Statements, truth values and logical connectives

A statement makes a claim with a fixed meaning

In classical two-valued logic, a statement, also called a proposition, is a declarative claim with exactly one of the truth values true and false once its meaning and context are fixed. The claim 14 − 6 = 8 is true. The claim 14 − 6 = 9 is false. Both are statements: being false does not stop a sentence from making a claim.

A question such as ‘Which number comes next?’ requests information. An instruction such as ‘Draw a square’ requests an action. Neither asserts something to which we assign a truth value. An isolated expression such as 5 + 2 also has a numerical value but makes no true-or-false claim until it appears in a statement, such as 5 + 2 = 7.

Fix the objects and the context before using a sentence as a logical building block. ‘This length is greater than four’ needs an identified length and a unit. ‘The marker is blue’ needs an identified marker and an agreed observation. The symbols below help organise precise claims; they cannot repair missing meanings by themselves.

A truth value and knowledge of that value are different

We write T for true and F for false. These are truth values, not numerical measurements, confidence scores or explanations. A proof can establish T and a counterexample can establish F, but the definition of a statement does not require that a particular observer already know which value it has.

Suppose a closed box has a fixed finite number of counters at this moment. The claim ‘The box contains exactly twelve counters’ has a definite truth value in that fixed situation, even while you have not counted them. ‘I have not checked’ describes your information; it is not a third truth value in the classical system used here. Opening the box could establish the value without changing the claim’s meaning.

An ambiguous sentence needs clarification before a truth table. An undefined mathematical expression needs its domain repaired before a truth value is assigned. Those problems differ from a well-defined claim whose value is unknown to the reader. Other systems of logic use different assumptions; all the laws and tables in this lesson use defined classical T/F inputs.

A free variable or an excluded input needs attention

The sentence x² < 10, with x a real variable and no value or quantifier supplied, is an open sentence or predicate P(x). It does not yet assert one closed proposition. Substitution gives statements: P(3) is true because 9 < 10, and P(4) is false because 16 is not less than 10. A later lesson will turn predicates into quantified claims about all or some inputs.

A predicate also has a domain on which its expressions are defined. For Q(x): 1/x > 0, use nonzero real inputs. Q(2) is true and Q(−2) is false. Q(0) is excluded because 1/0 is undefined in real arithmetic. It is not a false instance of the same real reciprocal comparison. Restricting the domain is part of stating the problem.

Writing ‘x ≠ 0 and 1/x > 0’ does not, by itself, turn an undefined atomic comparison at x = 0 into a T/F input for an ordinary propositional truth table. One option is to retain the nonzero domain. Another is to define a new total predicate explicitly: give it value F at x = 0, and value Q(x) otherwise. That is a stated extension, not the original reciprocal evaluated at zero. Programming rules that skip evaluating a second test are a separate issue.

Negation reverses a defined truth value

Let p name a fixed statement. Its negation ¬p means ‘It is not the case that p’. By definition, ¬p is F when p is T and T when p is F. The symbol applies to the whole expression immediately grouped after it. If p means a specified real number is greater than 5, its negation means that number is at most 5; ‘less than 5’ would wrongly omit equality.

Double negation returns the original value: when p is T, ¬p is F and ¬(¬p) is T; when p is F, the two reversals return F. These two cases exhaust the defined classical possibilities, so ¬(¬p) is logically equivalent to p for every classical statement p. Negation does not assign a value to an undefined expression.

‘And’ needs both; inclusive ‘or’ permits both

The conjunction p ∧ q is T exactly when both p and q are T. It is F whenever either is F. The inclusive disjunction p ∨ q is T when at least one input is T, including the case where both are T. It is F only when both are F. Mathematical ‘or’ has this inclusive meaning unless an exclusion is stated.

For example, let p mean ‘The identified tile is red’ and q mean ‘The identified tile is triangular’. A red triangular tile satisfies both p ∧ q and p ∨ q. If a rule requires exactly one of these two properties, write exclusive disjunction p ⊕ q, read ‘p or q, but not both’. It is T precisely when the two truth values differ.

