Interview with Dr. Henry Bauer - Part 1

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(2017-10-25, 10:05 AM)Chris Wrote: So for me the hypothesis "p implies q" just means "in all cases where p is true, q is also true".

OK, so, I can't resist. I'll bang on the drum a little more. I think what you've written above is a perfect example of "everyday" logic with respect to implication. I also think that you'd concede that formal logic goes a little further: it stipulates that also, "p implies q" means that "in all cases where 'not p' is true, the implication is true". This is where the beat becomes more intense... but I'm still waiting for somebody to start dancing to my drumming!
(This post was last modified: 2017-10-25, 11:40 AM by Laird.)
(2017-10-25, 11:13 AM)Laird Wrote: OK, so, I can't resist. I'll bang on the drum a little more. I think what you've written above is a perfect example of "everyday" logic with respect to implication. I also think that you'd concede that formal logic goes a little further: it stipulates that also, "p implies q" means that "in all cases where 'not p' is true, the implication is true". This is where the beat becomes more intense... but I'm still waiting for somebody to start dancing to my drumming!

I don't think I'd describe that as going further (because I think it's implicit in what I said), but I agree that the hypothesis "p implies q" is true (though not very useful) if the condition p is never true. So I think you could also say it's true for the subset of cases where p is not true. Though there may be a danger of confusion if we start talking about hypotheses being true for some cases and not others.
(2017-10-25, 12:37 PM)Chris Wrote: I don't think I'd describe that as going further (because I think it's implicit in what I said)

I disagree. It's not implicit in what you said. What you said had to do with "p" being true: this has no bearing - explicit or implicit - on what is the case when "p" is false.


(2017-10-25, 12:37 PM)Chris Wrote: but I agree that the hypothesis "p implies q" is true (though not very useful) if the condition p is never true.

"Not very useful" is one way of putting it. Further analysis is possible.


(2017-10-25, 12:37 PM)Chris Wrote: So I think you could also say it's true for the subset of cases where p is not true.

Quite so. This is really just a restatement of your immediately prior statement.


(2017-10-25, 12:37 PM)Chris Wrote: Though there may be a danger of confusion if we start talking about hypotheses being true for some cases and not others.

I'm not really sure what you mean by this.

But I feel that this post is quite cold - I don't mean it to be, I'm just trying to drum a beat that nobody wants to acknowledge or even hear. :-)
(This post was last modified: 2017-10-25, 12:48 PM by Laird.)
(2017-10-25, 12:46 PM)Laird Wrote: I disagree. It's not implicit in what you said. What you said had to do with "p" being true: this has no bearing - explicit or implicit - on what is the case when "p" is false.



"Not very useful" is one way of putting it. Further analysis is possible.



Quite so. This is really just a restatement of your immediately prior statement.



I'm not really sure what you mean by this.

But I feel that this post is quite cold - I don't mean it to be, I'm just trying to drum a beat that nobody wants to acknowledge or even hear. :-)

Overall, I don't think this discussion is one to which I can add anything of my own.

However, I wanted to say that as I read through it, your contributions are valuable, they help to provide clarity. Presumably they are useful to others too.
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Laird

I think we are looking at this in rather different ways. 

To my mind, to talk sensibly about whether the implication is true - "p implies q" - we really need to consider a collection of examples, or in the terms of the paper a whole set of cards, and then see whether the rule applies to all of them. That's why I'm a bit resistant to saying the implication is true in some cases and not others. Viewing all the cases as a whole, that would mean the implication wasn't true. 

Granted in theory we could say the implication is true for a subset of the cards, and we could even have a subset containing a single card, and so in theory we could say that for a card with (for example) "not p and not q" the implication "p implies q" is true. But it seems to me that that's not a very useful statement, and that the value of the implication is that it's a general statement that will apply to a whole set of different cards, so it would be better not to talk about the implication being true or false for individual cards.

The reason I feel that the statement about the "not p" cards was implicit in what I said is similar. I was thinking of "p implies q" being a general rule applying to all cards. So if it's restated as "in all cases where p is true, q is also true", then that implies that all cases where p is not true are consistent with the rule - the rule doesn't mention them, so they can't contradict it.
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(2017-10-25, 05:25 PM)Typoz Wrote: Overall, I don't think this discussion is one to which I can add anything of my own.

However, I wanted to say that as I read through it, your contributions are valuable, they help to provide clarity. Presumably they are useful to others too.

