Logic: Lecture 11, 8 September 2015
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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.

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Quick introduction to sequent calculus

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