theorem Th28:
  K is having_valuation & v.a < v.b implies v.a = v.(a+b)
  proof
    assume
A1: K is having_valuation & v.a < v.b;
A2: min(v.a,v.b) = v.a by A1,XXREAL_0:def 9;
A3: v.a <= v.(a+b) by A2,A1,Th27;
A4: a = a + 0.K by RLVECT_1:def 4;
A5: 0.K = b - b by RLVECT_1:15;
A6: a = (a + b) - b by A4,A5,RLVECT_1:28
    .= (a+b)+ -b;
A7: v.(-b) = v.b by A1,Th20;
A8: min(v.(a+b),v.b) <=v.a by A6,A7,A1,Th27;
    then min(v.(a+b), v.b) = v.(a+b) by A1,XXREAL_0:def 9;
    hence thesis by A3,A8,XXREAL_0:1;
  end;