Thanks, Typoz. I think though that I have made my share of missteps along the way!
(2017-10-25, 06:40 PM)Chris Wrote: Laird

I think we are looking at this in rather different ways. 

To my mind, to talk sensibly about whether the implication is true - "p implies q" - we really need to consider a collection of examples, or in the terms of the paper a whole set of cards, and then see whether the rule applies to all of them. That's why I'm a bit resistant to saying the implication is true in some cases and not others. Viewing all the cases as a whole, that would mean the implication wasn't true. 

Granted in theory we could say the implication is true for a subset of the cards, and we could even have a subset containing a single card, and so in theory we could say that for a card with (for example) "not p and not q" the implication "p implies q" is true. But it seems to me that that's not a very useful statement, and that the value of the implication is that it's a general statement that will apply to a whole set of different cards, so it would be better not to talk about the implication being true or false for individual cards.

The reason I feel that the statement about the "not p" cards was implicit in what I said is similar. I was thinking of "p implies q" being a general rule applying to all cards. So if it's restated as "in all cases where p is true, q is also true", then that implies that all cases where p is not true are consistent with the rule - the rule doesn't mention them, so they can't contradict it.

Thanks for that post, Chris. It's very clear and helpful.

I just want to add that as of a few posts ago I stopped suggesting that the rule can be "confirmed" by certain tests, recognising that instead all we can say is that we have, as the second paper calls it, an "ambiguous verification" - so we already saw eye to eye on that.
(2017-10-25, 06:40 PM)Chris Wrote: So if it's ["p implies q" --Laird] restated as "in all cases where p is true, q is also true", then that implies that all cases where p is not true are consistent with the rule - the rule doesn't mention them, so they can't contradict it.

And this is the heart of the quirk to which I've been alluding.

In everyday logic, if somebody said to us, "I am trying to decide (for some given 'p' and 'q') whether 'p' implies 'q', in other words, whether, in all cases where p is true, q is also true. How helpful are the cases where 'p' is false in making this decision?", we would probably respond: "Not at all. They can't help you to decide that the implication is true, and they can't help you to decide that it's false. They are simply irrelevant. Only the cases where 'p' is true can help you with that decision".

Unfortunately, in propositional logic, we can't put a couple of "irrelevant"s into the truth table - we are restricted to one of "true" or "false". I think you explain well in your post why the best choice out of these for the two "not p" rows is "true". But this does cause a break with everyday logic, and some very counter-intuitive results.

Rather than putting it into my own words, I've found a PDF online which explains this well. First off, on the issue just discussed - what the truth-table value should be for the material implication operator when the antecedent is false, given that we can't use a value like "irrelevant" - it puts it in an only slightly different way than you do, Chris:

Quote:Logicians have decided to take an “innocent until proven guilty” stance on this issue. An if—then statement is considered true until proven false. Since we cannot call the statement p implies q false when p is false, our only alternative is to call it true.

The paper goes on to describe some of the counter-intuitive results of this quirk:

Quote:We emphasize again the surprising fact that a false statement implies anything.

It provides several examples - the one I find most illustrative is this one:

Quote:(1) If elephants can fly, then the Cubs will win the World Series this year.
(2) If elephants can fly, then the Cubs will lose the World Series this year.
Both  statements  are  true—assuming,  of  course,  that  elephants can’t fly.

It also provides an example that doesn't flow from any case where "not p" is true - but rather follows from the fact that there is not necessarily a causal connection between "p" and "q" when both are true and thus "p implies q" is true, as we would require when saying in everyday logic that "p implies q":

Quote:(1) If hydrochloric acid (HCl) and sodium hydroxide (NaOH) are combined, then table salt (NaCl) will be produced.
(2) If March has 31 days, then dogs are mammals.

Both statements are true. The first statement is an example of cause and effect and reflects the chemical equation

HCl + NaOH = NaCl + H2O.

In the second statement, there is clearly no causal relation between days of the month and dogs being mammals.  Both the premise and the conclusion of Statement 2 are true, so according to the chart, the implication is true. The point here is that the use of implies in logic is very different from its use in ever[y]day language to reflect causality.

Anyhow. Hopefully that goes some way to explaining what I've been going on about re the unreliability of the material implication operator as a reflection of everyday-logic implication. It allows for such supposed entailments as we have discussed - e.g. (¬Q→¬P)→(P→Q) - that we would never accept in everyday logic.
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(2017-10-26, 11:11 AM)fls Wrote: I have no trouble admitting to and correcting errors. I do so all the time.

It's an eternal mystery to me why people take you seriously.
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