# When does material implication supposedly not "work?"

I am told that material implication does not always "work." I'm not sure what the means exactly. Can readers here give me "real-life" examples of it not working, e.g. logical propositions $A$ and $B$ such that $A$ is false and $B$ is true and the implication $A\implies B$ is false.

• A: I am a girl. B: I am a boy A does not imply B Commented Jan 18, 2018 at 21:54
• How do we know A is false? Commented Jan 18, 2018 at 21:57
• en.wikipedia.org/wiki/… Commented Jan 18, 2018 at 22:16
• @EricWofsey From your link: "The material conditional can yield some unexpected truths when expressed in natural language. For example, any material conditional statement with a false antecedent is true (see vacuous truth). So the statement 'if 2 is odd then 2 is even' is true. Similarly, any material conditional with a true consequent is true. So the statement 'if I have a penny in my pocket then Paris is in France' is always true, regardless of whether or not there is a penny in my pocket." Something to think about. Thanks. Commented Jan 18, 2018 at 22:50
• Right, "doesn't work" is a strange and misleading way to say it. It's more that you have some intuitive notion of what "implies" means and the intuitive meaning doesn't always match material implication. For example, intuitively "implies" has something to do with causality; material implication has nothing to do with causality. Commented Jan 18, 2018 at 23:20

Here is one absolutely standard line of thought (essentially due to Ramsey, some 90 years ago).

Jill is a little late home. Thoughts about various unlikely catastrophes pop unbidden into Jack's mind, such as $P$: Jill has had an accident. But Jack is good at keeping his worries in check. He in fact thinks it very probable that not-$P$.

Now, Jack knows that a disjunction is at least as likely to be true as its first disjunct (put it this way: there are at least as many ways the world might go which make $A$ or $B$ true as make $A$ true). So rational Jack realizes it is also very probable that not-$P$ or $Q$, for any second disjunct at all, including e.g. $Q$: Jill has been trampled by a herd of elephants. Since he knows that $(P \to Q)$ is by definition just not-$P$ or $Q$, Jack will also therefore give a very high degree of credence to the material conditional $(P \to Q)$ because of his very high confidence in the truth of not-$P$.

However, living as they do in a small English town, Jack will think it is very improbable indeed that, if the worst has happened and Jill has had an accident, then she has been tramped by a herd of elephants. In other words, Jack will give a very low degree of credence to the conditional if P then Q.

But how can this be, on the view $if$ = $\to$? If ordinary conditionals are no more and no less than unadorned material conditionals, rational Jack should give the same level of credence to if P then Q as to $(P \to Q)$. Since Jack's very different levels of credence are perfectly rational, 'if's aren't '$\to$'s.

• I didn't say anything about probabilities. You have $P$ = Jill was in an accident and $Q$ = Jill was trampled by elephants. If $P$ is false, i.e. we have $\neg P$, then $P\implies Q$ whether or not $Q$ is true. Commented Jan 19, 2018 at 1:46
• I'm not saying that you did say anything about probabilities. The point is that ordinary ifs and material conditionals behave differently in contexts -- the default for ordinary conversation! -- where credences are other than 0/1. There's a large literature on this. See e.g. Edgington's article in the SEP and the many references therein. (I'm not hereby endorsing this line: but you asked, I assumed, why do lots of serious writers on conditionals deny that ifs are material, and I'm pointing to one standard, widely accepted, non trivial, reason why not.) Commented Jan 19, 2018 at 7:58
• I am just pointing that where credences are only 0/1(as you put it), material implication seems to be inevitable. If you want to talk about causal or temporal considerations, ordinary true-or-false logic may not work. Commented Jan 19, 2018 at 12:42
• But @DanChristensen, that's a quite radical change of view on your part -- for credences aren't truth-values. Commented Jan 19, 2018 at 13:39
• I meant only that material implication seems to be inevitable for pairs of true-or-false logical propositions, e.g. if we have true-or-false logical propositions $A$ and $B$ then we can infer that $A\implies B$ knowing only that $A$ is false. Likewise, we can infer that $A\implies B$ knowing only that $B$ is true. Commented Jan 19, 2018 at 14:02

If Bertrand Russell was born in Paris, then Bertrand Russell was born in England.

• True, but nevertheless I think you would be thought perverse if you brought it up in casual conversation. Commented Jan 18, 2018 at 23:51
• But considerations about when it is "perverse"/inappropriate to utter conditrionals are precisely what some offer as arguments against identifying ordinary conditionals and material conditionals -- assertibiilty conditions for them differ (as per, e.g., Frank Jackson's theory). Commented Jan 18, 2018 at 23:58
• @DanChristensen I am not sure why it would be 'perverse' ... if I believe that Paris is in England, couldn't I be making a claim like that quite naturally? Commented Jan 19, 2018 at 0:17
• @PeterSmith It doesn't point to the need for a some different form of logic or that there are any serious limitations of material implications provided you are talking about a pair of logical, true-or-false proposition, not causality, passage of time or probabilities -- just things that are true or false in the moment. Commented Jan 19, 2018 at 2:02

I have noted the following fallacies that can spin from having $$A \Rightarrow B$$ available, but then jumping to conclusions. The following are all fallacies, means they are not tautological hence generally valid, they have all counter examples in propositional logic:

Affirming a Disjunct
$$(p \vee q) \wedge p \Rightarrow \neg q$$

Affirming the Consequent
$$(p \Rightarrow q) \wedge q \Rightarrow p$$

Commutation of Conditionals
$$(p \Rightarrow q) \Rightarrow (q \Rightarrow p)$$

Denying a Conjunct
$$\neg (p \wedge q) \wedge \neg p \Rightarrow q$$

Denying the Antecedent
$$(p \Rightarrow q) \wedge \neg p \Rightarrow \neg q$$

Improper Transposition
$$(p \Rightarrow q) \Rightarrow (\neg p \Rightarrow \neg q)$$