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I wonder if the following is consistent (in basic propositional logic): $\lnot p \land \lnot (p \implies q) \land q$.

Intuitively, it seems like it should be consistent. I believe you're not guilty of speeding ($\lnot p$), that being guilty of speeding doesn't mean having to go to jail ($\lnot(p \implies q)$), but that you should go to jail anyway ($q$) -- e.g. because you were driving drunk. No problem.

And yet, $p \implies q$ is equivalent to $\lnot p \lor q$, and so $\lnot(p \implies q)$ is (by De Morgan) equivalent to $p \land \lnot q$. That is indeed inconsistent with $\lnot p \land q$.

Where is my confusion?

giorgio farace
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  • According to Wolfram Alpha that proposition is false for any values of p and q, even if you only do $\lnot p \land \lnot (p \implies q)$ (leaving off the last part) or $\lnot (p \implies q) \land q$ (leaving off the first part). – smitop Jun 25 '22 at 00:06
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    This question is a variation of a common one. See e.g. https://math.stackexchange.com/questions/272664/validity-of-conditional-statement-when-the-premise-is-false – Karl Jun 25 '22 at 00:12
  • Here's a better duplicate: https://math.stackexchange.com/questions/232309/how-to-interpret-material-conditional-and-explain-it-to-freshmen – Karl Jun 25 '22 at 00:25
  • @Karl No, this formula is refutable in intuitionistic propositional logic. One refutation uses the fact that $q$ implies $p\to q$. A different refutation uses the fact that $\bot\to q$ is valid (where $\bot$ represents "contradiction") and therefore $p\to\bot$ (which is the intuitionistic meaning of $\neg p$) implies $p\to q$. – Andreas Blass Jun 25 '22 at 01:29

2 Answers2

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I believe you're not guilty of speeding ($\lnot p$), that being guilty of speeding doesn't mean having to go to jail ($\lnot(p \implies q)$), but that you should go to jail anyway ($q$) -- e.g. because you were driving drunk. No problem.

Problem: Implication is not causation in classical propositional logic.

Typically when we use implications in natural language, we are suggesting that there is some relevance to the connection. However the symbolic logic only requires the connected propositions to have appropriate truth values. It cares not for the reason they have those values.

($p\to q$ is true when $q$ is true or $p$ is false).

Should it be true that "you should go to jail" for whatever reason, it will also be true that "you should go to jail if you are guilty of speeding."

Should it be false that "you are not guilty of speeding", then the promise that "you should go to jail if you are guilty of speeding," will be vacuously true.

Graham Kemp
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$\lnot p \land \lnot (p \implies q) \land q$

$$\begin{array}{cc|c@{}c@{}c@{}cc@{}ccc@{}cc@{}ccc@{}c@{}c@{}ccc@{}c} p&q&(&(&(&\lnot&p&)&\land&(&\lnot&(&p&\rightarrow&q&)&)&)&\land&q&)\\\hline 1&1&&&&0&1&&0&&0&&1&1&1&&&&\mathbf{\color\red0}&1&\\ 1&0&&&&0&1&&0&&1&&1&0&0&&&&\mathbf{\color\red0}&0&\\ 0&1&&&&1&0&&0&&0&&0&1&1&&&&\mathbf{\color\red0}&1&\\ 0&0&&&&1&0&&0&&0&&0&1&0&&&&\mathbf{\color\red0}&0& \end{array}$$

The given formula is a contradiction, so its conjuncts compose an inconsistent set (that is, they are never simultaneously true).

Intuitively, it seems like it should be consistent. I believe you're not guilty of speeding ($\lnot p$), that being guilty of speeding doesn't mean having to go to jail ($\lnot(p \implies q)$), but that you should go to jail anyway ($q$) -- e.g. because you were driving drunk. No problem.

Stripping away the qualifications and fanciful tenses:

  1. Judy was not speeding ($\lnot p$),
  2. that Judy was speeding does not imply that she will go to jail ($\lnot(p \implies q)$),
  3. Judy will go to jail ($q$).

Suppose that #1 and #3 are true; then $(p\implies q)$ is true; then #2 is false. Hence, your claim of consistency is incorrect.

ryang
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