Proof of Theorem 3.2 Since H(z,w) and Q(z,w) are polynomials in w(z) with no common factors, Lemma 2.2 gives us
\begin{array}{rl}T(r,P(z,w))& =T(r,\frac{Q(z,w)}{H(z,w)})=max\{{deg}_{w}Q,{deg}_{w}H\}T(r,w)+S(r,w)\\ =\mathrm{\Lambda}T(r,w)+S(r,w).\end{array}
(4.1)
On the other hand, by the definition of P(z,w), we get from (2.2)
\begin{array}{rcl}N(r,P(z,w))& \le & {\mathrm{\Lambda}}_{0}N(r,w(z))+\sum _{i=1}^{n}{\mathrm{\Lambda}}_{i}N(r,w(z+{c}_{i}))+S(r,w)\\ =& \mathrm{\Lambda}N(r,w(z))+S(r,w)\end{array}
and from Lemma 2.1
\begin{array}{rcl}m(r,P(z,w))& \le & {deg}_{w}Pm(r,w(z))+m(r,\frac{P(z,w)}{{w}^{{deg}_{w}P}})\\ \le & {deg}_{w}Pm(r,w(z))+S(r,w).\end{array}
Combining (4.1) and the last two inequalities, we obtain
\mathrm{\Lambda}T(r,w)=\mathrm{\Lambda}N(r,w)+{deg}_{w}Pm(r,w)+S(r,w),
which is (\mathrm{\Lambda}{deg}_{w}P)m(r,w)=S(r,w), then m(r,w)=S(r,w) from (3.2). □
Proof of Theorem 3.3 We restrict p and q duo to the reasoning by Grammaticos et al. [6], where P(z,w) and Q(z,w) are polynomials in w with constant coefficients.
Applying Lemma 2.2 to (1.2), we get from Lemma 2.5
\begin{array}{rcl}{deg}_{w}RT(r,w)& \le & 2T(r,w)+T(r,\overline{w})+T(r,\underline{w})+S(r,w)\\ \le & (4+2\u03f5)T(r,w)+S(r,w).\end{array}
Then max\{p,q\}\le 4. Rewriting (1.2) gives
\overline{w}\underline{w}+\overline{w}w+w\underline{w}=\frac{P(z,w){w}^{2}Q(z,w)}{Q(z,w)}=:K(z,w).
(4.2)
Since P(z,w) and Q(z,w) have no common factors, the right side of (4.2) is irreducible. Applying Lemma 2.2 now to (4.2), we get from Lemma 2.5 and Lemma 2.7
\begin{array}{rcl}{deg}_{w}KT(r,w)& \le & T(r,w)+T(r,\overline{w})+T(r,\underline{w})+S(r,w)\\ \le & (3+2\u03f5)T(r,w)+S(r,w),\end{array}
so {deg}_{w}K\le 3. Thus q\le 3. If q=3, the degree of P{w}^{2}Q that was denoted by k would be 5 since p\le 4, a contradiction. Hence, q\le 2, p\le 4, and k\le 3.
If q=2, since the degree of P{w}^{2}Q, k\le 3, we have p=4 and the coefficients of the highest degree of P and Q are identical. □
Proof of Theorem 3.4 We get from (3.3)
where L(w)=\overline{w}\underline{w}+\overline{w}w+w\underline{w}+{w}^{2}, Q(z,w)=(w{a}_{1})(w{a}_{2}), and P(z,w)=(w{h}_{1})(w{h}_{2})(w{h}_{3})(w{h}_{4}). Then {deg}_{w}P=4, {deg}_{w}Q=2, {deg}_{w}L=2, and Λ of L(w) equals 4. It follows from Theorem 3.2 that
and
m(r,L(w))\le m(r,\frac{L(w)}{{w}^{2}})+m(r,{w}^{2})+S(r,w)=S(r,w),
thus N(r,L(w))=T(r,L(w))+S(r,w)=4T(r,w)+S(r,w). In a similar way to the proof of [[4], Lemma 4.2], we have
N(r,\frac{1}{w{a}_{m}})=T(r,w)+S(r,w)
(4.4)
for m=1,2. From Lemma 2.6, given \u03f5>0, there are at most S(r,w) points {z}_{j} where Q({z}_{j},w)={0}^{{k}_{j}}, but where L(w) has a pole of order greater that (1+\u03f5){k}_{j} or less than (1\u03f5){k}_{j} due to poles or zeros of P({z}_{j},w). The combined effect of all such points can be included in the error term, and so we only consider the rest of the zeros of Q in what follows.
