theorem Th18:
  m <> i & m <> k implies ((f,i):=(k,r)).m = f.m
proof
  assume that
A1: m <> i and
A2: m <> k;
  set fik = (f,i):=k;
  thus ((f,i):=(k,r)).m =fik.m by A2,FUNCT_7:32
    .=f.m by A1,FUNCT_7:32;
end;
