theorem Th76:
  a > 0 implies a #R (-c) = 1 / a #R c
proof
  assume
A1: a>0;
  then 1 = a #R (c+-c) by Th71
    .= a #R (-c) * a #R c by A1,Th75;
  then 1 / a #R c = a #R (-c) * (a #R c / a #R c) by XCMPLX_1:74
    .= a #R (-c) * 1 by A1,Lm9,XCMPLX_1:60;
  hence thesis;
end;
