) = T for at least one assignment vv (equivalently, PP is not a contradiction).

Since there are exactly 2n2^n truth assignments to nn atomic propositions, each of these properties is checked in principle by an exhaustive truth table with 2n2^n rows.

Intuition

Think of a compound proposition PP with nn atomic components as a function

fP:{T,F}n→{T,F}.f_P : \{T,F\}^n \to \{T,F\}.

The domain {T,F}n\{T,F\}^n has 2n2^n points, one for each row of the truth table. A tautology is a proposition whose function fPf_P is the constant function TT: every point in the domain lands on TT. A contradiction is the opposite extreme, the constant function FF. Everything else — the vast majority of compound propositions — is contingent: the function fPf_P takes both values somewhere on its domain. This picture explains why tautologies feel "empty" of factual content. Since fPf_P doesn't depend on which row of the table you're looking at, learning the truth value of PP tells you nothing new about the world — you already knew it. That is precisely why tautologies are the right notion of logically valid statement: their truth is guaranteed by the rules of logic, not by any fact about p1,…,pnp_1,\dots,p_n. Contradictions are the mirror image: they are guaranteed false, and deriving one from a set of assumed premises certifies that those premises cannot all be true together, which is the engine behind Proof by Contradiction.

Figure
The truth-value function fP:{T,F}n→{T,F}f_P:\{T,F\}^n\to\{T,F\} for the three categories, n=2n=2
Four rows of a 2-variable truth table, mapped to T/F p q Tautology p ∨ ¬p... (∨q) Contingency p → q Contra. p ∧ ¬p T T T F F T F F T T T T T F T T F F F F Constant-T column = tautology. Mixed column = contingency. Constant-F column = contradiction. Same 4 rows, three different "shapes" of output — this is the whole classification.
TheoremNegation Duality and Closure of Tautologies

Let PP and QQ be compound propositions built from atomic propositions among p1,…,pnp_1,\dots,p_n.

(a) Negation duality. PP is a tautology if and only if ¬P\neg P is a contradiction. Equivalently, PP is a contradiction if and only if ¬P\neg P is a tautology.

(b) Closure under conjunction. If PP and QQ are both tautologies, then P∧QP \wedge Q is a tautology.

(c) Modus-ponens closure. If PP is a tautology and P→QP \rightarrow Q is a tautology, then QQ is a tautology.

(d) Substitution invariance. If PP is a tautology (contradiction) and QQ is obtained from PP by uniformly replacing every occurrence of an atomic proposition pip_i with an arbitrary compound proposition RR, then QQ is also a tautology (contradiction).

Proof

We prove each part directly from the Definition above, using only the recursive semantics of the connectives — this is the standard technique for propositional logic: reduce a claim about all truth assignments to a case analysis on the two possible values T,FT, F.

(a) Negation duality. Recall the truth table for negation: v(¬P)=Tv(\neg P) = T exactly when v(P)=Fv(P) = F, and v(¬P)=Fv(\neg P) = F exactly when v(P)=Tv(P) = T, for every truth assignment vv. (⇒\Rightarrow) Suppose PP is a tautology, so v(P)=Tv(P) = T for every assignment vv. Then for every assignment vv, since v(P)=Tv(P) = T, the negation rule gives v(¬P)=Fv(\neg P) = F. Since this holds for every vv, by definition ¬P\neg P is a contradiction. (⇐\Leftarrow) Suppose ¬P\neg P is a contradiction, so v(¬P)=Fv(\neg P) = F for every assignment vv. Fix an arbitrary assignment vv. Since v(¬P)=Fv(\neg P) = F, the negation rule forces v(P)=Tv(P) = T (because if v(P)v(P) were FF, we'd get v(¬P)=Tv(\neg P) = T, contradicting v(¬P)=Fv(\neg P) = F). As vv was arbitrary, v(P)=Tv(P) = T for every assignment, so PP is a tautology. The second equivalence ("PP is a contradiction iff ¬P\neg P is a tautology") follows by applying the just-proved statement to the proposition ¬P\neg P in place of PP, and using the fact that ¬¬P\neg\neg P and PP receive the same truth value under every assignment (double-negation elimination, itself checked by the two-row truth table for ¬¬P\neg\neg P).

