
theorem ineq2:
for R being preordered non degenerated Ring,
    P being Preordering of R,
    a being Element of R
holds (-abs(P,a) <=P, a & a <=P, abs(P,a)) iff a is P-ordered
proof
let R be preordered non degenerated Ring, P be Preordering of R,
    a be Element of R;
hereby assume AS: -abs(P,a) <=P, a & a <=P, abs(P,a);
   now assume not a is P-ordered;
     then B: --1.R <=P, a & a <=P, -1.R by AS,av00;
     0.R <P, 1.R & -1.R <P, 0.R by c20;
     then -1.R <P, 1.R by c2,c3;
     then a <P, 1.R by B,c2,c3;
     hence contradiction by B,c2;
     end;
   hence a is P-ordered;
   end;
X: P + P c= P by REALALG1:def 14;
now assume a in P \/ -P;
  then per cases by XBOOLE_0:def 3;
  suppose A: a in P;
    then B: abs(P,a) = a by defa;
    then a + abs(P,a) in P + P by A;
    hence -abs(P,a) <=P, a & a <=P, abs(P,a) by X,B,c1;
    end;
  suppose a in -P;
    then A1: abs(P,a) = -a & -a in --P by defa;
    then B: - abs(P,a) = --a & -a in P;
    abs(P,a) + -a in P + P by A1;
    hence -abs(P,a) <=P, a & a <=P, abs(P,a) by X,B,c1;
    end;
  end;
hence thesis;
end;
