-1

How could you write NCpq in terms of moving the negation inside the conditional and keeping the conditional?

The question got posed here originally by someone else. I can't post there at the moment, so I thought I'd put it here.

If the only primitive connectives are negation N, and the material conditional C (all other connectives such as conjunction K, come as abbreviations, e. g. Kpq abbreviates NCpNq), can you write a formula logically equivalent to NCpq which has C as its first symbol in Polish notation? If so, what does it look like? If not, why not?

closed as off topic by Joseph Weissman Oct 12 '11 at 5:20

Questions on Philosophy Stack Exchange are expected to relate to philosophy within the scope defined by the community. Consider editing the question or leaving comments for improvement if you believe the question can be reworded to fit within the scope. Read more about reopening questions here. If this question can be reworded to fit the rules in the help center, please edit the question.

  • Is there a particular reason you want this? or is it simply out of curiosity whether it is possible? Are you limited in the operators you can use? What kind of normal form you want/what kind of simplification rules are allowed? – Mitch Oct 12 '11 at 2:16
  • @Mitch Just out of curiosity for me. The other questions you ask I think appropriate. It's not originally my question (link added). That said, does there exist a way to write an equivalent formula which is a conditional, using only conditionals and negations? That is, if you only have "C" and "N" as the only primitive connectives, can you write a formula equivalent to NCpq which has "C" as its primary connective? I would conjecture no, but I don't know how to prove this. – Doug Spoonwood Oct 12 '11 at 2:34
  • 1
    Posting another copy of the same question to a different stack is strongly discouraged. I am voting to close at this time pending some justification or reformulation of the question to specify the philosophical concern here. – Joseph Weissman Oct 12 '11 at 5:20
  • 1
    Just in passing, as per earlier meta consensus, this site is not intended to do logic for logics' sake. (The existing mathematics site already offers an enormous amount of domain knowledge in that particular area; this does not mean that logic should be ignored but that we should try to focus on philosophical concerns to the greatest extent possible or we will end up degrading our focus.) More generally, even if you believe two domains overlap in some case, this should be discussed and worked out on meta or in chat as far as possible before posting duplicates. – Joseph Weissman Oct 12 '11 at 16:05
  • 1
    @Doug, please use the @ for direct responses if you can -- it definitely helps facilitate discussion. I do not want to repeat myself here, so I would like to table this discussion here and raise further concerns about this up on meta or chat if possible; however, very quickly: duplicates are highly discouraged, but keep in mind this is a network-wide StackExchange guideline, not a local matter for us to resolve. – Joseph Weissman Oct 16 '11 at 6:21
1

If we write out an abbreviated truth table for NCpq with 0 as false, and 1 as truth, we get the following:

 N  C  p  q
 0  1  0  0
 0  1  0  1
 1  0  1  0
 0  1  1  1

So, since if p=1, and q=0, then NCpq holds. Otherwise, NCpq fails. So, CpNq ensures that NCpq holds. But, CpNq also holds when p=0. So, we need CpNq and p. Consequently, KpCpNq comes as logically equivalent to NCpq. Kxy comes as logically equivalent to NCxNy. So, KpCpNq with x=p, and y=CpNq, comes as logically equivalent to NCpNCpNq. In other words NCpq==NCpNCpNq. So, here a negation can get said to have gotten moved inside a conditional, even though the formula still qualifies as a negation.

Edit: The Sheffer Stroke D, define as D00=1, D01=1, D10=1, D11=0 comes as an adequate connective in itself, so to speak. Dxy==CxNy, and Dxx==Nx. So, NCpq comes as logically equivalent to DCpqCpq by Dxx==Nx, substituting x with Cpq. Since Dxy==CxNy, DCpqCpq comes as logically equivalent to CCpqNCpq, substituting x with Cpq, and y with Cpq. So, NCpq comes as logically equivalent to CCpqNCpq. Or in other symbolism ~(p->q) comes as logically equivalent to ((p->q)->~(p->q)).

Not the answer you're looking for? Browse other questions tagged or ask your own question.