Proving the propositional tableau sound and complete

1.1k Views Asked by At

There is something fundamental that stops me from understanding the proofs for the propositional tableau.

  • (1) soundness proves that all theorems that can be proved are valid
  • (2) completeness proves that all the valid statements are theorems.

When proving soundness and completeness of the tableau:

  • (3) For soundness: given a fully expanded open branch with root $A$, then there is a $v \vDash A$ [i.e.$A$ is satisfiable -- MA].
  • (4) For completeness: given a tableau with a [should be every -- PS] branch closed, then $A$ is unsatisfiable.

Now, I understand soundness and completeness on their own, and the practical proofs that we do for (3) and (4). But I can’t understand why we prove exactly those statements and I can’t draw the relation between (1) to (3) and (2) to (4). Why would those make it sound and complete?

Instead of (4) I was expecting the inverse of (3) like: given a $v \vDash A$, then the tree has an open branch.

Can someone clarify on how we can prove it sound and complete and why this would do it?

1

There are 1 best solutions below

14
On

It is the other way about: you use (4) to prove (1), and (3) to prove (2)! For remember, a wff $\varphi$ is declared proved by a tableau if the tree stating with the negation of $\varphi$ closes.

Thus, for soundness (1): We want to show that if a tableau starting with $\neg\varphi$ closes, so the tableau declares $\varphi$ to be proved, then $\varphi$ really is valid.

By the corrected version of (4), if a tree starting $\neg\varphi$ closes, then $\neg\varphi$ is unsatisfiable, so $\varphi$ is always true, i.e. is valid. QED.

For completeness (2): We want to show that if $\varphi$ is valid, a tableau starting $\neg\varphi$ closes. That is to say, contraposing, if some tableau starting $\neg\varphi$ stays open, then $\varphi$ is not valid.

But by (3) if some tableau starting $\neg\varphi$ stays open, i.e. has an open branch, there is a valuation which makes $\neg\varphi$ true, so $\varphi$ is not valid. QED.