A conditional is false in exactly one case: a true hypothesis with a false conclusion.
Choose the false case
For which combination is p → q false?
A conditional promises that q follows from p. The only way to break that promise is for p to happen and q not to: p true, q false. In both cases where p is false the promise was never tested, so it counts as unbroken and the conditional is true.
n simple propositions give 2 raised to n rows. Two letters, four rows.
| p | q | p ∧ q | p ∨ q | p → q |
|---|---|---|---|---|
| T | T | T | T | T |
| T | F | F | T | F |
| F | T | F | T | T |
| F | F | F | F | T |
Read the columns as personalities. And is strict: one T only. Or is generous: one F only. The conditional is also generous, and its one F is the row worth memorising.
"If 7 is an even number, then Baguio is a beach town" is a TRUE statement in logic. That feels wrong until you see that the promise was never put to the test. Only a broken promise makes a conditional false.
| p | q | ¬q | ¬p | ¬q → ¬p | p → q |
|---|---|---|---|---|---|
| T | T | F | F | T | T |
| T | F | T | F | F | F |
| F | T | F | T | T | T |
| F | F | T | T | T | T |
The last two columns are identical, so ¬q → ¬p (the contrapositive) and p → q are logically equivalent. Two statements are equivalent exactly when their final columns match row for row.
One more pair worth knowing: ¬p ∨ q. Row by row it gives F ∨ T = T, F ∨ F = F, T ∨ T = T, T ∨ F = T — the column T, F, T, T, the same as p → q. So a conditional can always be rewritten as "not the hypothesis, or the conclusion".
"It will rain or it will not rain" says nothing about the weather, and that is exactly what a tautology is: true whatever the facts, and therefore carrying no information.
The biconditional column: p ↔ q
p = T, q = T gives T; p = T, q = F gives F
true when the two match
p = F, q = T gives F; p = F, q = F gives T
both false also match
p ↔ q is the "same truth value" test, so both-false gives T. Reading it as "both true" is the slip.
One row of eight: when is (p ∧ q) → r false?
the conditional fails only as T → F
its one false case
p ∧ q = T needs p = T and q = T
the hypothesis true
r = F
the conclusion false
false only in the row T, T, F
true in the other 7
With three letters there are 8 rows, but we need not write them all: find the one case that breaks the conditional.
A hidden tautology: test (p ∧ q) → p
T T: T → T = T
p ∧ q true, p true
T F: F → T = T
false hypothesis
F T: F → F = T
false hypothesis
F F: F → F = T
false hypothesis
T in every row, so it is a tautology: if both hold, then certainly the first one holds.
2³ = 8 rows
every combination, once
p: TTTTFFFF
four, then four
q: TTFFTTFF
two, then two
r: TFTFTFTF
alternating
Halving the block length each time guarantees you list every combination exactly once. Inventing the rows freehand is how a row goes missing, and a missing row can turn a contingency into an apparent tautology.
Periodical exams ask for the number of rows, a completed truth table, the one false row of a conditional, or whether a statement is a tautology, contradiction or contingency; entrance tests like the conditional with a false hypothesis and spotting equivalent statements. List the rows by halving the blocks, then build one column at a time. For a 10-second check, count your rows (2 raised to the number of letters) and confirm a conditional's column has F only where T leads to F.
Classify the statement
What kind of statement is p ∧ ¬p?
p and ¬p can never both be true, so the conjunction is false in every row — a contradiction. A is p ∨ ¬p, which is true in every row. The two look alike on the page and behave as opposites.
Hint: try p true, then p false. If the result is the same both times, it is a tautology or a contradiction depending on which.
