In this section, we present some useful lemmas that are needed in the next section. The first two lemmas are well known.

**Lemma 2.1** (13). *Let f be any arithmetic function and n be a positive integer. Then* ${\sum}_{d|n}(f*\mu )(d)=f(n).$.

**Lemma 2.2** (24). *Let m , n be any positive integers and f be a multiplicative function. Then f*(*m*)*f*(*n*) = *f*((*m*, *n*))*f*([*m*, *n*]).

**Lemma 2.3**. *Let g be any arithmetic function and S be gcd closed. Then for any x* ∈ *S*, *we have*

$$\sum _{\begin{array}{c}y|x\\ y\in S\end{array}}{\displaystyle \sum _{\begin{array}{c}d|y,d\nmid z\\ z<y,z\in S\end{array}}g\left(d\right)={\displaystyle \sum _{d|x}g\left(d\right)}}}.$$(2)*Proof*. Clearly, the terms in the sum of the right-hand side of (2) are non-repetitive. Now we show that the terms in the sum of the left-hand side of (2) are non-repetitive. For this purpose, for any *y* ∈ *S* with *y* | *x*, we let *D*(*y*) = {*d* ∈**Z**^{+} : *d* | *y*, *d* ∤ *z*, *z < y*, *z* ∈ *S*}. Claim that *D*(*y*_{1}) ∩ *D*(*y*_{2}) = *ϕ* for any distinct elements *y*_{1} and *y*_{2} in the set *S* satisfying *y*_{1}| *x* and *y*_{2}| *x*. Otherwise, we may let *d* ∈ *D*(*y*_{1}) ∩ *D*(*y*_{2}) . Then *d* | *y*_{1} and *d* | *y*_{2}. So *d* |(*y*_{1}, *y*_{2}). But the assumption that *S* being gcd closed tells us that (*y*_{1}, *y*_{2}) = *y*_{3} for some *y*_{3} ∈ *S*. Hence *d* | *y*_{3}. On the other hand, we have *y*_{3} < *y*_{1} and *y*_{3} < *y*_{2} since *y*_{1} ≠ *y*_{2}. It then follows from *d* ∈ *D*(*y*_{1}) that *d* ∤ *y*_{3}. We arrive at a contradiction. The claim is proved. By the claim we know immediately that the terms in the sum of the left-hand side of (2) are non-repetitive.

For any term *g*(*d*) in the sum of the left-hand side of (2), one has *d* | *y*, *y*| *x* and *y* ∈ *S*. Thus *d* | *x*. This implies that *g*(*d*) is a term in the sum of the right-hand side of (2). To show that the converse is true, for any given positive integer *d* and *x* ∈ *S* with *d* | *x*, we let *I*(*d*, *x*) = {*u : d*| *u*, *u*| *x*, *u* ∈ *S*}. Then *I*(*d*, *x*) ≠ *ϕ* since *x* ∈ *I*(*d*, *x*) and *I*(*d*, *x*) is finite. Let *υ* = min(*I*(*d*, *x*)). Then *υ* | *x*, *υ* ∈ *S* and *d* |υ and *d* ∤ *z* for any *z* ∈ *S* with *z < υ*. It infers that the term *g*(*d*) in the sum of the right-hand side of (2) is also a term in the sum of the left-hand side of (2). So (2) is proved.

This ends the proof of Lemma 2.3. □

**Lemma 2.4**. *Let S be gcd closed. Then for any x* ∈ *S*, $\sum _{\begin{array}{c}y|x\\ y\in S\end{array}}{\alpha}_{S,f}(y)=f(x)}.$

Note that a special case of Lemma 2.3 is due to Beslin and Ligh 3 and a more general form is given in (3.4) of 10.

