reserve ADG for Uniquely_Two_Divisible_Group;
reserve a,b,c,d,a9,b9,c9,p,q for Element of ADG;
reserve x,y for set;
reserve AS for non empty AffinStruct;

theorem Th15:
  (ex a,b being Element of ADG st a<>b) implies (ex a,b being
Element of AV(ADG) st a<>b) & (for a,b,c being Element of AV(ADG) st a,b // c,c
holds a=b) & (for a,b,c,d,p,q being Element of AV(ADG) st a,b // p,q & c,d // p
  ,q holds a,b // c,d) & (for a,b,c being Element of AV(ADG) ex d being Element
of AV(ADG) st a,b // c,d) & (for a,b,c,a9,b9,c9 being Element of AV(ADG) st a,b
// a9,b9 & a,c // a9,c9 holds b,c // b9,c9) & (for a,c being Element of AV(ADG)
  ex b being Element of AV(ADG) st a,b // b,c) & (for a,b,c,b9 being Element of
AV(ADG) st a,b // b,c & a,b9 // b9,c holds b = b9) & for a,b,c,d being Element
  of AV(ADG) st a,b // c,d holds a,c // b,d
proof
  set A = AV(ADG);
  assume ex a,b being Element of ADG st a<>b;
  hence ex a,b being Element of A st a<>b;
A1: for a9,b9,c9,d9 being Element of A for a,b,c,d st a=a9 & b=b9 & c = c9 &
  d=d9 holds (a,b ==> c,d iff a9,b9 // c9,d9)
  by ANALOAF:def 2;
  thus for a,b,c being Element of A st a,b // c,c holds a=b
  proof
    let a,b,c be Element of A such that
A2: a,b // c,c;
    reconsider a9=a,b9=b,c9 = c as Element of ADG;
    a9,b9 ==> c9,c9 by A1,A2;
    hence thesis by Th8;
  end;
  thus for a,b,c,d,p,q being Element of A st a,b // p,q & c,d // p,q holds a,b
  // c,d
  proof
    let a,b,c,d,p,q be Element of A;
    reconsider a9=a,b9=b,c9 = c,d9=d,p9=p,q9=q as Element of ADG;
    assume a,b // p,q & c,d // p,q;
    then a9,b9 ==> p9,q9 & c9,d9 ==> p9,q9 by A1;
    then a9,b9 ==> c9,d9 by Th9;
    hence thesis by A1;
  end;
  thus for a,b,c being Element of A ex d being Element of A st a,b // c,d
  proof
    let a,b,c be Element of A;
    reconsider a9=a,b9=b,c9 = c as Element of ADG;
    consider d9 being Element of ADG such that
A3: a9,b9 ==> c9,d9 by Th10;
    reconsider d = d9 as Element of A;
    take d;
    thus thesis by A1,A3;
  end;
  thus for a,b,c,a9,b9,c9 being Element of A st a,b // a9,b9 & a,c // a9,c9
  holds b,c // b9,c9
  proof
    let a,b,c,a9,b9,c9 be Element of A;
    reconsider p=a,q=b,r=c,p9=a9,q9=b9,r9=c9 as Element of ADG;
    assume a,b // a9,b9 & a,c // a9,c9;
    then p,q ==> p9,q9 & p,r ==> p9,r9 by A1;
    then q,r ==> q9,r9 by Th11;
    hence thesis by A1;
  end;
  thus for a,c being Element of A ex b being Element of A st a,b // b,c
  proof
    let a,c be Element of A;
    reconsider a9=a,c9=c as Element of ADG;
    consider b9 being Element of ADG such that
A4: a9,b9 ==> b9,c9 by Th12;
    reconsider b=b9 as Element of A;
    take b;
    thus thesis by A1,A4;
  end;
  thus for a,b,c,b9 being Element of A st a,b // b,c & a,b9 // b9,c holds b =
  b9
  proof
    let a,b,c,b9 be Element of A;
    reconsider a9=a,p=b,c9=c,p9=b9 as Element of ADG;
    assume a,b // b,c & a,b9 // b9,c;
    then a9,p ==> p,c9 & a9,p9 ==> p9,c9 by A1;
    hence thesis by Th13;
  end;
  thus for a,b,c,d being Element of A st a,b // c,d holds a,c // b,d
  proof
    let a,b,c,d be Element of A;
    reconsider a9=a,b9=b,c9=c,d9=d as Element of ADG;
    assume a,b // c,d;
    then a9,b9 ==> c9,d9 by A1;
    then a9,c9 ==> b9,d9 by Th14;
    hence thesis by A1;
  end;
end;
