The conditional operator $\Rightarrow$ can be tricky to motivate in the cases where it is True. An approach taken by Elliott Mendelson in Number Systems and the Foundations of Analysis is to examine the expression $(C\land D)\Rightarrow D$. My understanding is as follows: Because English sentences that can be paraphrased with this schema are regarded as true by virtue of their structure, we axiomatically define any interpretation of the sentence letters in this schema to evaluate to a True truth value. For example, ``If I like ice cream and Au is the chemical abbreviation for gold, then I like cream'' is obviously a true sentence by it's structure, even if I hate ice cream. In particular, when $C$ is false and $D$ is true, we get $F\Rightarrow T$ evaluates to $T$.

The above makes sense to me, but then if I again try to use English to motivate the truth table of the conditional operator, I can run into issues. In particular, the following English structure sounds clearly false, irrespective of whether I like ice cream or not: ``If I like ice cream, then I don't like ice cream.'' Written as a schema, I would say that $C\Rightarrow\lnot C$ must always be false. But if I don't like ice cream, this would lead me to conclude that $F\Rightarrow T$ should evaluate to F, in contradiction with the previous conclusion.

In both instances, I think the English reasoning used to motivate how the truth table should look are equally sound. Yet they lead to contradicting results. What am I getting wrong here?

EDIT: After reading @ryang comments, my best attempt to explain what is going is as follows:

So we appeal to the notion that, when used in argumentation, the argument "if $C\land D$ then $D$ is regarded as valid, i.e. true all the time. This is by virtue of the structure of the statement, and we haven't stopped to consider at this point what the values of C and D are. Because the argument is valid, we can then go and axiomatically state that the conditional $C\land D\Rightarrow D$ should also be valid as well, i.e. true all the time. Then hashing out the special cases gives us 3 lines of the truth table.

Now if we rewind to the beginning and undo all of that, we could try and state that the argument "if C then not C" is clearly an invalid argument. However, we can't then continue the parallel with the other approach and claim that the conditional $C\Rightarrow \lnot C$ is also unsatisfiable. If we could, we would have conflicting truth tables from the two different approaches. But we can't, for some reason, and I'm going to try and give that reason.

A valid argument is true all the time, and it is then necessary that we assert that the truth-functional schema that represents it should also be truth-functionally valid. This is because our truth-functional calculus would be useless if it told us that true arguments had false truth values. On the other hand, an invalid argument, i.e. one that is false all the time, CAN be represented by a truth-functional schema that is true, because an argument is a different object than a schema; in particular, true statements can be thrown together in a way that doesn't constitute a true argument.

  • 2
    $\begingroup$ What specifically about $C \to \neg C$ sounds obviously false? Don’t you think inconsistent premises imply their negation, since they’re false? $\endgroup$
    – PW_246
    Mar 5 at 1:28
  • $\begingroup$ I'm not sure if 'implication' is relevant here, but I agree with what you're saying, truth-functionally. But I suppose it depends on what you view as the most important consideration when axiomatically defining the table. If it should represent English speech accurately, then I think sentences like $C\land D\Rightarrow D$, when uttered, sound just as True as sentences like $C\Rightarrow\lnot C$ sound False. $\endgroup$ Mar 5 at 1:38
  • $\begingroup$ You may be interested in the notion of connexive logics, which do interpret "$A\implies B$" as asserting some sort of compatibility between $A$ and $B$. $\endgroup$ Mar 5 at 1:50
  • $\begingroup$ I just read the short Wikipedia page on it, I had never heard of this before, thanks! Hmm... I guess the observation of such "paradoxes" might suggest that while it is helpful to use English to motivate the definition of the conditional operators truth table, taking the idea too seriously (as perhaps I am doing in this question) will inevitably lead to contradictory results. Would you agree with this characterization @NoahSchweber $\endgroup$ Mar 5 at 1:57
  • 1

3 Answers 3


This is really just an extended comment, but you are unlikely to get a satisfactory answer of how the logical properties of the ordinary English if should be understood. There is vast amount of discussion. For example, Humberstone, The Connectives, is a study of the four basic connectives (and, or, if, not), and weighs in at 1492 pages! For a short introduction to issues specifically about if, and a wide range of associated logics, see here.

  • $\begingroup$ I see, I didn't expect I was touching on such a vast area of study! Briefly, what would you regard as the heart of the problem here? It seems to me that using special cases of English speech to motivate the truth table, while seemingly helpful, is not an approach that converges to a single truth table because of fundamental contradictions in the ways that the "if/then" construction is used in English. Would you agree? $\endgroup$ Mar 5 at 3:54
  • 2
    $\begingroup$ It's almost certainly true that the way "if/then" is used in ordinary English doesn't pick out a single interpretation. Many of the uses don't correspond to any truth-table. One idea that might interest you is that $C$ somehow stands in the wrong relation to $\lnot C$ for "$C$ implies $\lnot C$" to be true when $C$ is false. This kind of idea is studied in relevance logic. $\endgroup$
    – David M
    Mar 5 at 7:28

the expression $(C∧D)⇒D$ is obviously a true sentence by it's structure

In other words: the sentence $$(C∧D)⇒D$$ is logically valid.

the following English structure sounds clearly false, irrespective of whether I like ice cream or not: ``If I like ice cream, then I don't like ice cream.'' Written as a schema, I would say that $C⇒¬C$ must always be false.

