Proof of Theorem <xref ref-type="statement" rid="thm1.1">3</xref>.
Let
(26)F=P[f]a, G=fa.
From the conditions of Theorem 3, we know that F and G share 1 IM. From (26), we have
(27)T(r,F)=O(T(r,f))+S(r,f),T(r,G)≤T(r,f)+S(r,f),(28)N¯(r,F)=N¯(r,G)+S(r,f),(29)N¯(r,F)=N¯(r,f)+S(r,f),N¯(r,G)=N¯(r,f)+S(r,f),(30)NE1)(r,1F-1)=NE1)(r,1G-1)+S(r,f),(31)N¯E(2(r,1F-1)=N¯E(2(r,1G-1)+S(r,f),(32)N¯L(r,1F-1)≤N¯(r,1F)+N¯(r,F)+S(r,f),(33)N¯(r,1F-1) =N¯(r,1G-1)+S(r,f) ≤NE1)(r,1F-1)+N¯E(2(r,1F-1) +NL(r,1F-1)+NL(r,1G-1)+S(r,f).

Let H be defined by (17). Suppose that H≢0. By Lemma 14, (18) holds.

From (17) and (28), we have
(34)N(r,H)≤N(2(r,1F)+N(2(r,1G)+N¯(r,G)+NL(r,1F-1)+NL(r,1G-1)+N0(r,1F′)+N0(r,1G′),
where N0(r,1/F′) denotes the counting function corresponding to the zeros of F′ which are not the zeros of F and F-1. Similarly, N0(r,1/G′) is defined.

From the second fundamental theorem, we have
(35)T(r,F)+T(r,G) ≤N¯(r,F)+N¯(r,1F)+N¯(r,G) +N¯(r,1G)+N¯(r,1F-1)+N¯(r,1G-1) -N0(r,1F′)-N0(r,1G′)+S(r,f).
Since F and G share 1 IM, we get from (33):
(36)N¯(r,1F-1)+N¯(r,1G-1) =2NE1)(r,1F-1)+2NL(r,1F-1) +2NL(r,1G-1)+2N¯E(2(r,1F-1).

From this, (18), and (34), we have
(37)N¯(r,1F-1)+N¯(r,1G-1) ≤N(2(r,1F)+N(2(r,1G)+N¯(r,G)+N0(r,1F′) +3NL(r,1G-1)+3NL(r,1F-1)+NE1)(r,1F-1) +2N¯E(2(r,1G-1)+N0(r,1G′)+S(r,f).
It is clear that
(38)NL(r,1F-1)+2NL(r,1G-1) +2N¯E(2(r,1G-1)+NE1)(r,1F-1) ≤N(r,1G-1) ≤T(r,G)+O(1).
Combining (37), and (38), we obtain
(39)N¯(r,1F-1)+N¯(r,1G-1) ≤N(2(r,1F)+N(2(r,1G)+N¯(r,G) +2NL(r,1F-1)+NL(r,1G-1)+T(r,G) +N0(r,1F′)+N0(r,1G′)+S(r,f).
Substituting (39) in (35) and using (28), we obtain
(40)T(r,F)≤3N¯(r,G)+N(r,1F)+N(r,1G)+2NL(r,1F-1)+NL(r,1G-1)+S(r,f).
Using (26) and (19), we get
(41)d¯(P)T(r,f) ≤3N¯(r,G)+(d¯(P)-d_(P))m(r,1f) +N(r,1f)+N(r,1fd¯(P)) +2NL(r,1F-1)+NL(r,1G-1)+S(r,f).

From (16), (20), and (26) we have
(42)2NL(r,1F-1)+NL(r,1G-1) ≤2N(r,1F′)+N(r,1G′) ≤2[N(r,1F)+N¯(r,F)]+N(r,1f) +N¯(r,f)+S(r,f) ≤2N(r,1P[f])+3N¯(r,f) +N(r,1f)+S(r,f) ≤2QN¯(r,f)+2N(r,1fd¯(P)) +2(d¯(P)-d_(P))m(r,1f) +3N¯(r,f)+N(r,1f)+S(r,f) ≤(2Q+3)N¯(r,f)+(2d¯(P)+1)N(r,1f) +2(d¯(P)-d_(P))m(r,1f)+S(r,f).
From (41) and (42), we get
(43)d¯(P)T(r,f) ≤(2Q+6)N¯(r,f)+(2+3d¯(P))N(r,1f) +3(d¯(P)-d_(P))m(r,1f)+S(r,f) ≤(2Q+6)N¯(r,f)+(2+3d_(P))N(r,1f) +3(d¯(P)-d_(P))T(r,1f)+S(r,f)(3d_(P)-2d¯(P))T(r,f) ≤(2Q+6)N¯(r,f)+(2+3d_(P))N(r,1f)+S(r,f) ≤{(2Q+6)(1-Θ(∞,f))+(2+3d_(P))(1-δ(0,f))} ×T(r,f)+S(r,f) ≤{(2Q+3d_(P)+8)-[(2Q+6)Θ(∞,f) +(2+3d_(P))δ(0,f)]}T(r,f)+S(R,f).
Therefore, we have
(44){(2Q+2d¯(P)+8)(2Q+6)Θ(∞,f)+(2+3d_(P))δ(0,f) -(2Q+2d¯(P)+8)}T(r,f)≤S(r,f),
which is a contradiction to our hypothesis (9).

