What would be an intuitive understanding of Peirce's law?

Question

Wikipedia describes Peirce's law as

In propositional calculus, Peirce's law says that ((P→Q)→P)→P. Written out, this means that P must be true if there is a proposition Q such that the truth of P follows from the truth of "if P then Q". In particular, when Q is taken to be a false formula, the law says that if P must be true whenever it implies falsity, then P is true. In this way Peirce's law implies the law of excluded middle.

What I am looking for is an intuitive understanding beyond that description and beyond any derivation.

At the moment I don't have an immediate, intuitive understanding why something simpler, like (P→Q)→P, won't do, although I am sure it won't. Also why not extend the law further to, say (((P→Q)→P)→P)→P, or beyond? I want this to be so understandably obvious that I could easily explain it to someone else.

References describing Peirce's law beyond derivations or truth tables would be useful to help with this intuitive understanding.

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Reference

"Peirce's law", Wikipedia https://en.wikipedia.org/wiki/Peirce%27s_law

Answer 1

Peirce himself notes that this is hardly "axiomatical", i.e. self-evident. But it helps to convert implications into derivations. Then (P→Q)→P becomes P→Q ⊢ P, which is obviously invalid because it is circular, we can not derive P from something that assumes P as a premise. On the other hand, (P→Q)→P ⊢ P assumes that P does come out of P→Q, which we know to be circular in general. If it is, nonetheless, true then P better be true by itself. And since ((P→Q)→P)→P is already a tautology, deriving from it is deriving from nothing at all. So ((P→Q)→P)→P ⊢ P is invalid, P can't follow from nothing, and hence so is (((P→Q)→P)→P)→P.

It is used as a stand-in for the law of excluded middle in systems where only implication is used. In such systems ¬P is expressed as P→⊥, P implies the falsehood. Taking Q=⊥, the law becomes: if ¬P implies P then P is true. But if the negation of P leads to P then it leads to ⊥, and is itself negated. In other words, negating ¬P we get P back, one of the forms of the excluded middle.

Answer 2

Let us take a concrete example for (((P→Q)→P)→P). Let P: it has rained and Q: be the streets are wet.

Bearing that In mind, the conditional, then, becomes

If (if it has rained, then the streets are wet) then (it has rained)

Notice the first conditional is true in 2 cases:

1. It has rained and the streets are actually wet.
2. It hasn't rained

Let us take case one, if we say the first conditional is true because (it has rained) and (the streets are wet), why wouldn't it follow that: If (if it has rained, then the streets are wet) then (it has rained) since we are not saying anything new but just reiterating: if it has rained then it has rained, which is a tautology.

Let us take case two, if we now say the first conditional is true because the antecedent is false, then the conditional:

(if it has rained, then the streets are wet) is vacuously true. Consequently, if you notice, [If (if it has rained, then the streets are wet) then (it has rained)] this conditional will be false, because we have a vacuously true antecedent (if it has rained, then the streets are wet) and a false consequent (it has rained). Therefore, in either case we will have a valid, and intuitive result. Notice, this conditional is nothing but: (P(F)→Q(?))→P(F), which is ~P→~P (Another tautology).

Similarly, you can follow the same reasoning to show the same intuition follows for the complete version: (((P→Q)→P)→P)...

Answer 3

Pierce's Law is not very intuitive. But using equivalences may give some insight.

((P → Q) → P) → P
is equivalent to:
~((P → Q) → P) v P       by implication equivalence
((P → Q) ^ ~P) v P       by implication negation
((~P v Q) ^ ~P) v P      by implication equivalence
which reduces to:
(~P) v P                 by absorption.
T                        by complementation (aka Law of Excluded Middle)

Answer 4

The statement P->Q can be used to prove that P is false (e.g. "If it were raining, the street would be wet, but the streets are not wet, therefore it's not raining"), but any attempt to prove that P is false would be fallacious (e.g. "If it were raining, the street would be wet, the streets are wet, therefore it's raining" is the fallacy of affirming the consequent.) So if someone says "I know that P->Q is true" in one part of a proof, and later in the proof says "Therefore, P is true", they must have proven P without using the fact that P->Q. That is, if we take the P->Q part out of their proof, their proof would still establish P. So if you claim (P->Q)->P, the only way that's true is if P has already been established. I.e. ((P->Q)->P)->P.

Suppose we have P = "Fluffy is a dog" and Q = "Fluffy is a mammal". Then "P->Q" is "If FLuffy is a dog, then Fluffy is a mammal". Then "(P->Q)-P" means "If Fluffy being a dog means it's a mammal, then Fluffy is a dog". But clearly that's false: if Fluffy is a cat, then "Fluffy being a dog means it's a mammal" is true (if Fluffy were a dog, it would be a mammal), but Fluffy is not a dog.

"((P->Q)-P)->P" can be translated as "The only time we can say 'If Fluffy being a dog means it's a mammal, then Fluffy is a dog' is when Fluffy is a dog". Note that to construct a counterexample for "(P->Q)->P", I had to assume that Fluffy is not a dog (in my example, I considered the possibility that Fluffy is a cat). If Fluffy isn't a dog, then "If Fluffy being a dog means it's a mammal, then Fluffy is a dog" is false. So if "If Fluffy being a dog means it's a mammal, then Fluffy is a dog" is true, then Fluffy is a dog.

"P->Q" is true if Fluffy is a mammal or not a dog. If Fluffy is not a dog, then "P->Q" is true, which means that "(P->Q)->P" has a true premise and a false conclusion, and so is false. So the only way it's true is if Fluffy is a dog. If P is false, then we have alternating truth values:

P : false
P->Q : true (all implications with false premises are true)
(P->Q)->P: false (true premise, false conclusion)
((P->Q)->P)->P: true (false premise)
(((P->Q)->P)->P)->P: false

This alternating pattern can be shown inductively: define S_1 = P and S_n = S(n-1)->P. If n is even, then S_(n-1) is false (inductive hypothesis), so we have a false premise, and thus the statement is true. If n is odd, then S_(n-1) is true, so we have a true premise giving a false conclusion, so the statement is false.

We can also see that once we accept ((P->Q)->P)->P is equivalent to ⊥,, we can simplify further iterations by substituting that in. (((P->Q)->P)->P)->P can simplified to ⊥->P, which is clearly false. But ((((P->Q)->P)->P)->P)->P simplifies to (⊥->P)->P, which is true.

Answer 5

Something that implies 'just anything' is obviously false. If P then pigs fly. So the innermost parenthesized part means P is false.

If P being false would imply it were true, then however you look at it, P must be true. Otherwise we would then use that P is false to deduce P was true.

So if "P implies 'just anything'" implies P, then P must be true.