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In other words, what you wrote is that every value when negated twice returns to itself. Here's a truth table of 3 values for the negation operator v | 0
by igravious 11y ago
In other words, what you wrote is that every value when negated twice returns to itself. Here's a truth table of 3 values for the negation operator
v | 0 | 1 | 2
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¬v | 1 | 0 | 2
So ¬¬0 ⇔ 0 and ¬¬1 ⇔ 1 and ¬¬2 ⇔ 2 which satisfies the principle of double negation but not LEM. Correct?
- IngoBlechschmid 11y agoYour counterexample could work, but it is not obvious that it does so. This is because you also have to give definitions of ∧, v, and ==>, and then check that your definitions satisfy the axioms of intuitionistic logic. As stated, your example only shows that one can define a function f on a set which contains more than two elements such that f . f = id. But this by itself is a trivial statement.
- igravious 11y agoOnce those definitions are given then my trivial example holds though, correct? Which are you saying; that it is, or, it is not possible to give those definitions?
- IngoBlechschmid 11y agoI didn't check it. If you want to do it (the key step being to verify that the axioms of intuitionistic logic holds for your definitions of AND and OR), you can check Wikipedia for the list of axioms to verify: https://en.wikipedia.org/wiki/Heyting_algebra https://en.wikipedia.org/wiki/Heyting_algebra