As per this paper by Grädel, Kolaitis and Moshe Vardi, they discuss computational complexity of satisfiability problem in $\mathrm{FO^2}$, In order to do this they use Scott's reduction. Which is the fact that any sentence in $\mathrm{FO^2}$ can be reduced to Scott's Normal form in polynomial time. The Scott's Normal form is given as $$\forall x \forall y \alpha(x,y) \land \bigwedge_{i=1}^{m} \forall x \exists y \beta_{i}(x,y) $$ Now, they also say that proving decidability for scott's reduction only proves decidability for $\mathrm{FO}^2$, and then they discuss scott's class for $\mathrm{FO}^2$ with equality.

My question: Is Scott's reduction sound for $\mathrm{FO}^2$ with equality, i.e every sentence can be reduced to the form presented above, where $\alpha$ and $\beta_i$ are all binary predicates, it is not obvious to me from there discussion?


1 Answer 1


Yes. Just employ the formula $\forall{x}\forall{y} \; R(x,y) \leftrightarrow (x=y)$ (for a fresh binary predicate $R$), which allows you to "hide" the equality inside the $\forall\forall$-part of the Scott normal form. Then you proceed as usual.

EDIT: I've noticed that you wrote that $\alpha$ in $\forall{x}\forall{y} \; \alpha$ is a binary predicate. It is not true and obviously not all $\textit{FO}^2$ formulae are reducible to such a form. The aforementioned $\alpha$ should be a quantifier-free formula (line 13 of page 60 of the attached pdf). Maybe this is what made you problems while understanding the Scott reduction.

  • $\begingroup$ Hey Bartosz, thanks for the great answer and added correction as well! Is it possible to have all $\beta_i$ such that they do not occur in $\alpha$ ? $\endgroup$
    – SagarM
    Commented Aug 12, 2020 at 11:38
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    $\begingroup$ Of course, you can have such a form, e.g. \forall x \forall y (a(x) \lor b(y)) \land \forall x \exists y R(x,y) but not all FO2 formulae can be rewritten into this form (you somehow lack the ability to put any constraints on the binary relations). $\endgroup$ Commented Aug 12, 2020 at 14:40
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    $\begingroup$ Writing a little bit more, in the form when betas do not appear in alpha you can always ignore them (e.g. if there is a model then there is also a model with the relations beta_i interpreted as self-loops (id relations)). $\endgroup$ Commented Aug 12, 2020 at 14:46

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