Given FOL=Turing machines, why is SOL different than FOL?

Question

[1] Every SOL (second order logic axiom system) has a corresponding Turing machine that verifies SOL statements, given a proof and axioms. (If this weren't the case, how could we be sure that our SOL proofs are correct?)

[2] FOL = Turing machines

[3] From [1] and [2] every SOL has a corresponding FOL that verifies its proofs (meaning for each SOL axiom system, there is a FOL that verifies that a (SOL proof) proves a (SOL statement)).

How then can SOL be really different than FOL? Is SOL just FOL in disguise? Can we re-formulate SOL as FOL?

FOL and SOL are the same syntactically but different semantically. — Oscar Cunningham, Apr 22 '20 at 07:41
@OscarCunningham That's not true - SOL has a whole additional class of variable symbols (the second-order variables). Every FOL sentence is an SOL sentence, but not conversely. — Noah Schweber, Apr 25 '20 at 05:09
@NoahSchweber I'm thinking that every SOL language can be translated into a FOL language with twice as many sorts. — Oscar Cunningham, Apr 25 '20 at 10:08
@OscarCunningham That's only with the Henkin semantics; the standard semantics for second-order logic is very different (e.g. compactness and Lowenheim-Skolem fail horribly). — Noah Schweber, Apr 25 '20 at 13:58
@NoahSchweber Which is why my original comment was that they were different semantically? — Oscar Cunningham, Apr 25 '20 at 14:18

score 1 · Answer 1 · answered Apr 21 '20 at 21:08

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Short answer: Second order logic with full semantics is not recursively axiomatizable. In other words, a given recursively axiomatized second order proof system (one whose proofs a Turing machine could check) always falls short of capturing all the valid second order logical consequences. That's why SOL is essentially different from FOL, and why e.g. second order Peano arithmetic is essentially stronger than first order Peano arithmetic.

Long answer: look at a standard treatment of SOL such as Stewart Shapiro's classic text in the Oxford Logic Guides -- or, for a headline version, this encyclopedia article.

answered Apr 21 '20 at 21:08

Peter Smith

54,743

"always falls short of capturing all the valid second order logical consequences." Is that equivalent to "there exist SOL statements that are syntactically valid but cannot be proved either way"? – Jesus is Lord Apr 22 '20 at 01:15
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No. It says that for any proof system for SOL there will be some semantically valid propositions that that system cannot prove. – Peter Smith Apr 22 '20 at 06:53

Given FOL=Turing machines, why is SOL different than FOL?

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