Prove that the law of excluded middle does not hold in some many-valued logic

[hatsoff]

Hi, all.

Wikipedia says:

In logic, the law of the excluded middle states that the propositional calculus formula "P ∨ ¬P" ("P or not-P") can be deduced from the calculus under investigation. It is one of the defining properties of classical systems of logic. However, some systems of logic have different but analogous laws, while others reject the law of excluded middle entirely.​

(emphasis added)

My question is, can we prove that the bolded claim is true? For some many-valued logic, can we show that for some condition, there is P with ¬(P∨¬P) ?

Thanks!

 

[xnull]

"can we show that for some condition, there is P with ¬(P∨¬P) ?"

There exists a P such that ¬(P∨¬P)
There exists a P such that ¬P Λ P

Lets prove this by contradiction. It seems really easy, although I'm not sure if it is valid since the law of excluded middle isn't meant to be applied to non-binary systems.

For every P, (P V ¬P)
In a multivalued system, this isn't true, because P can be something besides true or false.

Therefore the opposite, ¬P Λ P, is true.

Okay, that's my poor attempt at it. I would give it more of a mathematical go if I had more time. Actually, I'm procrastinating right now.

 

[CRGreathouse]

My question is, can we prove that the bolded claim is true? For some many-valued logic, can we show that for some condition, there is P with ¬(P∨¬P) ?

Surely the axiomatic logic with schema
¬(P∨¬P)
for all propositions P would meet your requirement, but I imagine that's not what you intend.

Also, be careful: there are logics which reject the law of the excluded middle (for all propositions P, P∨¬P) but accept the law of noncontradiction (for all propositions P, ¬(P∨¬P)). Intuitionistic logic would be an example.

 

[AUMathTutor]

When they say that some systems reject the law of the excluded middle, I believe what they're talking about is the constructivist viewpoint. Basically, if you are a constructivist, contradiction proofs go out the window.