(b) Closure under conjunction. Suppose PP and QQ are tautologies. Let vv be an arbitrary truth assignment to the atomic propositions occurring in P∧QP \wedge Q (this assignment restricts to an assignment on the atoms of PP and to one on the atoms of QQ). Since PP is a tautology, v(P)=Tv(P) = T. Since QQ is a tautology, v(Q)=Tv(Q) = T. By the truth table for conjunction, v(P∧Q)=Tv(P \wedge Q) = T whenever both conjuncts are TT. Hence v(P∧Q)=Tv(P\wedge Q) = T. Since vv was an arbitrary assignment, P∧QP \wedge Q is a tautology.

(c) Modus-ponens closure. Suppose PP is a tautology and P→QP \rightarrow Q is a tautology. Let vv be an arbitrary truth assignment to the atomic propositions of QQ (extended, if necessary, to also cover the atoms of PP; any such extension is still a legitimate truth assignment, and by hypothesis PP and P→QP\to Q are true under every assignment to their atoms, so in particular under this one). Since PP is a tautology, v(P)=Tv(P) = T. Since P→QP \rightarrow Q is a tautology, v(P→Q)=Tv(P \rightarrow Q) = T. Now examine the truth table for the conditional: v(P→Q)=Fv(P \rightarrow Q) = F exactly in the single case v(P)=Tv(P) = T and v(Q)=Fv(Q) = F. We are given v(P→Q)=Tv(P\to Q) = T, so this bad case does not occur. Since we already know v(P)=Tv(P) = T, the only way to avoid the bad case is v(Q)=Tv(Q) = T. Since vv was an arbitrary assignment to the atoms of QQ, we conclude QQ is a tautology.

(d) Substitution invariance. Let PP be a tautology in the atomic propositions p1,…,pnp_1,\dots,p_n, and let QQ be the result of replacing every occurrence of pip_i (for some fixed index ii) by a compound proposition RR built from atoms q1,…,qmq_1,\dots,q_m (possibly overlapping with p1,…,pnp_1,\dots,p_n). Let uu be an arbitrary truth assignment to the atomic propositions occurring in QQ, namely p1,…,pi^,…,pn,q1,…,qmp_1,\dots,\widehat{p_i},\dots,p_n,q_1,\dots,q_m (all atoms except pip_i, together with the atoms of RR). Define a truth assignment vv on p1,…,pnp_1,\dots,p_n by setting v(pj)=u(pj)v(p_j) = u(p_j) for j≠ij \neq i, and v(pi):=u(R)v(p_i) := u(R), the truth value that RR receives under uu (computed by the recursive semantics for RR). This vv is a bona fide truth assignment to the atoms of PP, since it assigns some definite value in {T,F}\{T,F\} to each pjp_j. A straightforward induction on the structure of PP (checking each connective rule: ¬,∧,∨,→,↔\neg, \wedge, \vee, \rightarrow, \leftrightarrow all commute with substitution because they are defined recursively from truth values of immediate subformulas) shows that u(Q)=v(P)u(Q) = v(P): replacing pip_i by RR everywhere and evaluating under uu gives exactly the same truth value as evaluating PP under the assignment that sends pip_i to u(R)u(R) and leaves the rest alone. Since PP is a tautology, v(P)=Tv(P) = T regardless of which vv we plug in — in particular for the vv constructed above. Hence u(Q)=Tu(Q) = T. Since uu was an arbitrary assignment to the atoms of QQ, QQ is a tautology. The parenthetical claim for contradictions follows by the identical argument with TT replaced by FF throughout, or by combining part (a) with the tautology case: PP is a contradiction iff ¬P\neg P is a tautology; substituting into ¬P\neg P and using part (a) again yields the contradiction case for QQ. ■\blacksquare

∎
CorollaryContradictions are closed under disjunction

If PP and QQ are both contradictions, then P∨QP \vee Q is a contradiction.

