Elementary question about consistency of propositions and implications

Question

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?

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
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

score 2 · Answer 1 · answered Jun 25 '22 at 13:41

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.

ryang · Accepted Answer · 2022-06-25T05:11:19.737

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

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

Since this formula is a contradiction, its conjuncts are not consistent with one another; in other words, in every context, whatever set of meanings you assign to its three atomic propositions, at least one of its three conjuncts will be false.

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.

Maybe this interpretation (without qualifications, window dressing, fanciful tenses) clearer:

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

Let it be the case that #1 and #3 are true; you are claiming that in some universe, #2 is also true.

Now, we're not dealing with predicates, and our atomic propositions have definite truth values, so “indeed does not imply” simply requires its antecedent to be true^҂ and its consequent false.

Thus, it must be that Judy was speeding and will not go to jail; this contradicts #1, so your claim is wrong.

^҂ Because “indeed implies” doesn't assert any conclusion for a false antecedent, a false antecedent is correspondingly irrelevant for its negation “indeed does not imply”.

Elementary question about consistency of propositions and implications

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