Transcription from SQL to FOL (First Order Logic)

[WWGD]

Hi All,
When a query is made in SQL , it is transcribed into FOL in the back end , and, if the transcription is a wff , the models, if any, are returned as the answer to the query. I have an idea of how the transcription works for basic statements, but, does anyone know the actual " transcription map" for more complicated queries, such as, e.g., joins? Do we use product models in this case, or do we union the models to form new ones?

 

[andrewkirk]

I'll have a go at this, since nobody else has.

If a table in the database is represented by an atomic wff, and we have tables ##p,q## represented by the a ##m##-ary predicate ##P## and a ##n##-ary predicate ##Q## respectively, then I think the Inner Join of the two tables on the first ##r## arguments to both tables, in the same order, will be represented by the following wff with ##m+n-r## free variables:
$$P(x_1,...,x_m)\wedge Q(x_1,..x_r,x_{m+1},...,x_{m+n-r})$$

 

[WWGD]

I'll have a go at this, since nobody else has.

If a table in the database is represented by an atomic wff, and we have tables ##p,q## represented by the a ##m##-ary predicate ##P## and a ##n##-ary predicate ##Q## respectively, then I think the Inner Join of the two tables on the first ##r## arguments to both tables, in the same order, will be represented by the following wff with ##m+n-r## free variables:
$$P(x_1,...,x_m)\wedge Q(x_1,..x_r,x_{m+1},...,x_{m+n-r})$$

Thanks, Andrew, how would the model of the union then be defined, as just a set union of the two models ( the models of the two wffs)?

 

[andrewkirk]

Thanks, Andrew, how would the model of the union then be defined, as just a set union of the two models ( the models of the two wffs)?

I think I'd try the following for

 select x1,x2,...,xr from p UNION select x1,x2,...,xr from q

$$\exists x_{r+1},x_{r+2},...,x_{m+n-r}:\ P(x_1,...,x_m)\vee Q(x_1,...,x_r,x_{m+1},...,x_{m+n-r})$$
This wff has ##r## free variables, corresponding to the ##r## columns in the tables ##p## and ##q## that match.