Proof

By part (a) of the theorem, PP contradiction ⇒¬P\Rightarrow \neg P tautology, and QQ contradiction ⇒¬Q\Rightarrow \neg Q tautology. By part (b), ¬P∧¬Q\neg P \wedge \neg Q is a tautology. De Morgan's law (itself verifiable by a 4-row truth table: v(¬(P∨Q))=Tv(\neg(P\vee Q)) = T iff v(P)=v(Q)=Fv(P)=v(Q)=F iff v(¬P∧¬Q)=Tv(\neg P \wedge \neg Q) = T, for every assignment vv) shows ¬(P∨Q)\neg(P \vee Q) and ¬P∧¬Q\neg P \wedge \neg Q agree on every truth assignment. Hence ¬(P∨Q)\neg(P\vee Q) is also a tautology (it takes value TT wherever ¬P∧¬Q\neg P \wedge \neg Q does, which is everywhere). By part (a) again, P∨QP \vee Q is a contradiction. ■\blacksquare

∎

Worked examples

Example 1 — an asymmetric three-variable tautology. Let

P  =  [(p→q)∧(q→r)]→(p→r).P \;=\; \big[(p \rightarrow q) \wedge (q \rightarrow r)\big] \rightarrow (p \rightarrow r).

This is the hypothetical syllogism schema. We verify it is a tautology by exhausting all 23=82^3 = 8 truth assignments to p,q,rp, q, r.

ppqqrrp→qp\to qq→rq\to r(p→q)∧(q→r)(p\to q)\wedge(q\to r)p→rp\to rPP
TTTTTTTT
TTFTFFFT
TFTFTFTT
TFFFTFFT
FTTTTTTT
FTFTFFTT
FFTTTTTT
FFFTTTTT

Every row's final column reads TT. To see why without checking each row mechanically: the only way the whole conditional PP could be false is if its antecedent (p→q)∧(q→r)(p\to q)\wedge(q\to r) is TT while its consequent p→rp \to r is FF. The consequent p→rp\to r is FF only when p=T,r=Fp = T, r = F. If also p→qp \to q is TT with p=Tp = T, then q=Tq = T is forced. But then q→rq \to r requires, with q=Tq = T, that r=Tr = T — contradicting r=Fr = F. So the "bad" combination never arises for any values of p,q,rp,q,r; this is exactly the case-analysis behind the table, and it confirms PP is a tautology.

Example 2 — an asymmetric contradiction, verified two ways. Let

Q  =  (p∨q)∧¬p∧(¬q∨r)∧¬r∧p.Q \;=\; \big(p \vee q\big) \wedge \neg p \wedge (\neg q \vee r) \wedge \neg r \wedge p.

This has four atoms p,q,rp,q,r (three distinct atoms, one repeated), so there are 23=82^3 = 8 rows. Method 1 (full table). Since QQ contains both pp and ¬p\neg p as conjuncts, and conjunction is only true when every conjunct is true, we can shortcut: v(Q)=Tv(Q) = T would require v(p)=Tv(p) = T (from the last conjunct) and v(¬p)=Tv(\neg p) = T, i.e. v(p)=Fv(p) = F (from the second conjunct), simultaneously. No truth value v(p)∈{T,F}v(p) \in \{T,F\} satisfies both v(p)=Tv(p)=T and v(p)=Fv(p) = F. Hence there is no assignment making QQ true — QQ is a contradiction, and we didn't even need to look at qq or rr. Method 2 (confirm via the theorem's duality, as a cross-check). Consider

¬Q=¬[(p∨q)∧¬p∧(¬q∨r)∧¬r∧p].\neg Q = \neg\big[(p\vee q)\wedge \neg p \wedge (\neg q \vee r) \wedge \neg r \wedge p\big].

By part (a) of the theorem, QQ is a contradiction iff ¬Q\neg Q is a tautology. Using De Morgan repeatedly,

¬Q≡¬(p∨q)∨p∨¬(¬q∨r)∨r∨¬p≡⋯∨p∨¬p∨…\neg Q \equiv \neg(p\vee q) \vee p \vee \neg(\neg q \vee r) \vee r \vee \neg p \equiv \dots \vee p \vee \neg p \vee \dots

and since ¬Q\neg Q is a disjunction containing both pp and ¬p\neg p as disjuncts, at least one of these two disjuncts is TT under every assignment (because v(p)v(p) and v(¬p)v(\neg p) are always complementary, one of the two is always TT). A disjunction with a disjunct that is always TT is itself always TT: it is a tautology. This confirms, via the duality theorem rather than brute enumeration, that QQ is a contradiction — illustrating how part (a) can replace an 8-row table with a one-line argument.

