Truth table for the conditional
An if-then statement makes one promise about two things. When exactly does it break that promise?
Where you are 3 in Truth tables
When you reach for it
“If n is divisible by 6, then n is even” is a single statement built out of two smaller ones, joined by if...then. With P for the first part and Q for the second, it is written P \to Q: a conditional. Like and and or, it has a truth value that depends only on the truth values of P and Q, and that dependence can be written down once, as a table.
Reach for the table whenever you need to decide, from the truth of P and the truth of Q alone, whether P \to Q holds; whenever a conditional sits inside a longer statement you are building a bigger table for; and whenever you want to see, in black and white, why a conditional with a false hypothesis still counts as true, which is the one part of this table that surprises almost everyone the first time.
The procedure
Two letters give four rows, in the usual order. P is true for the top half and false for the bottom half, and Q alternates within each half: TT, TF, FT, FF. Here is the finished table; the paragraphs after it say where each entry comes from.
| Row | P | Q | P \to Q |
|---|---|---|---|
| 1 | T | T | T |
| 2 | T | F | F |
| 3 | F | T | T |
| 4 | F | F | T |
Rows 1 and 2 come straight from the promise. “If P, then Q” promises that Q holds whenever P does. In row 1, P holds and Q holds, exactly as promised, so the statement is true. In row 2, P holds and Q fails: the promise is broken, so the statement is false.
Rows 3 and 4 are a choice, and the choice is forced. When P is false the promise makes no demand, and everyday language gives no clear verdict. Mathematics has to pick a value, and it picks T, because of how if-then statements are actually used. Take the sentence from the start, read as a claim about every whole number n: “if n is divisible by 6, then n is even.” Everyone accepts it as true, since no multiple of 6 is odd. A claim about every n is true only if it is true for each n in turn. At n = 6 both parts are true, row 1. At n = 2 the hypothesis is false and the conclusion true, row 3. At n = 3 both are false, row 4. If row 3 or row 4 held an F, the sentence would fail at n = 2 or at n = 3, and one of the most ordinary true statements in mathematics would come out false. So a conditional with a false hypothesis is true. This is vacuous truth: true because the statement demands nothing in that row, not because its two halves are connected. Any other filling of rows 3 and 4 would also make if-then a copy of something we can already say without it, as Exercise 17 shows for one of those fillings.
The column has three T’s and one F. The table for or has that shape too, with its F in row 4; the conditional has its F in row 2. That is no accident: the conditional written as an or shows that P \to Q says the same thing as \neg P \lor Q.
Worked examples
A promise about a library book
A library sign reads: “If a book is returned late, a fine is charged.” Let P be “the book is returned late” and Q be “a fine is charged.” Make the truth table for the sign’s promise, P \to Q, and say in words what each row means.
Solution| Row | P (returned late) | Q (fine charged) | P \to Q |
|---|---|---|---|
| 1 | T | T | T |
| 2 | T | F | F |
| 3 | F | T | T |
| 4 | F | F | T |
Row 1: late, and fined. The sign told the truth. Row 2: late, and not fined. The sign lied, the only row that does. Row 3: not late, but fined anyway, perhaps for a torn page. The sign never said anything about books returned on time, so nothing it said was false. Row 4: not late, not fined, exactly as expected outside the case the sign covers. Only row 2 contradicts the sign.
When there is nothing to promise about
Decide true or false, and explain: (a) “If 12 is a prime number, then 2 + 2 = 5.” (b) “If 12 is a prime number, then 6 is even.”
Solution12 = 2 \times 6, so 12 is not prime: P is false in both statements. A false P puts each statement in row 3 or row 4, and both of those rows are true, whatever Q says. In (a) the conclusion 2 + 2 = 5 is false, so (a) sits in row 4; in (b) the conclusion is true, so (b) sits in row 3. So both statements are true, even though (a)’s conclusion is false and neither conclusion has anything to do with 12.
This feels odd only because everyday “if...then” usually suggests a connection between its two halves. The table asks a narrower question: is this the case where the hypothesis holds and the conclusion fails? With P false, it is not, so the conditional is true.
A promise stated as a negative
A park’s sign reads: “If it is not raining, the pool is open.” Let P be “it is raining” and Q be “the pool is open,” so the sign says \neg P \to Q. Make its table, with a helper column for \neg P.
SolutionFill \neg P by flipping the P column, then apply the conditional rule to the \neg P column and the Q column, ignoring P itself:
| Row | P | Q | \neg P | \neg P \to Q |
|---|---|---|---|---|
| 1 | T | T | F | T |
| 2 | T | F | F | T |
| 3 | F | T | T | T |
| 4 | F | F | T | F |
The sign is broken only in row 4: it is not raining, and the pool is closed anyway. Rows 1 and 2 have \neg P false (it is raining), a case the sign never addresses, so both are true no matter what the pool is doing. Moving the negation changes which row is false, not how many rows are.
