reserve a,b,c for boolean object;
reserve p,q,r,s,A,B,C for Element of LTLB_WFF,
        F,G,X,Y for Subset of LTLB_WFF,
        i,j,k,n for Element of NAT,
        f,f1,f2,g for FinSequence of LTLB_WFF;
reserve M for LTLModel;

theorem Th19:
  (SAT M).[n,(B 'U' C)=>(('X' C)'or'('X'(B '&&'(B 'U' C))))]=1
 proof
  set sm=SAT M;
  A1: now assume that
    A2: sm.[n,(('X' C)'or'('X'(B '&&'(B 'U' C))))]=0 and
    A3: sm.[n,(B 'U' C)]=1;
   consider i such that
    A4: 0<i and
    A5: sm.[n+i,C]=1 and
    A6: for j st 1<=j & j<i holds sm.[n+j,B]=1 by A3,Def11;
   A7: not sm.[n,'X' C]=1 by A2,Th8;
   not sm.[n,'X'(B '&&'(B 'U' C))]=1 by A2,Th8;
   then sm.[n,'X'(B '&&'(B 'U' C))]=0 by XBOOLEAN:def 3;
   then A8: sm.[n+1,B '&&'(B 'U' C)]=0 by Th9;
   per cases by A4,NAT_1:25;
   suppose i=1;
    hence contradiction by A5,A7,Th9;
   end;
   suppose A9: i>1;
    A10: now let j;
     assume that
      A11: 1<=j and
      A12: j<i-' 1;
     j+1<i-' 1+1 by A12,XREAL_1:8;
     then A13: j+1<i-1+1 by A12,XREAL_0:def 2;
     1<=j+1 by A11,NAT_1:19;
     then sm.[n+(j+1),B]=1 by A6,A13;
     hence sm.[n+1+j,B]=1;
    end;
    A14: not sm.[n+1,B]=1 or not sm.[n+1,B 'U' C]=1 by A8,Th7;
    A15: i+(-1)>1+(-1) by A9,XREAL_1:8;
    n+i=n+1+(i-1)
     .=n+1+(i-' 1) by A15,XREAL_0:def 2;
    hence contradiction by A5,A6,A9,A15,A10,A14,Def11;
   end;
  end;
  sm.[n,(('X' C)'or'('X'(B '&&'(B 'U' C))))]=0 or sm.[n,(('X' C)'or'('X'(B
'&&'(B 'U' C))))]=1 by XBOOLEAN:def 3;
  then sm.[n,(B 'U' C)]=>sm.[n,(('X' C)'or'('X'(B '&&'(B 'U' C))))]=1 by A1,
XBOOLEAN:def 3;
  hence thesis by Def11;
 end;
