From a Wikipedia article about Negation Introduction:

Negation introduction states that if a given antecedent implies both the consequent and its complement, then the antecedent is a contradiction.

This is straight forward, except for what is named as "a contradiction". The antecedent is one entity with one truth value. As an entity, the antecedent only exists. That is has two possible states is not a contradiction, even if it somehow assumes both states at once. That is, of course, not possible. It would just be incorrect.

I do not know what might be the logical state that would be identified as incorrect. If you write 1 + 1 = 3, that is incorrect, but not because 1 + 1 is incorrect, or 3, for that matter. What is identified as incorrect is the equation, not the components.

How can one of anything be a contradiction? Why is it not the implication (or the pair of implications) that is the contradiction?

I think I need to ask this another way. I seem to be getting comments that explain a contradiction. I am not asking what is a contradiction. I am new to this, so my explanation of my question may be juvenile. Please bear with me. The statement $P \implies Q$ has three parts; an antecedent, a connector and a coincident. The statement $P \land Q$ also has three parts, but does not have an antecedent or coincident. I assume that in the statement $(P \implies Q) \land (P \implies \lnot Q)$, P is still the antecedent. This statement is, of course, a contradiction.

The text quoted above identifies P, the antecedent, as the contradiction rather than the statement. Is this just a convention or is there another reason for this?

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    $\begingroup$ Read it as "... is false" $\endgroup$ Mar 29, 2020 at 20:37
  • $\begingroup$ What would you call this antecedent: ($A \wedge {\rm not}\ A$)? $\endgroup$ Mar 29, 2020 at 20:40
  • $\begingroup$ @DavidG.Stork Are you suggesting that I can get from (P implies Q)and(P implies not Q) to (P and not P)? I will try it. $\endgroup$
    – user756686
    Mar 29, 2020 at 23:01
  • $\begingroup$ $1+1$ is not a statement: it is a term (i.e. a "name"). $1+1=3$ is a statement and it is implicitly a contradiction because we have that $1+1=2$ and we have that $2 \ne 3$. $\endgroup$ Mar 30, 2020 at 12:52
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    $\begingroup$ I agree with you that Wiki's explanation is not so good... The rule is: "if $P$ implies $Q \land \lnot Q$, then infer (derive) $\lnot P$." $\endgroup$ Mar 30, 2020 at 14:46

1 Answer 1


Standardly, “contradiction” is used widely in logic, to include self-contradictions, and any wff that comes out false on all valuations counts as a (self)-contradiction in this sense, even if not of the explicitly self-contradictory form $(\alpha \land \neg\alpha)$.

So for a start, any negation of a tautology -- for example, ‘$\neg(P \lor \neg P)$’ -- counts as a contradiction in this sense. So does ‘ $\bot$’, the primitive absurdity constant, although it is quite unstructured. Likewise for any wff $\gamma$ that entails both some wff and its negation; $\gamma$ can't possibly be true, comes out false on all relevant valuations, so counts as a contradiction in the wide sense.


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