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 Th25: for P be consistent complete PNPair,Q be Element of compn P st
A 'U' B in rng P`1 holds (B in rng Q`1 or A in rng Q`1 & A 'U' B in rng Q`1)
proof
  let P be consistent complete PNPair,Q be Element of compn P;
  set aub = A 'U' B,nb = 'not' B, na = 'not' A,nu = 'not' aub;
  assume aub in rng P`1;
  then A1: untn(A,B) in rng Q`1 by Th22;
  then A2: untn(A,B) in rng Q by XBOOLE_0:def 3;
  then A3: A in rng Q by Th8;
A4: aub in rng Q by A2,Th8;
A5: {}l |- Q^ => untn(A,B) by LTLAXIO3:28,A1;
    Q^ => untn(A,B) => (Q^ => (na '&&' nb) => ('not' (Q^))) is ctaut
    by LTLAXIO2: 50;then
    Q^ => untn(A,B) => (Q^ => (na '&&' nb) => ('not' (Q^))) in LTL_axioms
    by LTLAXIO1:def 17;then
    {}l |- Q^ => untn(A,B) => (Q^ => (na '&&' nb) => ('not' (Q^)))
    by LTLAXIO1:42;
    then A6: {}l |- Q^ => (na '&&' nb) => ('not' (Q^)) by LTLAXIO1:43,A5;
    Q^ => untn(A,B) => (Q^ => (nb '&&' nu) => ('not' (Q^))) is ctaut
    by LTLAXIO2:51;then
    Q^ => untn(A,B) => (Q^ => (nb '&&' nu) => ('not' (Q^))) in LTL_axioms
    by LTLAXIO1:def 17;then
    {}l |- Q^ => untn(A,B) => (Q^ => (nb '&&' nu) => ('not' (Q^)))
    by LTLAXIO1:42;
    then A7: {}l |- Q^ => (nb '&&' nu) => ('not' (Q^)) by LTLAXIO1:43,A5;
    assume that
A8: not B in rng Q`1 and
A9: not A in rng Q`1 or not aub in rng Q`1;
    B in rng Q by A2,Th8;
    then B in rng Q`2 by A8,XBOOLE_0:def 3;
    then A10: {}l |- Q^ => nb by LTLAXIO3:29;
    per cases by A9;
    suppose not A in rng Q`1;
      then A in rng Q`2 by A3,XBOOLE_0:def 3;
      then {}l |- Q^ => na by LTLAXIO3:29;
      then {}l |- Q^ => (na '&&' nb) by LTLAXIO2:52,A10;
      hence contradiction by LTLAXIO1:43,A6,LTLAXIO3:def 10;
    end;
    suppose not aub in rng Q`1;
      then aub in rng Q`2 by XBOOLE_0:def 3,A4;
      then {}l |- Q^ => nu by LTLAXIO3:29;
      then {}l |- Q^ => (nb '&&' nu) by LTLAXIO2:52,A10;
      hence contradiction by LTLAXIO1:43,A7,LTLAXIO3:def 10;
    end;
  end;
