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;
reserve T for pnptree of P,t for Node of T;

theorem Th30:
  for P be consistent complete PNPair,T be pnptree of P,t be path of T st
  A 'U' B in rng (T.(t.i))`2 holds for j st j > i holds
  (B in rng (T.(t.j))`2 or ex k st i < k & k < j & A in rng (T.(t.k))`2)
  proof
    set aub = A 'U' B;
    let P be consistent complete PNPair,T be pnptree of P,t be path of T;
    assume
A1: aub in rng (T.(t.i))`2;
    given j such that
A2: j > i and
A3: not B in rng (T.(t.j))`2and
A4: for k st i < k & k < j holds not A in rng (T.(t.k))`2;
A5: j >= i+1 by A2,NAT_1:13;
    per cases by A5,XXREAL_0:1;
    suppose j = i+1;
      then t.j in succ (t.i) by Def13;
      then ex n being Nat st t.j = (t.i)^<*n*> & (t.i)^<*n*> in dom T;
      then T.(t.j) in compn T.(t.i) by Th26;
      hence contradiction by Th24,A1,A3;
    end;
     suppose
A6:    j > i+1;
       i+1 >= 1 by NAT_1:11;
       then reconsider j1 = j - 1 as Element of NAT by XXREAL_0:2,A6,NAT_1:21;
       defpred P[Nat] means
       i < $1 & $1 < j implies aub in rng (T.(t.$1))`2;
       j = j1 + 1;
       then t.j in succ (t.j1) by Def13;
       then ex n being Nat st t.j = (t.j1)^<*n*> & (t.j1)^<*n*> in dom T;
       then A7: (T.(t.j)) in compn T.(t.j1) by Th26;
A8:    now
         let k be Nat;
         set k1 = k + 1;
         assume
A9:      P[k];
         thus P[k1]
         proof
           assume that
A10:       i < k1 and
A11:       k1 < j;
A12:       i <= k by NAT_1:13,A10;
           per cases by A12,XXREAL_0:1;
           suppose
A13:         i < k;
             set Pk1 = T.(t.k1);
             t.k1 in succ (t.k) by Def13;
             then ex n being Nat st t.k1 = (t.k)^<*n*> & (t.k)^<*n*> in dom T;
             then Pk1 in compn T.(t.k) by Th26;then
             A in rng Pk1`2 or aub in rng Pk1`2
             by Th24,A11,NAT_1:16,XXREAL_0:2,A13,A9;
             hence aub in rng (T.(t.k1))`2 by A4,A10,A11;
           end;
           suppose
A14:         i = k;
             set Pk1 = T.(t.k1);
             t.k1 in succ (t.i) by A14,Def13;
             then ex n being Nat st t.k1 = (t.i)^<*n*> & (t.i)^<*n*> in dom T;
             then Pk1 in compn T.(t.i) by Th26;
             then A in rng Pk1`2 or aub in rng Pk1`2 by Th24,A1;
             hence aub in rng (T.(t.k1))`2 by A4,A10,A11;
          end;
        end;
      end;
A15:  j + (-1) < j & j+(-1) > i+1+(-1) by XREAL_1:30, A6,XREAL_1:8;
A16:  P[0];
      for k being Nat holds P[k] from NAT_1:sch 2(A16,A8);
      then aub in rng (T.(t.j1))`2 by A15;
      hence contradiction by Th24,A7,A3;
    end;
  end;
