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 Th12: for k,n being Nat holds n <= k implies (con f).n = (con (f|k)).n
  proof
    let k,n be Nat;
    defpred P[Nat] means $1 <= k implies
    (con f).$1 = (con (f|k)).$1;
A1: now
      let i be Nat;
      assume
A2:   P[i];
      thus P[i+1]
      proof
        assume
A3:     i+1 <= k;
        then A4: 1 <= k by NAT_1:25;
A5:     i < k by A3,NAT_1:13;
        per cases;
        suppose
A6:       k <= len f;
          then A7: len (f|k) = k by FINSEQ_1:59;
A8:       i < len (f|k) by A5, FINSEQ_1:59,A6;
          then A9: i < len con (f|k) by Def2;
          i+1 <= len f by A6,A3,XXREAL_0:2;
          then A10: i < len f by NAT_1:13;
          then A11: i < len con f by Def2;
          per cases by NAT_1:25;
          suppose
A12:        i = 0;
            hence (con f).(i+1) = f.1 by Def2, A6, A3
            .= (f|k).1 by A4,FINSEQ_3:112
            .= (con (f|k)).(i+1) by Def2,A12, A7, A3;
          end;
          suppose
A13:        1 <= i;
            1 <= i+1 by XREAL_1:31;
            then A14: i+1 in Seg k by A3;
            k in Seg len f by A6,A4;
            then k in dom f by FINSEQ_1:def 3;
            then A15: f/.(i+1) = (f|k)/.(i+1) by A14,FINSEQ_4:71;
A16:        (con f)/.i = (con f).i by A11,A13,FINSEQ_4:15
            .= (con (f|k))/.i by A9,A13,FINSEQ_4:15,A2, A3,NAT_1:13;
            thus (con f).(i+1) = ((con f)/.i '&&' f/.(i+1)) by Def2,A10,A13
            .= (con (f|k)).(i+1) by Def2,A13,A8,A15,A16;
          end;
        end;
        suppose
          k > len f;
          hence (con f).(i+1) = (con (f|k)).(i+1) by FINSEQ_1:58;
        end;
      end;
    end;
     (con f).0 = 0 .= (con (f|k)).0;
     then
A17: P[0];
     for n being Nat holds P[n] from NAT_1:sch 2(A17,A1);
     hence thesis;
   end;