The expression (p ∨ q) ∧ ¬(p ∧ q) proves the exclusive rule directly. With neither property, its first part is F. With both, its second part is F. With exactly one, its first part and second part are both T. These four input combinations exhaust p and q, so this expression agrees with p ⊕ q on every valuation.

Conditional and biconditional are defined connectives

The material conditional p → q, read ‘if p, then q’, is F only when p is T and q is F. It is T in the other three cases. A false antecedent therefore makes this truth-functional conditional true regardless of q. This convention records the absence of a case with p true and q false; it does not assert that p happened, caused q or explains q.

Its complete table agrees with ¬p ∨ q: if p is F, ¬p makes the disjunction T; if p is T, the result equals q. This establishes the equivalence for all defined T/F inputs. Mathematical implications about whole domains require further attention to their variables and hypotheses; the later implication lesson develops that use.

The biconditional p ↔ q, read ‘p if and only if q’, is T when its two current values match and F when they differ. In particular F ↔ F is T. This is also the value of (p → q) ∧ (q → p): either mismatched pair falsifies one direction, while either matched pair satisfies both. A biconditional true in one situation does not establish logical equivalence of two formulas in every situation.

Parentheses say which operation acts on which claim

Use parentheses to specify a formula’s structure instead of guessing how an ungrouped sentence should be read. In (p ∨ q) ∧ ¬r, the final operation is ∧, with inputs p ∨ q and ¬r. First evaluate each of those inputs; then apply the conjunction. In p ∨ (q ∧ ¬r), the final operation is ∨. The same letters and connective symbols can therefore give different statements when grouped differently.

For p = T, q = F and r = T, the first formula is (T ∨ F) ∧ F = F. The second is T ∨ (F ∧ F) = T. One such disagreement is enough to show that the formulas are not logically equivalent. We keep mixed binary operations grouped explicitly throughout this lesson; negation applies to the atom or parenthesized formula immediately after ¬.

A valuation assigns one truth value to each distinct atom. Every occurrence of p in a formula uses that same assigned value. Thus p ∧ ¬p cannot use p = T in its first occurrence and p = F in its second. Likewise, if two letters were deliberately defined to mean the same statement, their actual values must agree. An unrestricted truth table for different letters lists all formal assignments, even if a chosen interpretation allows only some rows.

A complete table covers every Boolean valuation

For a formula with n distinct atoms, where n is a positive integer, there are 2ⁿ formal valuations: each atom independently receives T or F. To justify completeness, start with the two rows for one atom. Adding a new atom attaches T and F to each existing row, so the row count doubles and every extended assignment occurs exactly once. Repeating this construction yields all 2ⁿ assignments. Count distinct atoms, not their repeated occurrences.

For three atoms, an orderly list is TTT, TTF, TFT, TFF, FTT, FTF, FFT, FFF. Make a column for each necessary intermediate expression. Evaluate from the innermost expressions to the final connective, using the same row’s atomic values each time. Every connective’s defined rule gives one output for each permitted input, so this recursive procedure determines the formula uniquely on every row.

The table below evaluates (p ∨ q) ∧ ¬r. It is T exactly when r is F and at least one of p and q is T. There are three such rows. The eight rows constitute the whole three-atom Boolean table, not a selection of numerical examples. If extra assumptions constrain the atom meanings, inspect their allowed rows separately and state that restricted scope.

Compare the entire truth pattern

A tautology is T on every formal valuation. A contradiction is F on every valuation. A contingent formula has at least one T row and at least one F row. A formula is satisfiable if at least one valuation makes it T. Thus a tautology is satisfiable, a contingent formula is satisfiable, and a contradiction is not. These are properties of the whole formula, rather than descriptions of only the currently selected row.

For any defined classical p, exactly one of p and ¬p is T. Hence p ∨ ¬p is a tautology and p ∧ ¬p is a contradiction. The atom p itself is contingent as a propositional formula: its T and F assignments are both in its formal table. Substituting a particular known true statement for p does not remove the formula’s formal F valuation from its truth table.

Formulas A and B are logically equivalent, written A ≡ B here, if their values agree on every valuation of the union of their atoms. An atom absent from one formula simply does not affect that formula’s value. Equivalently, the formula A ↔ B is a tautology. Agreement on a single row proves only that row’s biconditional true; disagreement on one row disproves unrestricted equivalence.