For a point {z}_{j} where w({z}_{j})={a}_{m}({z}_{j}), define
L({z}_{j},w)=(\dots ,{z}_{j}1,{z}_{j},{z}_{j}+1,\dots )
to be the longest possible list of points such that each {z}_{j}+2n\in L({z}_{j},w) is a zero of w{a}_{m}, m=1,2, and each {z}_{j}+2n+1\in L({z}_{j},w) is a pole of w.
Suppose that w has more than S(r,w) poles that are not contained in any sequence L({z}_{j},w). Let {N}^{\ast}(r,w) be the integrated counting function of such poles; by assumption we have {N}^{\ast}(r,w)>CT(r,w) for some C>0 in a set of infinite logarithmic measure. By (3.3), {N}^{\ast}(r,L(w))=2{N}^{\ast}(r,w)+S(r,w), and so we get
\begin{array}{rcl}4T(r,w)& =& (N(r,L(w)){N}^{\ast}(r,L(w)))+{N}^{\ast}(r,L(w))+S(r,w)\\ \le & 4(N(r+1,w){N}^{\ast}(r+1,w))+2{N}^{\ast}(r+1,w)+S(r,w)\\ \le & (42C)T(r+1,w)+S(r,w),\end{array}
which implies that {\rho}_{2}(w)\ge 1 by Lemma 2.3. Therefore all except at most S(r,w) poles of w are in some sequence L({z}_{j},w).
We will call the total number of zeros of w{a}_{m} in L({z}_{j},w) divided by the total number of poles of w (both counting multiplicities) the {a}_{m}/\mathrm{pole} ratio of the sequence.
Consider a sequence L({z}_{j},w) that contains only one zero of w{a}_{m}. Then there are one or two poles in that sequence. With one pole we would have w({z}_{j})={a}_{m}({z}_{j})+{0}^{kj} and w({z}_{j}+1)={\mathrm{\infty}}^{{m}_{j}} or w({z}_{j}1)={\mathrm{\infty}}^{{m}_{j}}, where (1\u03f5){k}_{j}<{m}_{j}. If there are two poles, the situation is the same except that now we have L(w)({z}_{j})={\mathrm{\infty}}^{{m}_{j}}. In any case, in such a sequence the {a}_{m}/\mathrm{pole} ratio is at most \frac{1}{1\u03f5}=:\beta <2. We suppose that there are more than S(r,w){z}_{j}, L({z}_{j},w) contains only one zero of w{a}_{m}. Then there are more than S(r,w) sequences L({z}_{j},w) such that N(r,\frac{1}{Q(z,w)})\le \beta N(r+1,w). Hence, we have
N(r,L(w))=N(r,R)=2N(r,w)+N(r,\frac{1}{Q(z,w)})+S(r,w)<4T(r,w),
which is a contradiction, thus all except at most S(r,w) sequences L({z}_{j},w) contain at least two zeros of w{a}_{m}. This means that there must be at least T(r,w)+S(r,w) points {z}_{j} such that w({z}_{j}+1)=\mathrm{\infty} and one of the following holds:
w({z}_{j})={a}_{2}({z}_{j})\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}w({z}_{j}+2)={a}_{1}({z}_{j}+2),
(4.5)
w({z}_{j})={a}_{1}({z}_{j})\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}w({z}_{j}+2)={a}_{2}({z}_{j}+2),
(4.6)
w({z}_{j})={a}_{1}({z}_{j})\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}w({z}_{j}+2)={a}_{1}({z}_{j}+2),
(4.7)
w({z}_{j})={a}_{2}({z}_{j})\phantom{\rule{1em}{0ex}}\text{and}\phantom{\rule{1em}{0ex}}w({z}_{j}+2)={a}_{2}({z}_{j}+2).
