Talk:Logical assertion

The content of the Logical assertionLogical assertion page was merged into Judgment (mathematical logic) on 7 May 2018, and it now redirects there. For the contribution history and old versions of the merged article, please see its history.

Shouldn't it read "x (mod 2) \equiv 0"?

I've never heard of this term, and the article does not provide any references to this usage. It looks bogus to me. Any defenders? --- Charles Stewart 17:52, 19 May 2005 (UTC)Reply

I think this should be merged with judgment. 'Assertion' is introduced as a synonym for judgment in (Martin-Löf 1983) 'On the Meanings of the Logical Constants and the Justification of Logical Laws', and indeed the present content of this page expresses a special case of what can be found at judgment. Quiddital (talk) 22:26, 13 August 2016 (UTC)Reply

No. This is wrong. A judgment is a type judgment, as in type theory. Judgments are used to define logics, so you cannot make a logical assertion until a logic is defined, first. See natural deduction for examples of how to define a logic, and how to use judgments to create that definition. 67.198.37.16 (talk) 21:52, 18 December 2018 (UTC)Reply

This article is very misguided.

• It is not hypotheses that are asserted.
• It is provable formule that are asserted not truths (the set of things provable and the set of things true may not coincide).
• Assertions in programming languages such as C are quite different from assertions in the logical sense. 86.132.216.101 (talk) 17:45, 19 September 2016 (UTC)Reply

The current article contains this text:

For example, if p = "x is even", the implication

${\displaystyle (\vdash p)\rightarrow (x{\pmod {2}}\equiv 0)}$

which is clear-as-mud and/or wrong. I suspect the parenthesis is mis-placed. I suspect it should be

${\displaystyle \vdash (p\rightarrow (x{\pmod {2}}\equiv 0))}$

If I try to read the former, with the bad paren placement, I start with ${\displaystyle (\vdash p)}$ which can be readily recognized as a tautology: from nothing at all, from thin air, I can prove that p is true. Since its a tautology, p is always true. So this reduces to ${\displaystyle true\rightarrow (x{\pmod {2}}\equiv 0)}$, which is blatently wrong, unless x is restricted to be a member of the set of even natural numbers. For example, x must not belong to the set of quadruped furry animals, because ${\displaystyle x{\pmod {2}}}$ is not even well-defined for furry animals.

The second form with re-arranged parenthesis, makes slightly more sense, but is still ill-defined. There, the reading starts with ${\displaystyle p\rightarrow (x{\pmod {2}}\equiv 0)}$, so that p is now some proposition, might be true, might be false, who knows, but whenever p is true, then if follows that ${\displaystyle x{\pmod {2}}\equiv 0}$. Presumably, it works out that whenever p is true, then x is even.

Oh, hang on. It also says: ''For example, if p = "x is even",... and so perhaps this is meant to be the definition of what p is? In that case, simple substitution works. That is, perform the substitution ${\displaystyle p/(x\!\!\!\!{\pmod {2}}\equiv 0)}$ aka ${\displaystyle [p:=(x\!\!\!\!{\pmod {2}}\equiv 0)]}$. The first formula gives

${\displaystyle (\vdash (x{\pmod {2}}\equiv 0))\rightarrow (x{\pmod {2}}\equiv 0)}$

which is insane because ${\displaystyle (\vdash (x{\pmod {2}}\equiv 0))}$ is not a tautology. The second form gives

${\displaystyle \vdash ((x{\pmod {2}}\equiv 0)\rightarrow (x{\pmod {2}}\equiv 0))}$

which obviously is a tautology, and totally acceptable. 67.198.37.16 (talk) 15:48, 18 December 2018 (UTC)Reply