Reading the table backward
Suppose P \to Q is true, and separately you learn P is true. What follows about Q? Suppose instead all you learn is that P \to Q is false. What follows about P and about Q?
SolutionKnowing P \to Q is true rules out row 2, the only false one. Knowing P is true narrows the possibilities to row 1 or row 2. Only row 1 survives both facts, so Q must be true. This is the table’s version of modus ponens: from P \to Q and P, the conclusion Q follows, because it is the only row consistent with both.
For the second question, P \to Q false is row 2 and nothing else, so P is true and Q is false, pinned down completely by one fact.
Where the brackets go
Make one table for P \to (Q \to R) and (P \to Q) \to R, and find the rows where they differ.
SolutionThree letters give eight rows: P true for the top four, Q alternating in pairs, R alternating every row. Each statement needs its inner arrow worked out first, so there is a helper column for Q \to R and one for P \to Q.
| Row | P | Q | R | Q \to R | P \to (Q \to R) | P \to Q | (P \to Q) \to R |
|---|---|---|---|---|---|---|---|
| 1 | T | T | T | T | T | T | T |
| 2 | T | T | F | F | F | T | F |
| 3 | T | F | T | T | T | F | T |
| 4 | T | F | F | T | T | F | T |
| 5 | F | T | T | T | T | T | T |
| 6 | F | T | F | F | T | T | F |
| 7 | F | F | T | T | T | T | T |
| 8 | F | F | F | T | T | T | F |
Q \to R is false only where Q is true and R is false, rows 2 and 6. The outer arrow of P \to (Q \to R) then breaks only where P is true and Q \to R is false, which is row 2 alone; in row 6, P is false, so it is vacuously true. (P \to Q) \to R breaks wherever P \to Q is true and R is false. R is false in rows 2, 4, 6 and 8, and P \to Q is true in all of those except row 4, so it is false in rows 2, 6 and 8.
The two statements differ in rows 6 and 8. Take row 8, everything false: the first statement is true because its hypothesis P is false, while the second has a true hypothesis, P \to Q (true vacuously), and a false conclusion. So the brackets matter. With and and or a chain of three letters means the same thing however it is grouped; with the arrow it does not, which is why a chain of conditionals is always written with its brackets.
A conditional and its converse are different promises
Is “if you live in Paris, you live in France” the same claim as “if you live in France, you live in Paris”? Compare tables for P \to Q and its converse Q \to P.
SolutionLet P be “you live in Paris” and Q be “you live in France.”
| Row | P | Q | P \to Q | Q \to P |
|---|---|---|---|---|
| 1 | T | T | T | T |
| 2 | T | F | F | T |
| 3 | F | T | T | F |
| 4 | F | F | T | T |
The columns disagree in rows 2 and 3. Row 3 is someone living in Lyon: not in Paris, but in France. P \to Q is true there, and Q \to P is false. So they are different claims; how a conditional relates to its converse and to its other rewordings is the subject of the contrapositive.
Boundaries
Exactly one row is false, and it is not a test of relatedness. P \to Q asks only whether P true and Q false is the case. It does not ask whether P and Q are about the same subject, or whether one causes the other, which is why Example 2’s unrelated halves still make true statements.
The table judges one instance, not a claim about every n. It needs each letter to have a settled truth value, so it says nothing about “x is divisible by 6” until x is a particular number; until then that phrase is neither true nor false. A claim like “for every n, if n is divisible by 6 then n is even” is true only if every one of its instances lands in a true row, and showing that for infinitely many n takes a proof, not a table.
True or false only, never earlier or later. “If P then Q” carries no claim that P happens first or brings Q about. The table treats P and Q as two facts about the same moment, and whichever one is easier to check can be checked first.
Where this goes next
The four-row table is one of the small pieces a larger table is built from, the way the promise for “and” and the promise for “or” are. Paired with its own converse it becomes the biconditional, true exactly when both directions hold, and rewritten as \neg Q \to \neg P it becomes the contrapositive, which always agrees with the original, unlike the converse. Rewritten with no arrow at all, it becomes an or statement, \neg P \lor Q.
Reading a row backward, as in Example 4, is the whole content of modus ponens, and the same row-by-row reading is what checks whether an argument is valid. The language for talking about which half of the arrow is doing the work, sufficient or necessary, is developed in necessary and sufficient conditions.
Exercises
Reading the table
- In the row where P is true and Q is false, what is the value of P \to Q?
