Four slips account for many wrong answers when implications are evaluated, rewritten or used in proofs. 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.
Making P → Q false when P is false
When the antecedent is false, the conditional is felt to have "not happened", and it is marked F. The column that results is not the column of P → Q.
Wrong working: take the rows (P, Q) = (T, T), (T, F), (F, T), (F, F). Mark P → Q as T and F in the first two rows, then F in the row (F, T) because P did not happen, and T in the row (F, F) because both parts agree. The column T, F, F, T is the column of the biconditional P ↔ Q.
Right working: P → Q ≡ ¬P ∨ Q. In the row (F, T), ¬P is T, so ¬P ∨ Q is T. In the row (F, F), ¬P is again T, so ¬P ∨ Q is T. The column is T, F, T, T, with a single F at P = T, Q = F.
Why it happens: in everyday speech "if" is often heard as "if and only if". "If you finish your homework, you can go out" suggests that otherwise you cannot, and that extra promise is read into P → Q.
How to check: a conditional is F only when the hypothesis holds and the conclusion fails. Count the F entries in the column: P → Q has exactly one.
Negating Q instead of P
When P → Q is turned into a disjunction, the negation is put on the consequent instead of the antecedent, giving P ∨ ¬Q. That formula has a different truth table.
Wrong working: P → Q ≡ P ∨ ¬Q. At P = F, Q = T this gives F ∨ F, which is F.
Right working: P → Q ≡ ¬P ∨ Q. At P = F, Q = T this gives T ∨ T, which is T, matching P → Q, which is T because its antecedent is false. The formula P ∨ ¬Q is ¬Q ∨ P, the disjunction form of the converse Q → P.
Why it happens: the pattern "one part negated, joined by or" is remembered without the detail of which part carries the negation.
How to check: the disjunction must be F only at P = T, Q = F. There ¬P ∨ Q gives F ∨ F, which is F, while P ∨ ¬Q gives T ∨ T, which is T, so P ∨ ¬Q fails the test.
Reading P → ¬Q as the negation of P → Q
To say that "if P then Q" is false, the negation is moved onto the consequent and the arrow is kept. The result, P → ¬Q, is another implication, and like P → Q it is true in every row where P is false.
Wrong working: ¬(P → Q) ≡ P → ¬Q, so the denial of "if it rains, the ground is wet" would be "if it rains, the ground is not wet".
Right working: ¬(P → Q) ≡ ¬(¬P ∨ Q) ≡ ¬¬P ∧ ¬Q ≡ P ∧ ¬Q. At P = F, Q = T, the formula P → Q is T, so its negation must be F. P ∧ ¬Q is F there, but P → ¬Q is F → F, which is T. The denial is "it rains and the ground is not wet".
Why it happens: in speech, "it is not true that if it rains the ground is wet" sounds like a claim about what happens when it rains, so the "if" is kept.
How to check: a formula and its negation have opposite values in every row. P → Q and P → ¬Q are both T in the two rows with P = F, so neither is the negation of the other.
Negating without reversing
To form the contrapositive of P → Q, both parts are negated but left in their original order, giving ¬P → ¬Q. That is the inverse, which is not equivalent to P → Q.
Wrong working: with P = "n² is even" and Q = "n is even", the contrapositive is written as "if n² is odd then n is odd", which is ¬P → ¬Q. Proving it proves the inverse, and so the converse "if n is even then n² is even", not the original statement.
Right working: negate and swap: ¬Q → ¬P, "if n is odd then n² is odd". For (P, Q) = (T, T), (T, F), (F, T), (F, F), the formula ¬Q → ¬P takes the values T, F, T, T, the same as P → Q. The inverse ¬P → ¬Q takes T, T, F, T: at P = F, Q = T it is T → F, which is F, while P → Q is T.
Why it happens: "negate" is remembered and "reverse" is forgotten. The inverse also looks closer to the original, with the parts still in their familiar places.
How to check: the contrapositive must be F in the same row as the original, P = T, Q = F. There ¬Q → ¬P is T → F, which is F, as required; ¬P → ¬Q is F → T, which is T, so it is not the contrapositive.