# Modus tollens

In propositional logic, modus tollens[1][2][3][4] (or modus tollendo tollens and also denying the consequent)[5] (Latin for "the way that denies by denying")[6] is a valid argument form and a rule of inference.

The first to explicitly state the argument form modus tollens were the Stoics.[7]

The inference rule modus tollens, also known as the law of contrapositive, validates the inference from $P$ implies $Q$ and the contradictory of $Q$, to the contradictory of $P$.

The modus tollens rule can be stated formally as:

$\frac{P \to Q, \neg Q}{\therefore \neg P}$

where $P \to Q$ stands for "P implies Q", $\neg Q$ stands for "it is not the case that Q" (or in brief "not Q"). Then, whenever "$P \to Q$" and "$\neg Q$" each appear by themselves as a line of a proof, "$\neg P$" can validly be placed on a subsequent line. The history of the inference rule modus tollens goes back to antiquity.[8]

Modus tollens is closely related to modus ponens. There are two similar, but invalid, forms of argument: affirming the consequent and denying the antecedent.

## Formal notation

The modus tollens rule may be written in sequent notation:

$P\to Q, \neg Q \vdash \neg P$

where $\vdash$ is a metalogical symbol meaning that $\neg P$ is a syntactic consequence of $P \to Q$ and $\neg Q$ in some logical system;

or as the statement of a functional tautology or theorem of propositional logic:

$((P \to Q) \and \neg Q) \to \neg P$

where $P$ and $Q$ are propositions expressed in some logical system;

or including assumptions:

$\frac{\Gamma \vdash P\to Q ~~~ \Gamma \vdash \neg Q}{\Gamma \vdash \neg P}$

though since the rule does not change the set of assumptions, this is not strictly necessary.

More complex rewritings involving modus tollens are often seen, for instance in set theory:

$P\subseteq Q$
$x\notin Q$
$\therefore x\notin P$

("P is a subset of Q. x is not in Q. Therefore, x is not in P.")

Also in first-order predicate logic:

$\forall x:~P(x) \to Q(x)$
$\exists x:~\neg Q(x)$
$\therefore \exists x:~\neg P(x)$

("For all x if x is P then x is Q. There exists some x that is not Q. Therefore, there exists some x that is not P.")

Strictly speaking these are not instances of modus tollens, but they may be derived using modus tollens using a few extra steps.

## Explanation

The argument has two premises. The first premise is a conditional or "if-then" statement, for example that if P then Q. The second premise is that it is not the case that Q . From these two premises, it can be logically concluded that it is not the case that P.

Consider an example:

If the watch-dog detects an intruder, the watch-dog will bark.
The watch-dog did not bark
Therefore, no intruder was detected by the watch-dog.

Supposing that the premises are both true (the dog will bark if it detects an intruder, and does indeed not bark), it follows that no intruder has been detected. This is a valid argument since it is not possible for the conclusion to be false if the premises are true. (It is conceivable that there may have been an intruder that the dog did not detect, but that does not invalidate the argument; the first premise is "if the watch-dog detects an intruder." The thing of importance is that the dog detects or doesn't detect an intruder, not if there is one.)

Another example:

If I am the axe murderer, then I can use an axe.
I cannot use an axe.
Therefore, I am not the axe murderer.

## Relation to modus ponens

Every use of modus tollens can be converted to a use of modus ponens and one use of transposition to the premise which is a material implication. For example:

If P, then Q. (premise -- material implication)
If not Q , then not P. (derived by transposition)
Not Q . (premise)
Therefore, not P. (derived by modus ponens)

Likewise, every use of modus ponens can be converted to a use of modus tollens and transposition.

## Justification via truth table

The validity of modus tollens can be clearly demonstrated through a truth table.

p q p → q
T T T
T F F
F T T
F F T

In instances of modus tollens we assume as premises that p → q is true and q is false. There is only one line of the truth table—the fourth line—which satisfies these two conditions. In this line, p is false. Therefore, in every instance in which p → q is true and q is false, p must also be false.

## Formal proof

### Via disjunctive syllogism

Step Proposition Derivation
1 $P\rightarrow Q$ Given
2 $\neg Q$ Given
3 $\neg P\or Q$ Material implication (1)
4 $\neg P$ Disjunctive syllogism (2,3)

Step Proposition Derivation
1 $P\rightarrow Q$ Given
2 $\neg Q$ Given
3 $P$ Assumption
4 $Q$ Modus ponens (1,3)
5 $Q \and \neg Q$ Conjunction introduction (2,4)
6 $\neg P$ Reductio ad absurdum (3,5)