- P \to Q is known to be true, and Q is known to be false. What must P be?
- P \to Q is known to be false. What are the truth values of P and of Q?
- P \to Q is known to be true, and P is known to be true. What must Q be?
- In how many rows of the table for P \to Q is Q true while the whole statement is also true?
Building tables
- Make the truth table for Q \to P, the converse of P \to Q. In which row is it false?
- Make the truth table for P \to (Q \to P), with a helper column for Q \to P. In which rows is it false?
- Make the truth table for (P \land Q) \to R. In which row is it false?
- Make the truth table for P \to \neg Q, with a helper column for \neg Q. In which row is it false?
- A statement about P and Q is false only in the row where P is false and Q is true. Write it as a conditional.
Vacuous or not
- True or false, and why: “If 9 is a prime number, then 9 is even.”
- True or false, and why: “If 2 + 2 = 4, then 9 is a prime number.”
- True or false, and why: “If 15 is a prime number, then 15 is odd.”
Reasoning
- Explain why the table for P \to Q needs exactly four rows, no more and no fewer.
- Give your own example of a true if-then statement whose hypothesis is false, and explain why it counts as true.
- Priya says: “If a statement’s hypothesis is false, the statement is meaningless, not true or false.” Explain what is wrong with her claim.
- Suppose row 3 of the table held F instead of T. Which connective would the column then match, and what would go wrong with “for every whole number n, if n is divisible by 6, then n is even”?
- Explain why P \to Q is true in every row where Q is true, whatever P is.
- Show that whenever P \to Q and Q \to R are both true in a row, P \to R is true in that row too.
Answers
- F. This is row 2, the one row a conditional is false.
- P is false. Q false happens in rows 2 and 4; row 2 has the whole statement false, so the true one must be row 4, where P is false.
- P is true and Q is false. That is row 2, the only false row.
- Q is true. P true happens in rows 1 and 2; row 2 is false, so the true one is row 1.
- Two, rows 1 and 3. Both have Q true, and in both the conditional is also true.
- False only in row 3. Swapping the roles, Q \to P is T, T, F, T: it fails exactly when Q is true and P is false.
- It is never false: its column is T, T, T, T. The helper column Q \to P is T, T, F, T. In rows 1 and 2, P is true and so is Q \to P, so the outer arrow holds; in rows 3 and 4, P is false, so the outer arrow is vacuously true.
- False only in row 2 (T, T, F): P and Q both true and R false. P \land Q is true only in rows 1 and 2, and among those R fails only in row 2.
- False only in row 1: P true and Q true, so \neg Q is false and the promise is broken. The \neg Q column is F, T, F, T, and P \to \neg Q is F, T, T, T.
- One answer is Q \to P. Its column is T, T, F, T, false only where Q is true and P is false, which is the row described.
- True. 9 = 3 \times 3 is not prime, so the hypothesis is false, and a false hypothesis makes the whole statement true no matter what the conclusion says.
- False. 2 + 2 = 4 is true, and 9 is not prime, so this is the one row that breaks a conditional: hypothesis true, conclusion false.
- True. 15 = 3 \times 5 is not prime, so the hypothesis is false and the statement is true vacuously, even though its conclusion, that 15 is odd, happens to be true as well.
- P has two possible values and, independently, Q has two, giving 2 \times 2 = 4 combinations. Leaving one out would drop a case the rule has to cover; adding a row would repeat one already listed.
- One example: “If 100 is a prime number, then Paris is the capital of Germany.” 100 is not prime, so the hypothesis is false, which by the table makes the whole statement true, whatever the conclusion says and however unrelated it is.
- Her rule would break statements everyone accepts. “For every whole number n, if n is divisible by 6, then n is even” is true, and a claim about every n is true only when each instance is true. The instance n = 3 has a false hypothesis; if it were meaningless, the general claim would have a meaningless instance and could not be called true. Giving a false-hypothesis conditional the value T is exactly what keeps such claims true.
- The column would be T, F, F, T, the column of the biconditional P \leftrightarrow Q, true exactly when P and Q agree. The sentence about n would then fail at n = 2: 2 is not divisible by 6 but is even, which is row 3. A true claim would come out false, and if-then would say nothing that “if and only if” does not.
- P \to Q is false only when Q is false (row 2). In a row where Q is true, that case cannot occur, so the conditional is true: rows 1 and 3, one with P true and one with P false.
- Suppose, for contradiction, that some row has P \to Q and Q \to R true but P \to R false. Then P is true and R is false in that row. P true and P \to Q true force Q true, by Example 4’s reading; Q true and R false would then make Q \to R false, contradicting the assumption. So no such row exists, and P \to R is true wherever P \to Q and Q \to R both are.