(4.8)
Since N(r,1/(w{a}_{m}))=T(r,w)+S(r,w) holds for both choices of m=1,2, exactly one of the following is true:

(i)
Both (4.7) and (4.8) hold \u22d7S(r,w).

(ii)
Both (4.5) and (4.6) hold \u22d7S(r,w), (4.7) and (4.8) hold \u22d6S(r,w).

(iii)
Relation (4.6) holds \u22d7S(r,w), (4.5), (4.7), and (4.8) hold \u22d6S(r,w).

(iv)
Relation (4.5) holds \u22d7S(r,w), (4.6)(4.8) hold \u22d6S(r,w).
In what follows, we will derive some consequences separately for the conditions (i)(iv). We rewrite (3.3) as
\overline{w}w+\overline{w}\underline{w}+w\underline{w}=\frac{{b}_{3}{w}^{3}+{b}_{2}{w}^{2}+{b}_{1}w+{b}_{0}}{(w{a}_{1})(w{a}_{2})},
(4.9)
where {b}_{3}={a}_{1}+{a}_{2}{H}_{3} and {H}_{3}={\sum}_{j=1}^{4}{h}_{j}.
Case (i) holds. Now we have both (4.7) and (4.8) hold \u22d7S(r,w). If (4.7) holds \u22d7S(r,w), we get from (4.9)
\begin{array}{c}w({z}_{j}+2)w({z}_{j})+w({z}_{j}+2)w({z}_{j}+1)+w({z}_{j}+1)w({z}_{j})\hfill \\ \phantom{\rule{1em}{0ex}}=\left(\frac{{b}_{3}{w}^{3}+{b}_{2}{w}^{2}+{b}_{1}w+{a}_{0}}{(w{a}_{1})(w{a}_{2})}\right)({z}_{j}+1),\hfill \end{array}
which means {a}_{1}({z}_{j})+{a}_{1}({z}_{j}+2)={b}_{3}({z}_{j}+1) holds at more than S(r,w) points {z}_{j}, then {\underline{a}}_{1}+{\overline{a}}_{1}={b}_{3}. Similarly, if (4.8) holds \u22d7S(r,w), we get {\underline{a}}_{2}+{\overline{a}}_{2}={b}_{3}. In this case, we have
{b}_{3}={\underline{a}}_{1}+{\overline{a}}_{1}={\underline{a}}_{2}+{\overline{a}}_{2}.
(4.10)
Case (ii) holds. In the same way as above, we have
{b}_{3}={\underline{a}}_{1}+{\overline{a}}_{2}={\underline{a}}_{2}+{\overline{a}}_{1}.
(4.11)
Cases (iii) and (iv) hold. Assume that (iii) holds. As we know, L({z}_{j},w) contains at least two zeros of w{a}_{m}. From (4.6), there are exactly one zero of w{a}_{1} and one zero of w{a}_{2}, otherwise, (4.5), (4.7), and (4.8) will hold \u22d7S(r,w). Then all except at most S(r,w) poles of w must be contained in sequences of the form
({\mathrm{\infty}}^{{l}_{j}},{a}_{1}+{0}^{{k}_{j}},{\mathrm{\infty}}^{{m}_{j}},{a}_{2}+{0}^{{k}_{j+}},{\mathrm{\infty}}^{{l}_{j+}}),
where if {l}_{j\pm}<0, the corresponding endpoint of the sequence is a zero of order {l}_{j\pm}, and if {l}_{j\pm}=0, it is some nonzero finite value. By Lemma 2.6,
(1\u03f5){k}_{j\pm}<{m}_{j}+max\{{l}_{j\pm},0\}<(1+\u03f5){k}_{j\pm}
(4.12)
holds for both choices of the ± sign. Denote
U=(w{a}_{1})(\overline{w}{\overline{a}}_{2}).
