Logic: Lecture 11, 8 September 2015 ----------------------------------- A complete axiomatization (A1) All tautologies of PC (A2a) x = x (A2b) x = y => phi(x) = phi(y), where phi(u) is an atomic formula (A3) phi(t) => Ex phi(x) alpha, alpha => beta (MP) -------------------- beta phi(x) => psi (G) -------------------- x not in FV(psi) Ex phi(x) => psi (A1 at the level of prime formulas) Define X |- phi as usual. Can derive equality axioms (see notes) Soundness: If X |- phi then X |= phi Proof: By induction. Check that G preserves validity Assume phi(x) => psi is valid. For every I = (M,sigma), I |= phi(x) => psi Let I' = (M',sigma') such that I' |= Ex phi(x). Then, for some a in S', I'[x |-> a] |= phi(x). Since phi(x) => psi is valid, I'[x |-> a] |= psi. But x not in FV(psi), so I' |= psi as well. To show completeness First use PC and (G) to show that the following hold (1) If X |- phi -> psi and X |- ~phi -> psi then |- psi (2) If X |- (phi -> theta) -> psi then X |- ~phi -> psi and X |- theta -> psi (3) If x not in FV(psi) and X |- (Ey phi(y) -> phi(x)) -> psi, then X |- psi Proof of (3) Suppose X |- (Ey phi(y) -> phi(x)) -> psi, By (2), X |- ~Ey phi(y) -> psi, X |- phi(x) -> psi X |- phi(x) via (G) plus renaming gives X |- Ey phi(y) -> psi Use (A3) of PC to get X |- psi Now we have Completeness : If X |= phi then X |- phi Proof: X u {~phi} not satisfiable. Some finite subset Y of X U Phi_H U Phi_Q U Phi_Eq such that Y U {~phi} not satisfiable. List Y as {a1,..,am,b1,..,bk} where ai are from X, Phi_Q and Phi_Eq and b1 are from Phi_H in decreasing order of rank. |= (a1 -> (a2 -> .... (b1 -> (b2 -> ... ) -> phi) Obtain ai', bj' by replacing witnessing constants by distinct variables. |= (a1' -> (a2' -> .... (b1' -> (b2' -> ... ) -> phi') But phi' = phi, since phi is in L. Hence, by completeness of PC |- (a1' -> (a2' -> .... (b1' -> (b2' -> ... ) -> phi) ai' are from X (so ai' = ai) or from Phi_Q (axiom) or Phi_Eq (derivable), so eliminate by MP to get |- (b1' -> (b2' -> ... ) -> phi) Apply previous result part (3) m times to eliminate the bi's. ---------------------------------------------------------------------- Quick introduction to sequent calculus ----------------------------------------------------------------------