Material conditional: Why does the absence of the predicate validate the conditional? [closed]

Question

So yeah, basically from what i've read, and i've checked multiple sources on this, is that A -> B = ¬A V B.

So to really see where the confusion lies, I'll first state where it doesn't. Where A -> B, then this is saying that if A true, then B is true.

Thus 1 -> 1 is true, since this is just confirming the original statement.

1 -> 0 is false, since than A does not imply B.

Now, what I don't get is this:

A -> B | Truth value
0    0 |     1
0    1 |     1

Now, since the truth value lies in the confirmation or falsificion of A implying B, then a value of 1 means, 'yes, A implies B', and 0 means 'no, A does not imply B'. Now whilst this may sound blindingly obvious, this means that if A does not occur, this confirms (or perhaps reconfirms) that A implies B. A implies B given that A does not happen.

So for instance, if A = I am a qualified chef. B = I can cook well. Then if we let A -> B, then if I am not a qualified chef, then by me being an unqualified chef, this affirms the conditional that since I am a qualified chef, thus I can cook well. But I just said that I am NOT a chef.

I think it's the case that A could imply B in the absense of A, whether B occurs or not, but we just don't know. Why then don't we just say that the conditional leads to a partial truth table, where the only conclusions we verify are the one's where the antecedent occurs.

I mean, B is even called a consequent, one due to A, so we can't know about the consequences of A (or lack thereof) given that A does not occur.

And to be honest, I don't know why this in the norm. There's a reason why the way we handle the absense of A utterly confuses those new to logic (including myself).

Answer 1

Your confusion is understandable. Material implication is useful in mathematical contexts and in some scientific contexts where propositions are understood to be certainly true or false, but it is far less useful when it comes to representing ordinary everyday conditionals. One might even go so far as to say that it only works in the special case where the A and B are either certainly true or certainly false. In the real world this is seldom if ever true, and this is why material implication fails to represent ordinary conditionals very well.

As soon as things are uncertain, material implication gives completely the wrong answer to simple questions. Suppose I roll a regular 6-sided die and ask you, what credence do you attach to the conditional, "if it comes up even, it will be a six"? Nearly everyone will say one third. This of course is the value of the conditional probability P( six | even ). By contrast, the probability of the material implication P( even -> six ) is two thirds. The example generalises completely. Pick any typical conditional you like, just choose one where the A and B are not certainly true or false, and you will get the same result: the credence you attach to the conditional is the conditional probability, not the probability of the material implication. This has been tested experimentally in numerous trials conducted by cognitive psychologists: by and large we understand conditionals to mean that it is probably the case that B on the supposition of A. I have to qualify this with "by and large" because conditionals are very messy and unruly and there are many strange uses of them in English.

This approach to understanding conditionals was pioneered by Ernest Adams in his books "The Logic of Conditionals" and "A Primer on Probability Logic". He showed how this serves to explain the so-called paradoxes of implication, including the one you refer to in your question, i.e. that it is not generally plausible to deduce "if A then B" from ¬A. In my view, introductory logic textbooks do a disservice when they introduce material implication by not immediately warning the reader of its limitations.

Answer 2

This is not as outlandish as people who first encounter it formally often seem to think. It is the idea behind turns of phrase like "I'll do that when pigs fly". We all immediately understand such things, so it can't be deep. The phrase obviously means "I won't do that", because it directly implies that if I were going to, pigs would be flying, and they are not. But the statement gets to be true, and not a potential lie, because pigs do not, in fact, fly.

The reason to accept it formally in classical mathematics is that the focus is upon avoiding contradiction, and it never produces an inappropriate contradiction. It is convenient for the operator to be fully defined, and have appropriate values in all case, and this is the only value to give the operator in this case that is safe.

On the other hand, there are certainly versions of mathematics where this definition is not true. "Intuitionism" and other forms of "constructive" mathematics, which severely restrict the usage of negation in order to prompt more detailed and convincing proofs that more readily convert into applicable algorithms, do not accept this definition without reservations.

They balk at the introduction of the negation, which, in their theory, is ambiguous. In constructive terms, we cannot know something is provably false, just because it is not true -- it can lie in a kind of limbo between the two. And we certainly cannot presume it is false just because assuming so is not going to result in a contradiction. They expect us to prove that when do we assume it, we definitely get into trouble.

However, this results in some very difficult mathematical terrain. Evrett Bishop has gone through the first year course in calculus, and proved everything relevant in a way that is constructively acceptable. The result is about five times as long, and much of it is prohibitively complex. This seems excessive to many people, who do not necessarily want all of their mathematics to be directive of implementation in this way, but only need for it to be safe from contradictions instead.

Answer 3

It might help to think of it like this:

A: I am a qualified chef

B: I have great knife skills

I pick this example because it's absolutely impossible to be a decent chef, much less a qualified one, if you can't properly use a knife. You'll chop ingredients too slowly, cut yourself (or others) and not know when your knife is dull. I don't like your example because cooking well & being a qualified chef are less distinct from each other than knife skills and how good a chef one is.

Lets say A -> B = ¬A V B, as you did. In the absence of A, it still may be true that B. It's not hard to imagine an assassin or samurai who is an awful chef yet still has amazing knife skills (i.e. ¬A & B). We also know that ¬A & B -> ¬A V B (there's a technical name for this that escapes me).

In my case, where I exhibit both ¬A & ¬B (i.e. I'm both awful with knives and a poor chef) it is still true that A -> B = ¬A V B - it's just that in this case the OR operator is picking out ¬A instead of either one being a possible option. Since ¬A V B is still true, the material conditional itself is also true because of this equality.

Hopefully this clears things up for you.