Thus H≡0. By integration, we get from (17) that
(45)1G-1=AF-1+B,
where (A≠0) and B are constants. Thus
(46)G=(B+1)F+(A-B-1)BF+(A-B),F=(B-A)G+(A-B-1)BG-(B+1).
We discuss the following three cases.

Case 1. Suppose that B≠0,-1. From (46), we have
(47)N¯(r,1G-(B+1)/B)=N¯(r,F).
From this and second fundamental theorem, we have
(48)T(r,f)≤T(r,G)+S(r,f)≤N¯(r,G)+N¯(r,1G)+N¯(r,1G-(B+1)/B)+S(r,f)≤N¯(r,G)+N¯(r,1G)+N¯(r,F)+S(r,f)≤2N¯(r,f)+N¯(r,1f)+S(r,f)≤(2Q+6)N¯(r,f)+(2+3d_(P)N(r,1f)+S(r,f)≤{(2Q+3d_(P)+8)-[(2Q+6)Θ(∞,f)+(2+3d_(P))δ(0,f)]}T(r,f)+S(R,f).
Therefore, we have
(49){(2Q+6)Θ(∞,f)+(2+3d_(P))δ(0,f) -(2Q+3d_(P)+7)}T(r,f)≤S(r,f),
which is a contradiction to our hypothesis (9).

Case 2. Suppose that B=0, From (46), we get
(50)G=F+(A-1)A, F=AG-(A-1),
we claim A=1.

If A≠1 from (50), we obtain
(51)N(r,1G-(A-1)/A)=N(r,1F).
From this, second fundamental theorem, and (20), we have
(52)T(r,f)≤T(r,G)+S(r,f)≤N¯(r,G)+N¯(r,1G)+N¯(r,1G-(A-1)/A)+S(r,f)≤N¯(r,G)+N¯(r,1G)+N(r,1F)+S(r,f)≤N¯(r,f)+N(r,1f)+QN¯(r,f)+(d¯(P)-d_(P))m(r,1f)+N(r,1fd¯(P))+S(r,f)≤(Q+1)N¯(r,f)+(1+d_(P))N(r,1f)+(d¯(P)-d_(P))T(r,1f)+S(r,f)(1-d¯(P)+d_(P))T(r,f) ≤(2Q+6)N¯(r,f)+(2+3d_(P))N(r,1f) +S(r,f) ≤{(2Q+3d_(P)+8)-[(2Q+6)Θ(∞,f) +(2+3d_(P))δ(0,f)]}T(r,f)+S(R,f).
Hence, we have
(53){2Q+2d_(P)+d¯(P)+7(2Q+6)Θ(∞,f)+(2+3d_(P))δ(0,f) -(2Q+2d_(P)+d¯(P)+7)}T(r,f)≤S(r,f),
which is a contradiction to our hypothesis (9)

Thus, A=1.

From (50) we have F≡G.

Therefore, we have f≡P[f].

Case 3. Suppose that B=-1, from (46) we have
(54)G=A-F+A+1, F=(1+A)G-AG.
If A≠-1, we obtain from (54) that
(55)N(r,1G-A/(A+1))=N(r,1F).
By the same argument as in Case 2, we obtain a contradiction. Hence, A=-1.

From (54), we get
(56)FG≡1,
that is,
(57)f·P[f]≡a2.
From (57), we have
(58)N(r,f)+N(r,1f)=S(r,f).
Using (54), (57), Lemma 12, and first fundamental theorem, we get
(59)(d¯(P)+1)T(r,f) =T(r,1fd¯(P)+1) =T(r,1fd¯(P)f) =T(r,P[f]fd¯(P)a2)+S(r,f) =m(r,P[f]fd¯(P))+N(r,P[f]fd¯(P))+S(r,f) ≤(d¯(P)-d_(P))m(r,1f)+(d¯(P)-d_(P))N(r,1f) +Q[N¯(r,f)+N¯(r,1f)]+S(r,f) ≤(d¯(P)-d_(P))m(r,1f)+S(r,f) ≤(d¯(P)-d_(P))T(r,1f)+S(r,f).
From this, we have
(60)(d_(P))+1)T(r,f)≤S(r,f),
which is a contradiction. This completes the proof of Theorem 3.