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March 7

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A is the set of formulae that are conjunction of the following clauses:

  • clauses of the form or
  • clauses with the connector only (without )
  • clauses with the connectors (without ), such that appears in the clause exactly once

B is set of formulae defined just like A, but in the last kind of clauses they have also the connector .

Is that true that ?

I think I can solve this somehow with a technique similar to the one used for functional completeness proofs, but I can't figure out how exactly. עברית (talk) 19:53, 7 March 2018 (UTC)[reply]

Pretty sure I'm not understanding something, but it looks to me like A⊋B, so you could just take φ to be ψ. Maybe explain the context more? --RDBury (talk) 14:14, 8 March 2018 (UTC)[reply]
It's the opposite: B⊋A, since B has the extra connector. Is my question clear now? עברית (talk) 16:44, 8 March 2018 (UTC)[reply]

However, I found a better formulation of what I need:

We say that a formula is 3-symmetric if (where is the symmetric group)

is the set of k-variable formulae that are conjunction of expressions (or, clauses) that satisfy the following property: each expression uses the connectors (but not ), such that appears in the clause exactly once

is defined just like , but in that case using the connector is also allowed.

is the set of k-variable formulae that satisfy the 3-symmetry.

Is the following statement true?

 
 

Motivation: This is a definition of equivalence between two formulae (of course, it's different from the usual definition). I would like to define them as equal if either both have a "true" in their truth table or both don't have. But this definition is too weak, so I strengthened the definition: Even after "conjuncting" with any other formula, both have or both don't have a "true" in their truth tables. Finally, the question is whether the formulae in A are equivalent to the formulae in B. That is, whether adding the connector enlarges the group also under this definition of equivalence. עברית (talk) 18:39, 8 March 2018 (UTC)[reply]

I misinterpreted what you said before, but I'm still not entirely sure I understand the difference between A and B or what difference "conjuncting" with other expressions makes. In any case, maybe you should start with the simplest non-trivial case. I think here it would be m=3, n=0, φ(p, q, r) = p→(q∨r), ψb = True. What is the A-equivalent form of φ in this case? Once you have a few examples to go on, you will need to construct some kind of inductive argument, presumably based on the length of φ. This seems like a lot of work, but it would help to define A and B recursively as they do for formal languages. --RDBury (talk) 08:21, 9 March 2018 (UTC)[reply]
Could you clean up the statement you're asking about? You've clearly made mistakes in stating it. For example, phi has arity n, but you've fed it a tuple of length n + k + m.2601:245:C500:754:4831:C848:4E82:B552 (talk) 12:48, 9 March 2018 (UTC)[reply]
Thank you. I fixed this, and cleaned up my statement.
I thought about some solution: we can just copy all of the variables in the disjunction to separated clauses: for example goes to .
However, in such a way the number of variables grows exponentially, so I added a new requirement that k is polynomial in n,m. עברית (talk) 07:06, 10 March 2018 (UTC)[reply]

I've done a few experiments with Mathematica, but have not found the general result.

Am I correct that the set form an orthogonal set of polynomials under ? If so, what is the result for ?--Leon (talk) 21:19, 7 March 2018 (UTC)[reply]

I get
which is non-zero. The integral is zero if one of and is odd and one even, since the integrand is then odd in . --catslash (talk) 12:38, 8 March 2018 (UTC)[reply]
I am not sure why you think that the first derivatives of associated Legendre polynomials form an orthogonal set? And what is n?Ruslik_Zero 20:25, 8 March 2018 (UTC)[reply]
n is the degree, as is clear from the condition that it be no less than the order - so was intended. Orthogonality is not an entirely daft supposition, and could still be the case with the introduction of a suitable weight function. --catslash (talk) 13:15, 9 March 2018 (UTC)[reply]
My goals is a "complete" set of polynomials with the following properties.
  1. .
  2. if and only if .
  3. if and only if .
When I says "complete", I mean being able to express all polynomials for which is true, not all polynomials full stop.
I tried something with sines and cosines instead if polynomials, but whilst it "worked", it lacked other properties that I wanted.--Leon (talk) 13:33, 9 March 2018 (UTC)[reply]
You can read this or this. Ruslik_Zero 19:28, 9 March 2018 (UTC)[reply]