For an example of full equivalence, ¬(p ∧ q) is T precisely when the conjunction is F, which happens precisely when p is F or q is F. That is exactly the condition for ¬p ∨ ¬q to be T. The four-row table confirms both directions exhaustively. Similarly ¬(p ∨ q) is T precisely when both inputs are F, exactly the condition for ¬p ∧ ¬q. These two De Morgan laws preserve the grouping of the original negation; negating each atom while keeping the same binary connective generally gives a different result.

Explain what the check actually establishes

A complete truth table proves an equivalence between two fixed finite propositional formulas because the connectives depend only on the atoms’ truth values and every such valuation has been covered. A partial table cannot establish that unrestricted equivalence. A single countervaluation can refute it because the definition requires agreement on every row.

An arithmetic claim over all integers has a different input set. For instance, (n − 2)(n − 4)(n − 6) = 0 is true at n = 2, 4 and 6, but at n = 8 its left side is 6 × 4 × 2 = 48. Those three successes were not an exhaustive table of all integer inputs. Writing the whole arithmetic claim as one atom does not prove that atom true; its internal mathematical content still needs a justification.

When reporting a check, name the formula, the domains, the allowed assignments and the scope of the conclusion. ‘This row is T’, ‘this formula has a T witness’ and ‘the two formulas agree on every formal valuation’ are different results. The next lessons develop quantifiers, implication and proof methods so that claims about whole mathematical domains can be stated and established precisely.

Explore statements, grouped formulas and complete tables

Use the definitions above to separate a closed true-or-false claim from an open sentence, a request or an excluded input. Before evaluating a compound formula, identify its atoms and check that each selected instance has a defined classical truth value. An excluded reciprocal input should retain its domain warning rather than receive a fabricated F value.

Compare the connective rules, follow the intermediate values of explicitly grouped formulas, and then inspect every valuation required by the formula’s distinct atoms. Keep a selected row’s result separate from classifications such as tautology or contingency. When comparing two formulas, a differing row is a countervaluation; an equivalence verdict requires the entire table over their combined atom set.

A finite interactive choice illustrates the selected formula and interpretation. The completeness argument for 2ⁿ Boolean valuations and the connective definitions explain why a genuinely exhaustive table establishes the stated propositional result. They do not turn a few arithmetic inputs into a proof about all numbers.

Worked examples

Try it yourself

1. Classify 4² = 16, 4² = 15, ‘Is 4² equal to 16?’, ‘Square 4’, y² = 16 with real y unassigned, and 1/0 > 0 in real arithmetic. Evaluate the open sentence at y = −4 and y = 3.

   Show the answer and reasoning

   The two closed equalities are T and F. The question and instruction are not propositions. The unassigned y sentence is open; its instances at −4 and 3 are T and F. The reciprocal comparison is undefined in real arithmetic and is not a defined T/F atom.

   A claim can be false and remain a statement, whereas a request or question does not assert a claim. Substitution gives (−4)² = 16 and 3² = 9 ≠ 16. There is no real value of 1/0 to compare with zero, so that expression needs a corrected domain or a separately stated total definition before entering a classical truth table.

2. Let p = T, q = F and r = T. Find ¬p, p ∧ q, p ∨ q, p ⊕ r, q → r and q ↔ r. Give each result from its definition.

   Show the answer and reasoning

   In the stated order: F, F, T, F, T, F.

   Negation reverses p. The conjunction has a false q, while the inclusive disjunction has a true p. Exclusive disjunction is false when p and r are both true. The conditional has a false antecedent and is therefore true in its truth table. The biconditional is false because q and r differ.

3. Let p mean ‘The identified card has a star’ and q mean ‘The card has a border’. Express the rule ‘exactly one of a star and a border’ using only ¬, ∧ and ∨. State its values on TT, TF, FT and FF.

   Show the answer and reasoning

   (p ∨ q) ∧ ¬(p ∧ q), with values F, T, T, F in the requested order.

