Five slips account for many wrong answers when tautologies are checked or used as inference rules. Each is marked where it happens in the sections above; here each is worked through in full, with the wrong and the right working side by side.
Checking only some rows
A formula is evaluated on a few rows, comes out T in each of them, and is declared a tautology. A single F row anywhere in the table is enough to show that it is not one.
Wrong working: for P → Q, take P = T, Q = T: T → T is T. Take P = F, Q = F: F → F is T. Both rows give T, so P → Q is a tautology.
Right working: two variables give four rows. For (P, Q) equal to (T, T), (T, F), (F, T), (F, F), the formula P → Q takes the values T, F, T, T. The row (T, F) gives F, so P → Q is not a tautology. By contrast, (P ∧ Q) → P is T in all four rows.
Why it happens: the rows where the variables share a value are the easiest to evaluate, and a run of T values feels like evidence. A truth table settles the question only when it is complete.
How to check: count the rows first, 2ⁿ for n variables, and fill in every one. To refute a claimed tautology, search directly for an F row; for an implication, that is a row where the antecedent is T and the consequent is F.
Chaining through a shared consequent
Two implications that end in the same proposition are joined as if they formed a chain. P → Q and R → Q both lead into Q, so nothing in them leads from P to R.
Wrong working: "If it rains, the ground is wet. If the sprinkler runs, the ground is wet. So if it rains, the sprinkler runs." In symbols, ((P → Q) ∧ (R → Q)) → (P → R).
Right working: at P = T, Q = T, R = F, the premise P → Q is T → T, which is T, and R → Q is F → T, which is T, so the antecedent is T. The conclusion P → R is T → F, which is F. The formula is F in that row and is not a tautology. Hypothetical syllogism needs Q → R, with Q as the antecedent of the second link; then ((P → Q) ∧ (Q → R)) → (P → R) is T in all eight rows.
Why it happens: both premises mention Q, and the shared letter looks like the link between them. In syllogistic logic the same pattern is known as the fallacy of the undistributed middle.
How to check: write the links one under the other; the consequent of each line must be the antecedent of the next. Or test a row with P = T and R = F, where the conclusion is F: if both premises can be T there, the chain fails.
Affirming the consequent
From a rule P → Q and the fact Q, the conclusion P is drawn. Modus ponens works only from the antecedent; knowing the consequent says nothing about P.
Wrong working: "If it rains, the ground is wet. The ground is wet. So it rained." In symbols, ((P → Q) ∧ Q) → P.
Right working: at P = F, Q = T, the premise P → Q is F → T, which is T, and Q is T, so the antecedent is T, while the conclusion P is F. The formula is F in that row, so it is not a tautology: the ground can be wet from a sprinkler. Modus ponens, ((P → Q) ∧ P) → Q, is T in all four rows.
Why it happens: both arguments combine an implication with one of its parts, and the rule is remembered as "the implication plus one part gives the other part".
How to check: look for a row where the premises are T and the conclusion is F. For a conclusion P, try P = F: with Q = T both premises hold, so the argument is invalid.
Affirming a disjunct
From P ∨ Q and P, the conclusion ¬Q is drawn, as if "or" meant "one but not both".
Wrong working: "The light is on or the fan is on. The light is on. So the fan is off." In symbols, ((P ∨ Q) ∧ P) → ¬Q.
Right working: at P = T, Q = T, the premise P ∨ Q is T and P is T, so the antecedent is T, while the conclusion ¬Q is F. The formula is F in that row. The connective ∨ is inclusive, so P ∨ Q allows both disjuncts to be true. Disjunctive syllogism uses a false disjunct instead: ((P ∨ Q) ∧ ¬P) → Q is T in all four rows.
Why it happens: everyday "or" often means one or the other but not both, as in "tea or coffee", so knowing one alternative seems to rule out the other.
How to check: test the row where both disjuncts are T. If the conclusion fails there, the argument relied on an exclusive or.
Proving only one direction
To prove P ↔ Q, only the forward direction P → Q is proved, and the biconditional is claimed.
Wrong working: from "if n is divisible by 4 then n is even", conclude "n is divisible by 4 if and only if n is even".
Right working: at P = F, Q = T (for example n = 6, which is even but not divisible by 4), the formula P → Q is F → T, which is T, while P ↔ Q is F. The backward direction Q → P is T → F, which is F in that row, and it is the part that was never proved. Only P → Q and Q → P together give P ↔ Q, since (P → Q) ∧ (Q → P) has the same truth table as P ↔ Q.
Why it happens: the forward direction is often the one stated first, or the easier one, and once it is done the proof feels complete.
How to check: before writing "if and only if", make sure the proof has two parts, one for each arrow. A quick test: look for a case where Q is true and P is false, such as n = 6 in the example.