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 Th5:
  for i being Element of NAT holds
  (SAT M).[i,p 'U' q] = 1 iff
  ex j st j > i & (SAT M).[j,q] = 1 &
  for k st i < k & k < j holds (SAT M).[k,p] = 1
  proof
    let i be Element of NAT;
    set s = SAT M;
    hereby
      assume s.[i,p 'U' q] = 1;
      then consider j being Element of NAT such that
A1:   0 < j & s.[i+j,q] = 1 and
A2:   for k st 1 <= k & k < j holds s.[(i + k),p] = 1 by LTLAXIO1:def 11;
      set m = i+j;
      now
        let k;
        assume that
A3:     i < k and
A4:     k < m;
        reconsider k1 = k-i as Element of NAT by A3,NAT_1:21;
        i + (-i) < k + (-i) by A3,XREAL_1:8;
        then A5: 1 <= k1 by NAT_1:25;
        k + (-i) < m + (-i) by A4,XREAL_1:8;
        then s.[i+k1,p] = 1 by A5,A2;
        hence s.[k,p] = 1;
      end;
      hence
      ex j st j > i & s.[j,q] = 1 & for k st i < k & k < j holds s.[k,p] = 1
      by A1,NAT_1:16;
    end;
    given j such that
A6: j > i and
A7: s.[j,q] = 1 and
A8: for k st i < k & k < j holds s.[k,p] = 1;
    reconsider n = j - i as Element of NAT by A6,NAT_1:21;
A9: now
      let k;
      assume 1 <= k & k < n;
      then k+i < n+i & i < i+ k by XREAL_1:8,NAT_1:16;
      hence s.[i+k,p] = 1 by A8;
    end;
    j + (-i) > i + (-i) & s.[i+n,q] = 1 by A6,XREAL_1:8, A7;
    hence s.[i,p 'U' q] = 1 by A9,LTLAXIO1:def 11;
  end;