   The first part requires at least one property, and the negated conjunction rules out both. TT fails the second part, FF fails the first, and each single-property row satisfies both. This exhausts all four formal valuations and proves agreement with exclusive disjunction.

4. Build the complete table for K = ¬(p ∨ q) ∨ p on the rows TT, TF, FT and FF. Classify K and decide whether it is equivalent to p ∨ ¬q.

   Show the answer and reasoning

   K has the column T, T, F, T. It is contingent and satisfiable, and K ≡ (p ∨ ¬q).

   On TT and TF, p makes the final disjunction true. On FT, the inner disjunction is T and p is F, so K is F. On FF, the inner disjunction is F and its negation makes K true. The column of p ∨ ¬q is also T, T, F, T. All four valuations of the combined atom set have been covered, so equivalence is proved. TT is a satisfying witness and FT is a falsifying witness.

5. For a fixed real x, let p mean x > 2 and q mean x < 5. State ¬(p ∧ q) without a negated group. Evaluate it at x = 2, 3 and 5, and distinguish this full logical law from just checking the three numbers.

   Show the answer and reasoning

   The equivalent condition is x ≤ 2 or x ≥ 5. Its values at 2, 3 and 5 are T, F and T.

   ¬p includes the equality boundary x = 2, and ¬q includes x = 5. De Morgan’s complete two-atom law gives ¬p ∨ ¬q for every defined valuation, hence for every real x under this interpretation. At the three requested numbers, the original between-condition is respectively F, T and F, so its negation has the stated values. The law establishes the general equivalence; the three substitutions only illustrate it.

6. At p = q = F, evaluate p → q and p ↔ q. Are the formulas p → q and p ↔ q logically equivalent? Then classify p ∨ ¬p and p ∧ ¬p without assigning the repeated p different values.

   Show the answer and reasoning

   Both requested current values are T, but the conditional and biconditional formulas are not equivalent. The disjunction is a tautology and the conjunction is a contradiction.

   A false antecedent makes the conditional T, and matching false inputs make the biconditional T. At p = F, q = T, however, the conditional is T while the biconditional is F; that row refutes equivalence. For each of the only two values of p, exactly one of p and ¬p is T. Their disjunction is therefore always T and their conjunction always F, with one consistent assignment reused at each occurrence.

7. A formula uses p twice and q once. How many rows does its unrestricted truth table need? A learner also checks (n − 2)(n − 4)(n − 6) = 0 at n = 2, 4 and 6 and claims it holds for every integer. Compare these checks and give a counterexample to the integer claim.

   Show the answer and reasoning

   Two distinct atoms need four rows: TT, TF, FT and FF. The three integer samples do not cover all integers; n = 8 gives 48 rather than 0 and refutes the general arithmetic claim.

   Occurrences are not independent atoms: the second p must use the first p’s assigned value. The four-row Boolean table exhausts every formal assignment to p and q. The arithmetic domain is the whole integer set, and the three tested inputs only make one factor zero each time. At 8, the product is (8 − 2)(8 − 4)(8 − 6) = 6 × 4 × 2 = 48. An observed pattern of successes cannot replace the missing domain-wide argument.

What to carry forward

• A classical proposition is a precise claim with one defined T/F value. False statements remain propositions; questions, requests and standalone numerical expressions serve different roles.
• Not knowing a well-defined claim’s truth value is different from ambiguity, a free variable or an undefined expression. Retain original domains, including reciprocal exclusions.
• Negation reverses a value, conjunction needs both true inputs, and inclusive disjunction permits both. An exactly-one rule requires explicit exclusive disjunction or its grouped equivalent.
• The material conditional is false only at T → F; a biconditional is true when current values match. Neither a true conditional nor a true current biconditional establishes causation or unrestricted equivalence.
• Parentheses determine formula structure, and every occurrence of one atom shares its assigned value. A formula with n distinct atoms has 2ⁿ formal valuations.
• Tautology, contradiction, contingency and satisfiability describe the complete truth pattern. Logical equivalence requires agreement on every valuation of the combined atom set; one countervaluation can refute it.
• A genuinely complete finite Boolean table can prove a propositional result. A few numerical samples do not establish a claim over an entire infinite domain; explain the hypotheses and scope of every check.