Counterexamplewhy "true for one assignment" is not enough to be a tautology

The theorem's definition insists a tautology be true under every truth assignment — dropping the word "every" and settling for "true under some assignment" collapses the useful distinction between tautology and mere satisfiability, and breaks the modus-ponens closure result (part (c)). Take P=pP = p (the atomic proposition itself) and P→QP \rightarrow Q where Q=p∧qQ = p \wedge q, i.e. consider the conditional

p→(p∧q).p \rightarrow (p \wedge q).

Check: is P=pP = p "true under some assignment"? Yes — take v(p)=Tv(p) = T; then v(P)=Tv(P) = T. Is P→QP \rightarrow Q "true under some assignment"? Yes — take v(p)=T,v(q)=Tv(p) = T, v(q) = T; then v(P∧Q)=v(p∧q)=Tv(P\wedge Q)=v(p\wedge q) = T and the conditional is TT. If we (incorrectly) accepted "true under some assignment" as enough to call PP and P→QP \to Q "tautologies," the conclusion of part (c) would claim Q=p∧qQ = p\wedge q is a tautology too. But p∧qp \wedge q is not a tautology at all — it is contingent: under the assignment v(p)=T,v(q)=Fv(p) = T, v(q) = F, we get v(p∧q)=Fv(p\wedge q) = F. So the "theorem" fails outright once we weaken "every assignment" to "some assignment" in the hypotheses. The genuine theorem survives exactly because it demands v(P)=Tv(P) = T and v(P→Q)=Tv(P\to Q) = T for every single assignment vv simultaneously, which (as the real proof shows) forces v(Q)=Tv(Q) = T for that same vv, for every vv — not merely for some lucky assignment where things happen to align. This is why the universal quantifier "for every truth assignment" in the definition of tautology cannot be weakened to an existential one without destroying the theorem.

Scope, edge cases & common traps

SituationWhat to checkCommon error
Zero atomic propositions (a closed sentence with no variables)The "truth table" has one row; the proposition is simply either always TT or always FF, hence trivially a tautology or contradictionForgetting this degenerate case exists when counting 2n2^n rows with n=0n=0
Large nnTable has 2n2^n rows; still finite, so tautology/contradiction status is always decidable in principleAssuming decidability implies efficiency — the table can be astronomically large; this is the seed of the P vs NP-adjacent SAT problem
Confusing "not a tautology" with "a contradiction"A proposition can be a contingency: false somewhere, true elsewhere, neither a tautology nor a contradictionAssuming ¬(tautology)\neg(\text{tautology}) status automatically means "contradiction" — true only when the proposition really is a tautology or contradiction to begin with, not for a generic contingent PP
Substitution (part d)Substitution must replace every occurrence of the chosen atom uniformlySubstituting different formulas for different occurrences of the same atomic letter breaks the invariance
Checking equivalence of two propositions via a single rowOne matching row of a truth table proves nothing about tautology/equivalenceVerifying only a convenient row (e.g. all-true) and declaring victory, instead of checking all 2n2^n rows

The mathematician behind it

Origins · 1 minGeorge Boole and how this was discovered
GB
George Boole
1815–1864 · Mid-19th-century England, self-taught mathematician and logician

George Boole had almost no formal higher education, working as a schoolteacher before eventually becoming the first professor of mathematics at Queen's College, Cork. His 1854 treatise An Investigation of the Laws of Thought recast logical reasoning as an algebra over the values 00 and 11, in which the connectives "and," "or," and "not" became operations obeying precise algebraic laws. This was the decisive step that made it possible to ask, mechanically and exhaustively, whether a compound statement is guaranteed true (a tautology, in Boole's algebra: identically equal to 11) or guaranteed false (a contradiction: identically equal to 00), for every assignment of 00s and 11s to its variables — exactly the truth-table method used throughout this article.

History
- 1854 — George Boole publishes The Laws of Thought, founding Boolean algebra and the arithmetic treatment of logical truth and falsity.
- 1879 — Gottlob Frege's Begriffsschrift introduces a formal symbolic system for propositional and predicate logic, sharpening the syntax needed to state tautologies precisely.
- 1921 — Emil Post gives a systematic truth-table analysis of propositional connectives and proves functional completeness results, formalizing exhaustive verification of tautologies.
- 1921 — Ludwig Wittgenstein's Tractatus Logico-Philosophicus popularizes the term "tautology" as a proposition true under every possible state of affairs, independent of empirical fact.
- 1936 — Alonzo Church and Alan Turing independently show that while propositional tautology-checking is decidable (via truth tables), the analogous question for first-order logic is not, sharpening the boundary this concept sits on.
Is p→pp \to p a tautology, and why does this matter?

