reserve A,B,p,q,r,s for Element of LTLB_WFF,
  i,j,k,n for Element of NAT,
  X for Subset of LTLB_WFF,
  f,f1 for FinSequence of LTLB_WFF,
  g for Function of LTLB_WFF,BOOLEAN;

theorem (for i be Nat st i in dom f holds {}LTLB_WFF |- p => f/.i) implies
  {} LTLB_WFF |- p => (con f)/.(len con f)
 proof
   assume
A1: for i be Nat st i in dom f holds {}l |- p => f/.i;
    per cases;
    suppose
A2:   len f = 0;
      p => TVERUM is ctaut by Th22;
      then A3: p => TVERUM in LTL_axioms by LTLAXIO1:def 17;
      f = {} by A2;
      hence thesis by A3,Th10,LTLAXIO1:42;
    end;
    suppose
A4:   len f > 0;
      defpred P3[Nat] means $1 <= len f implies{}l |- p => ((con f)/.$1);
A5:   now
        let k being non zero Nat such that
A6:     P3[k];
        thus P3[k+1]
        proof
          set a = (con f)/.k, b = f/.(k+1);
          assume
A7:       k+1 <= len f;
          1 <= k by NAT_1:25;then
A8:       (con f)/.(k+1) = (con f)/.k '&&' f/.(k+1) by A7,NAT_1:13,Th7;
          1 <= k+1 by NAT_1:25;
          then A9: {}l |- p => f/.(k+1) by FINSEQ_3:25,A7,A1;
          p => a => (p => b => (p => (a '&&' b))) is ctaut by Th40;then
          p => a => (p => b => (p => (a '&&' b))) in LTL_axioms
          by LTLAXIO1:def 17;then
          {}l |- p => a => (p => b => (p => (a '&&' b))) by LTLAXIO1:42;then
          {}l |- p => b => (p => (a '&&' b)) by LTLAXIO1:43, A6, A7,NAT_1:13;
          hence {}l |- p => (con f)/.(k+1) by LTLAXIO1:43,A9,A8;
        end;
      end;
A10:  P3[1]
      proof
        assume
A11:    1 <= len f;
        then {}l |- p => f/.1 by FINSEQ_3:25,A1;
        hence thesis by A11,Th6;
      end;
A12:  for k being non zero Nat holds P3[k] from NAT_1:sch 10(A10,A5);
      len f = len con f by A4,Def2;
      hence {}l |- p => kon(f) by A12, A4;
    end;
  end;
