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What's the difference between a negation and a contrapositive?

I keep mixing them up, but it seems that a contrapositive is a negation where the terms' order is changed and where there is an imply sign.

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Could you explain how this question differs from your other question – Trevor Wilson Nov 1 '12 at 22:39
@Trevor: negation is not the same as contradiction, though the conjunction ($\land$) of a statement and its negation constitutes a contradiction. – amWhy Nov 1 '12 at 22:42
@amWhy It is true that the two words mean different things in logic. However, both questions seem a bit vague to me, and it is not clear that the two words are used to mean different things in the questions. Note that "negation" and "contradiction" are interchangeable in many English sentences. – Trevor Wilson Nov 1 '12 at 22:48
up vote 11 down vote accepted

Consider the statement: $$P \implies Q.\tag{1}$$ Then, as you know, the contrapositive of (1) is $$\lnot Q \implies \lnot P.\tag{2}$$

Now, note that statements (1) and (2) are equivalent. I.e., $$\;P \implies Q\equiv \lnot Q \implies \lnot P.$$

One additional observation: Note that the contrapositive of statement $(2)$ is $(1)$, so a contrapositive need not contain any negation symbol $(\lnot)$ at all!

Negating statement (1) (and hence negating (2)) would be a statement to the effect that it is NOT the case that $P \implies Q$. I.e. the negation of statement (1) is given by $$\lnot (P \implies Q)\tag{3}$$ and since $(P \implies Q) \equiv (\lnot P \lor Q)$, we can write (3) as follows: $$\lnot(\lnot P \lor Q)\tag{4}$$ By DeMorgan's, (4) is equivalent to: $$\lnot(\lnot P) \land \lnot Q\tag{5}$$ which yields $$(P \land \lnot Q)\tag{6}$$ So (3), (4), (5) and (6) are all equivalent expressions, each negating statement (1).

Put another way, the contrapositve of a statement is equivalent to the statement [both a statement and its contrapositive have the same truth-value], while the negation of the statement negates or reverses the truth-value of the original statement.

One last comment: When one speaks of the contrapositive of a statement, one is usually speaking about the contrapositive of an implication, whereas one can negate any logical statement whatsoever simply by enclosing the original statement in parentheses and negating it, as in statement (3) above.

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$\lnot \lnot P \implies P$ only under the assumption of excluded middle, not in general, correct? – alancalvitti Nov 2 '12 at 3:54
Post edit, referring to (5) --> (6) – alancalvitti Nov 2 '12 at 14:50
The $\implies$ direction is the problem. Could you be more explicit on "accepted more often than rejected"? What accounts for the "more often" is that most mathematicians work with classical or non-constructive systems. It is not accepted in general, see eg Goldblatt Topoi. Similarly, mathematicians prior to Gauss and Lobachevsky considered Euclidean geometry to be natural. – alancalvitti Nov 2 '12 at 22:34
Hey man, either $\lnot \lnot P \implies P$ or it doesn't. There's no in-between - oh wait, maybe there is. As justice Scalia wrote, most aspects of the real world are graded. – alancalvitti Nov 3 '12 at 0:35
What about the rest of the comment? How to write that in propositional logic? – alancalvitti Nov 3 '12 at 0:58

When you negate a true statement you get a false statement, but the contrapositive of a true statement is still a true statement, formally if $P \Rightarrow Q$ is your statement then $\sim Q \Rightarrow \sim P$ is the contrapositive, where $\sim$ denotes logical NOT and $\Rightarrow$ denotes logical implication. You could use the truth table to convince yourself that they are equivalent, i.e., $(P \Rightarrow Q) \Longleftrightarrow (\sim Q \Rightarrow \sim P)$. Consider the following statement: If today is Monday then we are tired. This is the same as: If we are not tired then today is not Monday.

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