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if P and Q are boolean, we can just consider the four possible combinations of P and Q and show that Pierce's Law is true for all four:
P=0,Q=0: ((0→0)→0)→0 = (1→0)→0 = 0→0 = 1.
P=0,Q=1: ((0→1)→0)→0 = (1→0)→0 = 0→0 = 1.
P=1,Q=0: ((1→0)→1)→1 = (0→1)→1 = 1→1 = 1.
P=1,Q=1: ((1→1)→1)→1 = (1→1)→1 = 1→1 = 1.
just remember that the implication operator (→) only takes on the value of 0 when 1→0, in all other cases, even 0→0, it's true. Digfarenough 18:10, 2 January 2006 (UTC)
The → as in ((P→Q)→P)→P does not stand for the implication operator, so it is not obvious why this proof should work, with my limited mathematical knowledge. However, I think I can still point out that, this proof assumes "the law of the excluded middle", while Peirce's Law in the context implies "the law of the excluded middle". The two are different, because the proof did not show that Peirce's Law would be false without the law of the excluded middle.218.250.191.237 (talk) 15:14, 28 March 2008 (UTC)
What do you mean by "... → ... does not stand for the implication operator ..."? Of course, it does. What else could it mean here? JRSpriggs (talk) 01:42, 30 March 2008 (UTC)
Material conditional, surely. "Implication" ain't conditional. 96.245.85.214 (talk) 05:39, 18 April 2018 (UTC)
Current Explanation for Laymen is Useless
I've been involved in Math and Science for some time, and I couldn't understand any of it. We need an example that does NOT somehow involve a contradiction. (If there were Kiwi and no Pears, then there were no Pears. A and B implies A, for goodness sake. It's got to go!)
Admittedly, finding an intuitive case where (P->Q)->P isn't too easy, but that's what needs to be done. Ideas are welcome. But I'm removing the whole kiwi and pear mess.
--Obsidian-fox 01:29, 22 February 2006 (UTC)
Unfortunately for us, we have been drafted into the navy. Life at sea is no fun, but at least the lunch line serves pears. We like pears, so we're glad that they're served often. The navy has a very odd custom. Sometimes when we go through a line, every pear is glued to a kiwi. We call these "glue lines". The custom says that every glue line must have pears. Customs often have the force of laws, and this custom is so old that it's as strong as the law of gravity. Today, we come upon a line which has no pears. How sad for us. On the other hand, a funny thing happens when there are none of something: nearly any claim about it is true. For instance, all of the pears in this line are purple with orange polka dots. They are also all three feet long and taste like steak. And they are all very sweet pears. More to the point, each of the pears in this line is glued to a kiwi, so this must be a glue line. But every glue line has pears, which means that this line had pears after all. Odd that we didn't notice them earlier, isn't it? Well, now that we see some, they aren't purple with orange polka dots after all, nor are they three feet long or taste like steak. They are, however, very sweet. Hmm... my pear has a kiwi glued to it. Did you notice if this line was a glue line? No? Me neither.
Let's put this back into our logic context now. Okay, so P means that the line has pears and Q means that it has kiwi. (Well, can you come up with a common fruit starting with Q? Quinces are not very good raw.) Now, let's say that P->Q represents a glue line, since it's certainly true in a glue line. That means (P->Q)->P says that glue lines always have pears. Now, if we find a line with no pears (i.e. ¬P), then that line is a glue line (¬P->(P->Q)), but glue lines always have pears ((P->Q)->P), so not having pears means that the line has pears (¬P->P), which doesn't make sense (contradiction). That means that we can't find a line without pears, or rather that all lines have pears (P). If we put that all together, we see that the fact that glue lines always have pears means that all lines have pears, but not necessarily that all lines are glue lines (((P->Q)->P)->P, but not ->(P->Q)).
Article fixed
Bah, I forget about my pet article for a bit and someone comes through and removes a vital section while someone else shows up with a question.
(duplicated at User talk:68.49.135.47)
Basically, what's going on is that we need to show a contradiction, e.g today is Tuesday and today is Wednesday. It's a bit bizarre. Let's start with these assumptions:
1. Today is Tuesday.
2. We never eat meat on Tuesday.
3. We ate meat today.
Notice that 2 and 3 together mean that today is not Tuesday:
1. Today is Tuesday.
2+3. Today is not Tuesday.
This is a contradiction, so something was wrong. In this case, one of our three assumptions is wrong, but we don't know which. Perhaps today is Wednesday. Perhaps we do sometimes eat meat on Tuesday. Perhaps we ate meat yesterday and not today. All of those are valid ways to fix our assumptions.