(4.13)
Next, we will show that U is a small function with respect to w. From (4.3), we get m(r,U)=S(r,w). From the definition of U and the fact that all but at most S(r,w) poles of w are in sequences of the above form, it follows that if U has more than S(r,w) poles, then there are more than S(r,w) sequences where {l}_{j\pm}>0.
For the sequences with {l}_{j}>0, we may assume that {l}_{j}/{m}_{j}\ge s>0 for all such sequences, otherwise the poles with {l}_{j}>0 will only have a small effect (at most S(r,w)) on N(r,U). The {a}_{m}/\mathrm{pole} ratio for the sequences in consideration is
\frac{{k}_{j}+{k}_{j+}}{{m}_{j}+{l}_{j}+max\{{l}_{j+},0\}}<\frac{1}{1\u03f5}\frac{2{m}_{j}+{l}_{j}+max\{{l}_{j+},0\}}{{m}_{j}+{l}_{j}+max\{{l}_{j+},0\}}
using (4.12). Take d such that
Then d\in (1/2,1). For a fixed j, there exists an {\u03f5}_{j} satisfying
{\u03f5}_{j}<1\frac{2{m}_{j}+{l}_{j}+max\{{l}_{j+},0\}}{2d({m}_{j}+{l}_{j}+max\{{l}_{j+},0\})}.
Define \u03f5={inf}_{j}{\u03f5}_{j}. Since
\begin{array}{c}2d({m}_{j}+{l}_{j}+max\{{l}_{j+},0\})2{m}_{j}{l}_{j}max\{{l}_{j+},0\}\hfill \\ \phantom{\rule{1em}{0ex}}\ge (2d2){m}_{j}+(2d1){l}_{j}\hfill \\ \phantom{\rule{1em}{0ex}}\ge {m}_{j}(2d2+(2d1)s)\hfill \\ \phantom{\rule{1em}{0ex}}>0.\hfill \end{array}
Noting that {m}_{j}\ge 1, then ϵ is well defined. Thus we conclude that if {l}_{j}>0, then in such sequences the {a}_{m}/\mathrm{pole} ratio is at most some 2d<2. We will get a contradiction as the above. If {l}_{j+}>0, we will get a contradiction similarly. Therefore, U\in S(w), and so (3.3) becomes the Riccati difference equation
\overline{w}=\frac{{\overline{a}}_{2}wU{a}_{1}{\overline{a}}_{2}}{w{a}_{1}}.
The same reasoning works for case (iv) as we exchange the roles of {a}_{1} and {a}_{2}.
In the case that {a}_{1}={a}_{2}, the proof is similar. The condition (i) is the only possibility. □
Restriction of {h}_{n}. We claim that for each n=1,2,3,4,
{h}_{n}+{\overline{h}}_{k}=0\phantom{\rule{1em}{0ex}}\text{or}\phantom{\rule{1em}{0ex}}{h}_{n}+{\underline{h}}_{k}=0
(4.14)
for some k\in \{1,2,3,4\}. Following the same reasoning as in the proof of [[4], Lemma 4.2], we have
N(r,\frac{1}{w{h}_{n}})=T(r,w)+S(r,w),
(4.15)
provided {h}_{n} is not a solution of (3.3), i.e., {h}_{n} does not satisfy {h}_{n}+{\overline{h}}_{n}=0. In the following, we may assume that w{h}_{n} has a large number of zeros, or (4.14) holds as desired.
Consider the points {z}_{j} where w({z}_{j})={h}_{n}({z}_{j})+{0}^{{k}_{j}}. From (3.3) we get
w({z}_{j})+w({z}_{j}\pm 1)=0
for all except at most S(r,w) points {z}_{j} and for either or both choices of ±.
We assume the condition (i) or (ii) is true, otherwise, w satisfies a Riccati difference equation. We discuss the following three cases:
Case 1. More than S(r,w) points {z}_{j} satisfy
w({z}_{j})+w({z}_{j}+1)={0}^{{m}_{j+}}.
