I understand that second order logic does not have a sound and complete proof system but I was wondering, how this lack of sound and complete proof system affects the usefulness of the SoL.

As far as I understand SoL is a very strong logic that allows us to pin down the structure of Natural numbers. However, not having a sound and complete proof system means that there will be statements in the language that cannot be proved using finite steps via computers as opposed to FoL. Is my understanding correct?

My question is that how not being able to prove things using finite steps affect the usage of SoL? Is it deemed useless because proofs cannot be checked by computers or is it still useful and popular among mathematicians?

  • $\begingroup$ Second-order logic is not sound? $\endgroup$ Jul 3, 2017 at 22:41
  • $\begingroup$ Thanks. What do you mean it is not sound? Do you mean it does not have a sound proof system? If yes the question is that what is the consequence of not being sound? $\endgroup$
    – abk
    Jul 3, 2017 at 22:48
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    $\begingroup$ You were the one who said second-order logic wasn't sound. Generally, the statement is second order logic is not both consistent and complete. Consistency, as far as I understand, is different than soundness. $\endgroup$ Jul 3, 2017 at 22:52
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    $\begingroup$ @MichaelMcGovern abk said that "second-order logic does not have a sound and complete proof system," but that's different from second-order logic not being sound. The thing that's not sound is any purported complete proof system. $\endgroup$ Jul 3, 2017 at 22:56
  • $\begingroup$ @MichaelMcGovern Sorry didnot see the question mark. SoL doesnot have a sound and complete proof system. Yes soundness is different from consistency. I have not seen anywhere saying that SoL is not consistent. I guess consistency is to do with having a model or not and not to do with SoL or FoL. $\endgroup$
    – abk
    Jul 3, 2017 at 22:58


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