Now, in the case of our Pierce's Law example, we have two assumptions and one hypothesis:
1. A line is a glue line if every pear is glued to a kiwi.
2. A glue line has pears.
H. Today, we saw a line without pears.
Through our little mental experiment, that becomes:
H'. Today, we saw a line with every pear glued to a kiwi, but no pears.
So, we combine 1 and H'. Now we know this:
1+H'. Today we saw a glue line, but it had no pears.
2. A glue line has pears.
That gives us our contradiction, and also your confusion. You jumped here to start, and that didn't make any sense. Instead, our contradiction means that something was wrong. We trust our two assumptions, but the hypothesis could be wrong. It was wrong, meaning that the line we saw had pears. So we alter our hypothesis:
1. A line is a glue line if every pear is glued to a kiwi.
2. A glue line has pears.
H. Today, we saw a line with pears.
This is consistent, meaning that there is no contradiction. Notice that we can't combine 2+H:
2+H. Today, we saw a glue line.
is wrong.
Hope that helps! -FunnyMan 21:01, 2 January 2006 (UTC)
The explanation using the pear story does not seem to match the encyclopedic style surrounding it. Maybe we should replace it with exerpts from this discussion. Teply 20:03, 7 February 2006 (UTC)
Confusion
Dear FunnyMan, Thanks for your interest. Please be more specific as to what you find confusing and we can all discuss it here on the talk page, as that is likely to be the best policy for improving the article. In the meantime, though, I think that pinning a "scarlet letter" on the main article page is likely to be counter-productive, leading the casual reader of popflock.com resource to lose confidence in the enterprise as a whole. So now that you've drawn initial attention to your concerns, maybe you would consider removing it. Many regards, Jon Awbrey 16:00, 7 January 2006 (UTC)
Why oh why...
1) You don't need to prove the "contrapositive." The article on Contraposition that you refer to is a repulsive sesquipedalian hodgepodge, *and* it gives a wrong example for the conversion to "contrapositive," thus completely useless as an explanation.
2) It would be more helpful, but still completely unnecessary, to prove the "obverted contrapositive."
3) In fact, noone needs a proof so far-fetched its author could barely keep track of it (or he wouldn't have missed the ¬ in (2) in the first place.)
4) Even if you absolutely had to prove it this way, if just to make everyone squint, you could have made it clear which and what follows from where. I mean that it took me half an hour to wade through through all your s...tuff, even though I already knew what you were taking about. Imagine what it would take for someone who didn't.
Andrew Cooke (Feb 1 2006) - I just want to say thanks for the above. I'm learning logic and have been trying to understand this law. The proof on the front page had me completely lost, but the above is great; I can follow each step with no problems. I'm just learning, so don't feel competent to make changes myself, but I'd really, really encourage people to put the above on the front page rather than what's there now.
Rendering of the law in English — quantification
On 23 October 2007, Nortexoid (talk·contribs) rewrote the second paragraph of the lead to "correct" the explanation of the law in English. He made it say:
In propositional calculus, Peirce's law says that ((P->Q)->P)->P. Written out, this says that P must be true if you can show that P implying arbitraryQ implies P is true.
I corrected the word "arbitrary" to "some", making it:
In propositional calculus, Peirce's law says that ((P->Q)->P)->P. Written out, this says that P must be true if you can show that P implying someQ implies P is true.
On my talk page, Nortexoid said:
In the statement of Peirce's law, you've got the quantifiers in the metatheory wrong. It's for all P and Q, not all P and some Q. Your statement becomes trivial by letting Q be P.
If we insert quantifiers over propositions into the law to make its content more explicit, it says:
$\forall P\forall Q(((P\to Q)\to P)\to P)$
If I understand Nextroid's wording, it says:
$\forall P([\forall Q((P\to Q)\to P)]\to P)$
whereas my version is intended to say:
$\forall P([\exists Q((P\to Q)\to P)]\to P)$
which is correct. JRSpriggs 00:55, 25 October 2007 (UTC)
?our version is wrong, as it is trivial by letting Q be P. The statement of the law is equivalent to its universal closure in the metatheory, if we take P, Q to be propositional variables. The theory is closed under substitution, so we may replace Q with an arbitrary wff. Nortexoid 09:51, 25 October 2007 (UTC)
May the following help in your discussion? $\forall x(A(x)\to B)\leftrightarrow (\exists xA(x)\to B)$. This is a tautology (if B does not depend on x). It reminds us about the following: what one is thinking of as a for all, some other may think of as exists. It depends on your parenthesis. If we say $\forall Q(\ldots \to P)$ we might instead equivalently say $(\exists Q\ldots )\to P$. Jesper Carlstrom 13:07, 25 October 2007 (UTC)
Yes, as Jesper suggests, the main difficulty is in the English language. It is difficult to clearly specify the scope of the quantification over Q. The scope must be among: ((P->Q)->P)->P, (P->Q)->P, P->Q, and Q only. Only the first two are usable. Putting universal quantifiers over Q on either of the last two reduces Peirce's law to the equivalent of ((¬P)->P)->Pwhich is weaker (in intuitionistic logic) than Peirce's law (I think).