Then w({z}_{j}+1)={h}_{n}({z}_{j})+{0}^{{m}_{j+}}. By
(w({z}_{j}+2)+w({z}_{j}+1))(w({z}_{j}+1)+w({z}_{j}))=\frac{P({z}_{j}+1)}{Q({z}_{j}+1)},
we have w({z}_{j}+2)=\mathrm{\infty}. If {h}_{n}\ne {\overline{a}}_{m} for m=1,2, then {h}_{n}({z}_{j})\ne {a}_{m}({z}_{j}+1). It follows that w({z}_{j}+3)={a}_{m}({z}_{j}+3) (note the conditions (i) and (ii)). Then from
(w({z}_{j}+3)+w({z}_{j}+2))(w({z}_{j}+2)+w({z}_{j}+1))=\frac{P({z}_{j}+2)}{Q({z}_{j}+2)},
we get {a}_{m}({z}_{j}+3){h}_{n}({z}_{j})=b({z}_{j}+2), which means
{\overline{\overline{\overline{a}}}}_{m}{h}_{n}=\overline{\overline{b}}.
Combining the above equation with (4.10) or (4.11), we have {h}_{n}={\overline{a}}_{m} for m=1 or m=2, which is a contradiction.
Thus
{h}_{n}={\overline{a}}_{m}
(4.16)
holds for m=1 or m=2.
Case 2. More than S(r,w) points {z}_{j} satisfy
w({z}_{j})+w({z}_{j}1)={0}^{{m}_{j}}.
By a similar reasoning to the above, we obtain
{h}_{n}={\underline{a}}_{m}
(4.17)
holds for m=1 or m=2.
Case 3. More than S(r,w) points {z}_{j} satisfy
w({z}_{j})+w({z}_{j}+1)={0}^{{m}_{j+}}.
Also, more than S(r,w) points {z}_{j} satisfy
w({z}_{j})+w({z}_{j}1)={0}^{{m}_{j}}.
In this case, we have both (4.16) and (4.17).
Note the above three cases, all except at most S(r,w) points {z}_{j} are contained in one of forms the sequences of
({h}_{n}({z}_{j})+{0}^{{k}_{j}},{h}_{n}({z}_{j})+{0}^{{m}_{j+}},{\mathrm{\infty}}^{{s}_{j+}},{a}_{m}({z}_{j}+3),\phi ),
(4.18)
(\phi ,{a}_{m}({z}_{j}3),{\mathrm{\infty}}^{{s}_{j}},{h}_{n}({z}_{j})+{0}^{{m}_{j}},{h}_{n}({z}_{j})+{0}^{{k}_{j}}),
(4.19)
({h}_{n}({z}_{j})+{0}^{{t}_{j}},{a}_{m}({z}_{j}+1)+{0}^{{m}_{j}},{\mathrm{\infty}}^{{s}_{j}},{a}_{m}({z}_{j}+3)+{0}^{{l}_{j}},{h}_{n}({z}_{j}+4)+{0}^{{r}_{j}}),
(4.20)
where {s}_{j+}\approx {k}_{j}+{m}_{j+}, {s}_{j}\approx {k}_{j}+{m}_{j}, {s}_{j}\approx {t}_{j}+{m}_{j}\approx {l}_{j}+{r}_{j}, and φ is a pole or a finite value but not the zero of w{h}_{n}. In fact, if φ is the zero of w{h}_{n}, it will be a starting point of another sequence.
If (4.20) holds for more than S(r,w) points, there are more than S(r,w) sequences L({z}_{j},w) such that N(r,\frac{1}{Q(z,w)})\le \beta N(r+1,w), where \beta <2. This is a contradiction. Hence, (4.20) holds at most S(r,w) points. Then almost all the zeros of w{h}_{n} are in the sequences (4.18) and (4.19).
However, noting that {s}_{j+}\approx {k}_{j}+{m}_{j+} and {s}_{j}\approx {k}_{j}+{m}_{j}, we have N(r+2,w)>(1+\u03f5)N(r,\frac{1}{w{h}_{n}})+S(r,w)=(1+\u03f5)T(r,w)+S(r,w), which contradicts with Lemma 2.3.