Yes. Under v(p)=Tv(p) = T: the conditional T→TT \to T is TT. Under v(p)=Fv(p) = F: the conditional F→FF \to F is TT (a conditional with a false antecedent is always true, regardless of the consequent). Both rows give TT, so by definition p→pp \to p is a tautology — the simplest nontrivial example there is. It matters because it is the propositional shadow of the logical law of identity: any proposition implies itself, and no fact about the world is needed to establish this, only the shape of the conditional connective.

How is a tautology different from a proposition that just happens to be true?

A proposition that "happens to be true," like "7 is prime," depends on specific mathematical facts beyond the reach of a truth table — its truth is not determined merely by which connectives appear in it. A tautology's truth is guaranteed by its logical form: no matter what truth values you plug into its atomic components, the output is always TT. You can verify tautology status by exhaustively checking all 2n2^n rows of a truth table using only the connective rules, without knowing anything about primes, arithmetic, or any other subject matter. That structural, content-independent guarantee is exactly what the word "tautology" captures.

Can a proposition be both a tautology and a contradiction?

No, provided it has at least one atomic proposition, so that at least one truth assignment exists. A tautology requires v(P)=Tv(P) = T for every assignment vv; a contradiction requires v(P)=Fv(P) = F for every assignment vv. If both held, then for any single assignment vv we'd need v(P)=Tv(P) = T and v(P)=Fv(P) = F simultaneously, which is impossible since v(P)v(P) is a single, well-defined truth value in {T,F}\{T,F\}. The only edge case is a proposition with zero atomic components (a "0-ary" formula), where there is exactly one (trivial) assignment; even then, that one assignment yields either TT or FF, never both, so tautology and contradiction remain mutually exclusive.

Why do we care about tautologies if they carry "no information"?

Precisely because they carry no information about the subject matter, tautologies carry maximal information about the logic. Every valid rule of inference (modus ponens, modus tollens, hypothetical syllogism, proof by cases) corresponds to a tautology, and checking that correspondence is exactly how we certify an argument form as universally trustworthy, independent of which particular propositions are plugged into it. In a proof, you're not usually asserting a tautology as your final conclusion — you're using tautologies as the licensed "moves" that let you get from true premises to a true conclusion without ever risking falsehood.

How does this connect to Proof by Contradiction?

In a proof by contradiction, you assume the negation of your goal, ¬Q\neg Q, alongside your other premises, and derive some proposition of the form R∧¬RR \wedge \neg R. Such a proposition is a contradiction: it is false under every truth assignment, because one of R,¬RR, \neg R is always false. Since your premises plus ¬Q\neg Q jointly imply something that's guaranteed false, the premises and ¬Q\neg Q cannot all be true together — and since the original premises are given as true, ¬Q\neg Q must be the culprit, forcing QQ to be true. The whole method leans on the fact that a contradiction is false in every conceivable scenario, with no exceptions to worry about.

Is checking whether a formula is a tautology always fast to do?

Not necessarily. The truth-table method always terminates and always gives a correct yes/no answer — it's fully decidable — but the number of rows is 2n2^n for nn atomic propositions, which grows exponentially. For 2020 variables that's over a million rows; for 5050 variables it's over 101510^{15}. In practice, tautology-checking (equivalently, its complement, unsatisfiability-checking) is believed to require exponential time in the worst case, and it sits at the heart of the P vs. NP question in computer science. Clever algorithms (SAT solvers) do far better than brute force on most real-world formulas, but no polynomial-time method is currently known to work on all formulas.

What's the difference between a tautology and a "logically true" mathematical theorem, like the Pythagorean theorem?

The Pythagorean theorem is true because of the specific axioms of Euclidean geometry — change the axioms (e.g. move to spherical geometry) and it can fail. A propositional tautology like p∨¬pp \vee \neg p is true purely from the meanings of "or" and "not," with no geometric or arithmetic content at all; it would remain true under literally any reinterpretation of the atomic proposition pp, since the argument only used the truth table for negation and disjunction. Tautologies are truths of logical form; theorems like Pythagoras's are truths that also depend on the specific subject-matter axioms in force.

Before this

Where this leads