Regarding Nortexoid's counter-example: (P->P)->P indeed implies $\exists Q((P\to Q)\to P)$. However, (P->P)->P is equivalent to P intuitionistically. It is not a tautology theorem of intuitionistic logic. Perhaps he was thinking of P->P? JRSpriggs 00:51, 26 October 2007 (UTC)
Actually what I was thinking of concerned the explicit definition of the axiom schema in the metatheory using quantifiers, on the one hand, and then applying that to the intuitive informal reading of the schema. This was an accident. (Whoops.) Nortexoid 23:05, 27 October 2007 (UTC)
The rule ((¬P)->P)->P added to intuitionistic logic gives full classical logic, so it is not weaker than Peirce's law in intuitionistic logic. Jesper Carlstrom 08:19, 26 October 2007 (UTC)
Perhaps you meant minimal logic. There ((¬P)->P)->P is indeed weaker than Peirce's law. To see this, we can use the so called A-translation: define ¬P as P->A for a choosen propositional variable A. Consider minimal logic with the additional axiom A. In this logic ((¬P)->P)->P is derivable. But ((P->Q)->P)->P is not derivable, because the system can be translated into pure minimal logic by replacing A by A->A. Under this interpretation, ((P->Q)->P)->P is interpreted as itself (provided P and Q are different from A), and it is known not to be derivable in minimal logic. Jesper Carlstrom 09:52, 26 October 2007 (UTC)
To Jesper: Thank you for the correction. To Nortexoid: OK. We all make mistakes. As you see, I made some in my last message. JRSpriggs 01:43, 28 October 2007 (UTC)
Question concerning Axioms
I am not an expert on Peirce's logic but I know that (at least in one point of his career) he was hesitant to all axioms, owing to his empiricism/fallibilism/ pragmaticism.
In the second sentence of the introduction, the wiki states:
"It was taken as an axiom in his first axiomatisation of propositional logic."
Whereas, in the excerpt in his own description Peirce writes:
"This is hardly axiomatical."
Are there two notions of "axiom" being used? Or is there an error?
In any case the current explanation is rather confusing. -- Preceding unsigned comment added by Quaquaquaqua (talk o contribs) 17:46, 30 May 2013 (UTC)
Remember this is not an article about Peirce or his system of logic. It is merely an article about one formula in the language of propositional logic. JRSpriggs (talk) 06:19, 31 May 2013 (UTC)
Dubious affirmation
In part 1 (History), I read that:
"[a->x]->[((x->y)->a)->x] is a tautology."
Isn't it plain wrong? For me a tautology would be some theorem in minimal logic without any axiom. I cannot prove the previous statement in this framework...
Sedrikov (talk) --Preceding undated comment added 10:02, 20 November 2013 (UTC)
Usually, a tautology is defined as a sentence of propositional logic which has the valuation "true" for any assignment of values (from {"true", "false"}) to the atomic propositions contained within it. To verify this one usually uses a truth table such as the one which appears at Tautology (logic)#Verifying tautologies. In this case, that gives:
a
x
y
a->x
x->y
(x->y)->a
((x->y)->a)->x
[a->x]->[((x->y)->a)->x]
T
T
T
T
T
T
T
T
T
T
F
T
F
T
T
T
T
F
T
F
T
T
F
T
T
F
F
F
T
T
F
T
F
T
T
T
T
F
T
T
F
T
F
T
F
T
T
T
F
F
T
T
T
F
T
T
F
F
F
T
T
F
T
T
Is that clear? JRSpriggs (talk) 10:11, 21 November 2013 (UTC)
Arrow?
Since this expression is true for all values of P and Q, it might benefit from being written using the double arrow of material implication ((P->Q)->P)=>P. How about it? Is there a reason this isn't used?
No. Is there some reason why we should confuse the readers by using two different symbols (arrows) for material implication? JRSpriggs (talk) 14:18, 17 June 2016 (UTC)