The argument $$C, \text{ therefore } \lnot C$$ is certainly invalid. This means that as $C$'s assigned meaning varies, the argument's underlying conditional $$C⇒¬C\tag1$$ is at least sometimes false—not that it is always false! As a matter of fact, sentence $(1)$ is a satisfiable: when $C$ stands for the false assertion "I like ice-cream", sentence $(1)$ is true.

Don't conflate the assertions "sentence $(1)$ is invalid" and "sentence $(1)$ must be false". A conditional with contradictory premise and conclusion can be true; $$\text{I enjoy Science}⇒\text{I enjoy Literature}$$ is another true sentence that corresponds to an invalid argument.

  • $\begingroup$ If we have yet to define the conditional's truth table, then stating $C\land D\Rightarrow D$ is logically valid is an axiom motivated by the validity of the English sentences that are paraphrased by the schema. Correct? And if we are choosing an English sentence structure to motivate the truth table of the conditional, why is saying axiomatically that $C\Rightarrow \lnot C$ is invalid not as acceptable as the former axiom? $\endgroup$ Mar 5 at 2:52
  • $\begingroup$ I am trying to think about the conditional from the perspective of someone who has not yet accepted the standard truth table. So when you stated that sentence (1) is satisfiable, don't you necessarily assume the standard truth table when doing so? I agree that if you accept the conventional definition then sentence (1) is clearly satisfiable. $\endgroup$ Mar 5 at 2:57
  • 1
    $\begingroup$ @MaanDoabeDa My first paragraph already points out that true by virtue of structure is exactly what logical validity means. $\quad$ You claim that C⇒¬C (whether symbolically or expressed in natural language) must be false, as you implicitly recognise that it represents an invalid argument, however an invalid argument doesn't preclude its conditional from being true. $\quad$ You claim that this invalidity is fundamental/axiomatic, but this invalidity is in fact premised on the truth table, not vice versa. $\quad$ Finally, and this is probably ultimately most pertinent: $\endgroup$
    – ryang
    Mar 5 at 13:46
  • 1
    $\begingroup$ note that the material conditional underpins merely some—not all—natural-language conditionals. $\quad$ P.S. "because the If makes it clear that it is hypothetical, meaning the truth value of the statements is not what is under examination, only the structure of the statement." $\quad$ No and no: whether we examine a sentence's structure/validity is independent of whether it is a conditional, and to examine a conditional's structure is not to disregard the truth values of its antecedent and consequent. $\quad$ P.P.S. The bottom of $\endgroup$
    – ryang
    Mar 5 at 13:47
  • 1
    $\begingroup$ this answer illustrates how, without quantification, propositional logic sometimes feels weird and unintuitive. $\endgroup$
    – ryang
    Mar 5 at 13:47

$C\implies \neg C~~$ is "problematic" only if $C$ is true.

It is trivial to prove: $~~(C \implies \neg C)\iff \neg C$

The truth table:

enter image description here

Using a form of natural deduction (not using the truth table):

enter image description here enter image description here

Plain text version:

1   C => ~C

    2   C

    3   ~C
        Detach, 1, 2

    4   C & ~C
        Join, 2, 3

5   ~C
    Conclusion, 2

6   C => ~C => ~C
    Conclusion, 1

7   ~C

    8   C

        9   C

        10  ~C & C
            Join, 7, 8

    11  ~C
        Conclusion, 9

12  C => ~C
    Conclusion, 8

13  ~C => [C => ~C]
    Conclusion, 7

14  [C => ~C => ~C] & [~C => [C => ~C]]
    Join, 6, 13

15  C => ~C <=> ~C
    Iff-And, 14

Re: The meaning of "implies in natural language vs. propositional logic

Example: Consider the implication, "If it is raining, then it is cloudy." In propositional logic, this does not mean that rain causes cloudiness. Or that it is always cloudy when it is raining, sunshowers being a counterexample. It means only that, at present, it is not both raining and not cloudy.

  • $\begingroup$ I don't think I invoked implication anywhere, did I? I don't think that was my confusion. More importantly, I am trying to motivate the definition of the conditionals truth table, whereas you have simply used it. So I'm afraid I am no closer to resolving my confusion. $\endgroup$ Mar 5 at 3:02
  • $\begingroup$ @MaanDoabeDa I took your if-then constructs to be implications. The validity of the truth table for logical implications is justified at dcproof.wordpress.com/2017/12/28/if-pigs-could-fly from what might be called "first principles." $\endgroup$ Mar 5 at 3:29
  • 1
    $\begingroup$ ... The above proof does not make use of the truth table in any case. The only properties of logical implication that it uses are (1) conditional proof on lines 6, 12 and 13, and (2) the biconditional on line 15. $\endgroup$ Mar 5 at 4:51

You must log in to answer this question.

Not the answer you're looking for? Browse other questions tagged .