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 Th19: (VAL g).((con f)/.(len con f)) = 1 iff
  for i be Nat st i in dom f holds (VAL g).(f/.i) = 1
  proof
    set v = VAL g;
    defpred P[Nat] means $1 <= len f implies v.kon(f|$1) = 1;
    hereby
      assume
A1:   v.kon(f) = 1;
      given i be Nat such that
A2:   i in dom f and
A3:   not v.(f/.i) = 1;
A4:   i <= len f by A2,FINSEQ_3:25;
      f/.i = f.i & 1 <= i by PARTFUN1:def 6,A2,FINSEQ_3:25;then
      f = ((f | (i -' 1)) ^ <*(f/.i)*>) ^ (f /^ i) by A4,FINSEQ_5:84;then
A5:   v.kon(f) = v.kon((f | (i -' 1)) ^ <*(f/.i)*>) '&' v.kon(f /^ i) by Th17
      .= v.kon(f | (i -' 1)) '&' v.kon(<*(f/.i)*>) '&' v.kon(f /^ i) by Th17
      .= v.kon(f | (i -' 1)) '&' v.kon(f /^ i) '&' v.kon(<*(f/.i)*>);
      v.kon(<*f/.i*>) = v.(f/.i) by Th11
      .= 0 by A3,XBOOLEAN:def 3;
      hence contradiction by A5,A1;
    end;
    assume
A6: for i be Nat st i in dom f holds v.(f/.i) = 1;
A7: now
      let k being Nat;
      assume
A8:   P[k];
      thus P[k+1]
      proof
A9:     1 <= k+1 by NAT_1:11;
        assume
A10:    k+1 <= len f;
        then f|(k+1) = (f|k)^<*(f.(k + 1))*> by NAT_1:13,FINSEQ_5:83
        .= (f|k)^<*(f/.(k + 1))*> by FINSEQ_4:15, NAT_1:11,A10;
        hence
        v.kon(f|(k+1)) = v.kon(f|k) '&' v.kon(<*(f/.(k + 1))*>) by Th17
        .= v.kon(f|k) '&' v.(f/.(k + 1)) by Th11
        .= 1 by A6, A9,FINSEQ_3:25,A10,A8, NAT_1:13;
      end;
    end;
A11: P[0] by Th10,Th4;
A12: for k be Nat holds P[k] from NAT_1:sch 2(A11,A7);
     f = f|len f by FINSEQ_1:58;
     hence v.kon(f) = 1 by A12;
   end;
