Talk:Logical equivalence

This article is rated Start-class on Wikipedia's content assessment scale. It is of interest to the following WikiProjects:

Mathematics Low‑priority Philosophy: Logic High‑importance This article is within the scope of WikiProject Philosophy, a collaborative effort to improve the coverage of content related to philosophy on Wikipedia. If you would like to support the project, please visit the project page, where you can get more details on how you can help, and where you can join the general discussion about philosophy content on Wikipedia.PhilosophyWikipedia:WikiProject PhilosophyTemplate:WikiProject PhilosophyPhilosophy High This article has been rated as High-importance on the project's importance scale. Associated task forces: /  Logic

Semantically, p and q are equivalent if they have the same truth value in every model... Would it not be more appropriate to write if they have the same truth value in every interpretation? I thought that a model is just a set and relations for each n-ary symbol, we can not say whether a formula (containing free variables) is "true" in a model (?). 147.231.6.9 (talk) 13:30, 9 May 2012 (UTC)[reply]

It is common to use "model" and "interpretation" essentially as synonyms. In first-order logic, each model has a canonical truth function which tells whether a given formula is true or false in the model. Thus models are a particular way of getting interpretations. Conversely, the completeness theorem shows that every interpretation of a set of sentences is in fact the interpretation obtained from some model.

Many treatments of first-order logic do include a way to tell whether an arbitrary formula is true or false in a given model. One common way they do this is by including a variable assignment in the model, which is used to give values to free variables (this is the method used by Enderton's book, for example). — Carl (

CBM · talk) 12:00, 9 April 2013 (UTC)[reply

]

I added something. Really, I think this should just be cited in the main section where it's treated in the article. Most of the lede is just a summary of other parts of the article, and as such the lede doesn't need duplicative references (cf.

CBM · talk) 14:37, 31 January 2018 (UTC)[reply

]

The statement given in the example is not necessarily correct, as French Guiana is in France but is not in Europe by most definitions (unless you count EU membership as in Europe) Fozzzyyy (talk) 21:09, 3 August 2018 (UTC)[reply]

The article writes:

statements ${\displaystyle p}$ and ${\displaystyle q}$ are said to be logically equivalent if they are provable from each other under a set of axioms, or have the same

model

.<ref>{{Cite book|title=Introduction to Mathematical Logic|url=https://archive.org/details/introductiontoma00mend|url-access=limited|last=Mendelson|first=Elliott|year=1979|edition=2|pages=[https://archive.org/details/introductiontoma00mend/page/n63 56]|isbn=9780442253073}}</ref>

However, the book Introduction to Mathematical Logic actually defines "logically equivalence" semantically (same truth value in every interpretation). There's no mentioning of "provab[ility] from each other under a set of axioms" at all. And this totally makes sense, because for e.g., an unsound proof system, "semantically equivalent" does not coincide with "provably equivalent", so what it currently states is technically incorrect. --Nullzero (talk) 18:34, 11 January 2022 (UTC)[reply]

Miaumee (talk · contribs) added the mentioned sentence in Special:Diff/927818765 which actually cites mathvault.ca. However, this website was determined to be unacceptable (low quality / spam), so the reference was removed. (See Wikipedia:Administrators'_noticeboard/IncidentArchive1048#Recent_bot-like_reverts_of_a_specific_user and Wikipedia_talk:WikiProject_Spam/2021_Archive_Nov#Mass_spam_of_mathvault.ca for details about this problematic website). However, the accompanying content was not removed.

Therefore, I think it makes sense to remove this sentence, as it's not backed up by any source (and is in fact inaccurate). --Nullzero (talk) 19:03, 11 January 2022 (UTC)[reply]

Citing GeeksforGeeks was also done by Miaumee (talk · contribs). The website has low quality, so I will remove it as well. --Nullzero (talk) 19:17, 11 January 2022 (UTC)[reply]