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Why in first order logic theories are not possible a demonstration with infinite steps?

  1. Oct 7, 2015 #1
    Why in first order logic theories are not possible a demonstration with infinite steps?
     
  2. jcsd
  3. Oct 7, 2015 #2

    MathematicalPhysicist

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    Since sentences (closed wff) are finitely formed.
     
  4. Oct 7, 2015 #3
    I don´t understand your answer. Could it be you most widespread explanation? Sorry for my bad english. I think that transfinite steps in a proof are imposible, cause the induction rules uses one or two hypothesis only (modus ponens and introducing quantizer), but, what about non standard natural numbers from first order logic. Thanks anyway
     
  5. Oct 7, 2015 #4

    MathematicalPhysicist

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    So you meant why do syntactical proofs have finite steps and not infinite steps in their proof; well at some point you need to infer your conclusion, and this occurs after finite steps, otherwise you cannot construct a proof, since the conclusion is the last step in a proof, you cannot construct an infinite steps' proof, since there's no last step where you conclude your conclusion.
    This should also work in non standard analysis.
     
  6. Oct 7, 2015 #5
    Ok. But the problem I see is that demostration could have non standard steps, because for first order logic is impossible let she the non standard models of the a theory in first order language
     
  7. Oct 8, 2015 #6
    No infinite step, only finite steps but infinity pases
     
  8. Oct 8, 2015 #7

    MathematicalPhysicist

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    It's hard to understand you, what do you mean by: infinity pases?
     
  9. Oct 8, 2015 #8
    infinite steps, sorry. But I refer to a demostration in a number non standard natural numbers finite of steps
     
  10. Oct 8, 2015 #9

    Mark44

    Staff: Mentor

    I recognize that English is not your native language, but frankly, what you have written is pretty much incomprehensible, especially "infinity pases" and "a number non standard natural numbers finite of steps". These make no sense.
     
  11. Oct 8, 2015 #10
    Other way: why not a formal demostration in a non standard natural numbers steps?
     
  12. Oct 8, 2015 #11

    Mark44

    Staff: Mentor

    I still don't get what you're asking. What do you mean by "non standard natural numbers steps"?
     
  13. Oct 9, 2015 #12
    In the first order version of Peano axioms, there are the number to count real objects....1,2,3....but there are models for the PA theory in first order that contains non standard numbers that metamathematically are infinite, but in first order language are "finite". There is no way to free of non standard models to theories expressed in first order languages. If we work inside a first order theory, ¿how can we drop the non standards models, cause the adjective "non standard" is inexpressible in first order language theories formulates ?
     
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