reserve a,b for object, I,J for set;

theorem Th8A:
  for m,n,x,y being bag of I st x divides m & n = (m-'x)+y holds
  x-'(m-'n) = y-'(n-'m)
  proof
    let m,n,x,y be bag of I;
    assume Z0: x divides m;
    assume Z1: n = (m-'x)+y;
    let a; assume a in I;
A1: n.a = (m-'x).a+y.a = (m.a-'x.a)+y.a & x.a <= m.a
    by Z0,Z1,PRE_POLY:def 5,def 6,def 11;
    thus (x-'(m-'n)).a = x.a-'(m-'n).a by PRE_POLY:def 6
    .= x.a-'(m.a-'n.a) by PRE_POLY:def 6
    .= y.a-'(n.a-'m.a) by A1,Th5A
    .= y.a-'(n-'m).a by PRE_POLY:def 6
    .= (y-'(n-'m)).a by PRE_POLY:def 6;
  end;
