
theorem Th14:
for F being Field
for m being Ordinal
for p,q being Polynomial of F holds Poly(m,p+q) = Poly(m,p) + Poly(m,q)
proof
let F be Field, m be Ordinal, p,q be Polynomial of F;
set r1 = Poly(m,p), r2 = Poly(m,q), r = Poly(m,p+q),
    n = card(nonConstantPolys F);
I: dom r = Bags n by FUNCT_2:def 1 .= dom(r1+r2) by FUNCT_2:def 1;
now let o be object;
  assume o in dom r;
  then reconsider b = o as bag of n;
  per cases;
  suppose support b = {};
    then for i being object st i in n holds b.i = {} by PRE_POLY:def 7;
    then H: b = EmptyBag n by PBOOLE:6;
    then r.b = (p+q).0 by defPg
            .= p.0 + q.0 by NORMSP_1:def 2
            .= r1.b + q.0 by H,defPg
            .= r1.b + r2.b by H,defPg
            .= (r1+r2).b by POLYNOM1:15;
    hence r.o = (r1 + r2).o;
    end;
  suppose H: support b = {m};
    then r.b = (p+q).(b.m) by defPg
            .= p.(b.m) + q.(b.m) by NORMSP_1:def 2
            .= r1.b + q.(b.m) by H,defPg
            .= r1.b + r2.b by H,defPg
            .= (r1+r2).b by POLYNOM1:15;
    hence r.o = (r1 + r2).o;
    end;
  suppose H: support b <> {} & support b <> {m};
    then r.b = 0.F + 0.F by defPg
            .= r1.b + 0.F by H,defPg
            .= r1.b + r2.b by H,defPg
            .= (r1+r2).b by POLYNOM1:15;
    hence r.o = (r1 + r2).o;
    end;
  end;
hence thesis by I;
end;
