Talk:Logical assertion

message: 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)
thread_title: Is this article encyclopedic?

message: 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)
replies: text: 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)
thread_title: Merge

message: 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)
thread_title:  Wrong wrong wrong and wrong! 

message: The current article contains this text:
:For example, if p = "x is even", the implication
:: (\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
: \vdash (p\rightarrow(x \pmod 2 \equiv 0))
If I try to read the former, with the bad paren placement, I start with (\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 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 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 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 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 p/ (x\!\!\!\! \pmod 2 \equiv 0) aka [p := (x\!\!\!\! \pmod 2 \equiv 0)]. The first formula gives
: (\vdash (x \pmod 2 \equiv 0))\rightarrow(x \pmod 2 \equiv 0)
which is insane because  (\vdash (x \pmod 2 \equiv 0)) is not a tautology.  The second form gives
: \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)
thread_title:  How should this be understood?