reserve A,B,p,q,r for Element of LTLB_WFF,
  M for LTLModel,
  j,k,n for Element of NAT,
  i for Nat,
  X for Subset of LTLB_WFF,
  F for finite Subset of LTLB_WFF,
  f for FinSequence of LTLB_WFF,
  g for Function of LTLB_WFF,BOOLEAN,
  x,y,z for set,
  P,Q,R for PNPair;

theorem Th6: (SAT M).[n,(con f)/.(len con f)] = 1
  iff for i st i in dom f holds (SAT M).[n,f/.i] = 1
  proof
   set s = SAT M;
   defpred P[Nat] means for f st len f = $1 holds
   s.[n,kon(f)] = 1 iff (for i st i in dom f holds s.[n,f/.i] = 1);
A1: now
      let k be Nat;
      assume A2: P[k];
      thus P[k + 1]
      proof
        let f;
A3:     1 <= k+1 by NAT_1:11;
        set fk = f|k;
        assume
A4:     len f = k+1;
A5:     dom fk c= dom f by RELAT_1:60;
        per cases;
        suppose
A6:       k > 0;
          then A7: 1 <= k by NAT_1:25;
A8:       kon(f) = (con f)/.(k+1) by LTLAXIO2:def 2,A4
          .= ((con f) /. k) '&&' (f /. (k + 1)) by LTLAXIO2:7,A7, NAT_1:16,A4;
A9:       k < len f by NAT_1:16,A4;
          then A10: len fk = k by FINSEQ_1:59;
A11:      (con f)/.k = (con (f | k))/.k by A7,A9,LTLAXIO2:13
          .= kon(fk) by A10,LTLAXIO2:def 2,A6;
          hereby
            assume A12: s.[n,kon(f)] = 1;
            then A13: s.[n,kon(fk)] = 1 by A8,LTLAXIO1:7,A11;
            let i;
            assume
A14:        i in dom f;
            then A15: 1 <= i by FINSEQ_3:25;
A16:        i <= len f by A14,FINSEQ_3:25;
            per cases by A16,XXREAL_0:1;
            suppose
              i < len f;
              then i <= k by A4,NAT_1:13;
              then A17: i in dom fk by A10,FINSEQ_3:25,A15;
              then s.[n,fk/.i] = 1 by A2,A10,A13;
              hence s.[n,f/.i] = 1 by FINSEQ_4:70,A17;
            end;
            suppose
              i = len f;
              hence s.[n,f/.i] = 1 by A4, A12,A8,LTLAXIO1:7;
            end;
          end;
          assume
A18:      for i st i in dom f holds s.[n,f/.i] = 1;
          now
            let i;
            assume
A19:        i in dom fk;
            then s.[n,f/.i] = 1 by A5,A18;
            hence s.[n,fk/.i] = 1 by FINSEQ_4:70,A19;
          end;
          then A20: s.[n,((con f) /. k)] = 1 by A2,A10,A11;
          s.[n,(f /. (k + 1))] = 1 by A18, FINSEQ_3:25,A4,A3;
          hence s.[n,kon(f)] = 1 by LTLAXIO1:7,A20,A8;
        end;
        suppose
A21:      k = 0;
          then A22: kon(f) = (con f)/.1 by A4,LTLAXIO2:def 2
          .= f/.1 by LTLAXIO2:6,A4;
          hereby
            assume
A23:        s.[n,kon(f)] = 1;
            let i;
            assume i in dom f;
            then 1 <= i & i <= len f by FINSEQ_3:25;
            hence s.[n,f/.i] = 1 by NAT_1:25, A21,A4, A23, A22;
          end;
          assume
          for i st i in dom f holds s.[n,f/.i] = 1;
          hence s.[n,kon(f)] = 1 by A21,A4,FINSEQ_3:25,A22;
        end;
      end;
    end;
A24: P[0]
     proof
       let f;
       assume len f = 0;
       then f = {};
       hence thesis by LTLAXIO1:6,LTLAXIO2:10;
     end;
     for k being Nat holds P[k] from NAT_1:sch 2(A24,A1);
     then P[len f];
     hence thesis;
   end;
