the truth table of the sentence $$(p \rightarrow q) \vee (q \rightarrow p)$$

is \begin{array}{ c c l } p & q & (p \rightarrow q) \vee (q \rightarrow p) \\ \hline T & T & \, \, T \; T \> T \> \> \mathbf{T} \> \> T \; T \> T\\ T & F & \, \, T \; F \> F \> \> \mathbf{T} \> \> F \; T \> T\\ F & T & \, \, F \; T \> T \> \> \mathbf{T} \> \> T \; F \> F\\ F & F & \, \, F \; T \> F \> \> \mathbf{T} \> \> F \; T \> F\\ \end{array}

where the italic truth values are of subclauses and the boldface truth values are of the whole statement. From the truth table it is seen that the given statement is a tautology. So far so good. For me the problem arises when I verbally think of the statement. It can be translated into metalanguage as,

the statement "either $p$ implies $q$, or $q$ implies $p$", is a tautology.

This tautology means that, given two arbitrary statements $p$ and $q$, if $p$ does not imply $q$, then $q$ must imply $p$, and the other way around. Or in other words, it is not the case that none of them implies the other. This doesn't make too much sense to me. Why it is always the case that, given two arbitrary statements, one should be implying the other for sure? Why two random statements should bound in such way?

  • $\begingroup$ Because the conditional does not mean that there must be "a link" about the to sentences. $\endgroup$ Oct 18, 2014 at 17:56
  • 1
    $\begingroup$ This is one of the many paradoxes of the material conditional; it is similarly odd that $p\to q$ should be true whenever $p$ is false, even when $p$ and $q$ seem unrelated. The oddity that you've pointed out is one of the reasons why the $\to$ operator, as usually defined, may not be a good way to model implication as it is normally understood. You may be interested to look into the understanding of $\to$ in constructive or intuitionistic logic, where $(p\to q)\land (q\to p)$ does not hold for all $p$ and $q$. $\endgroup$
    – MJD
    Oct 18, 2014 at 17:56
  • $\begingroup$ @MJD thanks for the comment. It answered my question. $\endgroup$ Oct 18, 2014 at 18:28
  • $\begingroup$ A "terminist" way of reading [(p→q)∨(q→p)] goes "every p is q, or every q is p". So, if some p is non-q, then every q is p. For example "every horse is an animal, or every animal is a horse." So, if some horse is not an animal, then every animal is a horse. As another example "every fish is a horse, or every horse is a fish." So, if some fish is not a horse, then every horse is a fish. $\endgroup$ Oct 18, 2014 at 23:48

2 Answers 2


In classical logic, there is nothing wrong in the above "un-natural" tautology.

You can see this recent post (with many links) about the "correct" reading of the truth functional aspects of $\rightarrow$.

You can check with truth table that :

$p \rightarrow q$ and $\lnot p \lor q$

are equivalent.

Thus, we can rewrite :

$(p→q)∨(q→p)$ (called also : Dumemtt's law)

as :

$(\lnot p \lor q) \lor (\lnot q \lor p)$.

Now we can rearrange the disjuncts to get :

$(\lnot p \lor p) \lor (\lnot q \lor q)$.

Now, the tautology "sounds" much more ... tautological.

As per comment above, you can avoid this counter-intuitive aspects leaving classical logic and adopting, for example, the intuitionsitic one; but you have to remember that the truth functional properties of the connectives (i.e. the truth table) are nor more valid in it.

  • $\begingroup$ [(p→q)∨(q→p)] is a tautology for Lukasiewicz three-valued logic. $\endgroup$ Oct 18, 2014 at 23:27

Use the equivalences $$[A\implies B] \equiv \neg[A\land \neg B]\equiv [\neg A\lor B]$$ to show that $$[[p\implies q]\lor [q\implies p]]\equiv[[p\land \neg q]\implies[\neg q \lor p]] $$


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