*Proof*. Letting *g = f* * *μ* in Lemma 2.3 gives us that

$$\sum _{\begin{array}{c}y|x\\ y\in S\end{array}}{\displaystyle \sum _{\begin{array}{c}d|y,d\nmid z\\ z<y,z\in S\end{array}}(f*\mu )(d)}={\displaystyle \sum _{d|x}(f*\mu )(d).}$$Then the desired result follows from the definition of *α*_{S}, _{f} (*d*) and Lemma 2.1. This completes the proof of Lemma 2.4. □

We need the following definition to state Lemma 2.6 below.

**Definition 2.5**. *Let S* = {*x*_{1}, …, *x*_{n}} *be a set of positive integers and S* = {*y*_{1}, …, *y*_{m}} *be the minimal gcd-closed set containing S. Then we define the n × m matrix E*(*S*) = (*e*_{ij}) *by*

$${e}_{ij}:=\{\begin{array}{l}1,if\text{\hspace{0.17em}\hspace{0.17em}}{y}_{j}|{x}_{i},\\ 0,\text{\hspace{0.17em}\hspace{0.17em}}otherwise.\end{array}$$*For* 1 ≤ l ≤ *m , we define E*_{l} (*S*) *to be the n* × (*m* – 1) *matrix obtained from E* (*S*) *by deleting its lth column*.

We can now use the gcd-closed set to describe the structure of the matrix (*f*(*S*)) on any set *S* of positive integers.

**Lemma 2.6**. *Let f be an arithmetic function and S* = {*x*_{i}, …., *x*_{n}} *be a set of distinct positive integers and S* = {*y*_{1}, …. , *y*_{m}} *be the minimal gcd-closed set containing S. Then* (*f*(*S*)) = *E*(*S*) · diag(*α*_{S, f} (*y*_{1}), …, *α*_{S, f} (*y*_{m})) · *E*(*S*)^{T}.

*Proof*. Let *S* = {*x*_{1}, …, *x*_{n}} and ∆ = diag(*α*_{S, f}(*y*_{1}), *…*, *α*_{S, f} (*y*_{m})). Then for any integers *i* and *j* with 1 ≤ *i*, *j* ≤ *n*, we have

$${(E(S)\Delta E{(S)}^{T})}_{ij}={\displaystyle \sum _{k=1}^{m}{e}_{ik}{\alpha}_{\overline{S},f}({y}_{k}){e}_{jk}}={\displaystyle \sum _{\begin{array}{c}{y}_{k\text{\hspace{0.17em}}}|{x}_{i}\\ {y}_{k}|{x}_{j}\end{array}}{\alpha}_{\overline{S},f}({y}_{k})}={\displaystyle \sum _{{y}_{k}|({x}_{i},{x}_{j})}{\alpha}_{\overline{S},f}({y}_{k})}.$$Since *S* is the minimal gcd-closed set containing *S*, one has (*x*_{i}, *x*_{j}) ∈ *S*. Then there exists one element *y*_{h} ∈ *S* such that *y*_{h} = (*x*_{i}, *x*_{j}). It follows that

$${\left(E\left(S\right)\Delta \text{\hspace{0.17em}}E{\left(S\right)}^{T}\right)}_{ij}={\displaystyle \sum _{{y}_{k}|{y}_{h}}{\alpha}_{\overline{S,}f}}\left({y}_{k}\right)$$(3)But Lemma 2.4 together with the fact that *S* being gcd closed implies that

$$\sum _{{y}_{k}|{y}_{h}}{\alpha}_{\overline{S},f}\left(yk\right)}=f\left(yh\right)=f\left({x}_{i},{x}_{j}\right).$$(4)Thus by (3) and (4), one has (*E*(*S*)∆*E*(*S*)^{T})_{ij} = (*f*(*S*))_{ij} as desired. This completes the proof of Lemma 2.6. □

Li 21, Hong 12 and Mattila and Haukkanen 22 made use of the Cauchy-Binet formula to the Smith’s matrices. Now we use this renowned formula to show the following lemma.

**Lemma 2.7**. *Let f be an arithmetic function and S* = {*x*_{1}, …., *x*_{n}} *be a set of n distinct positive integers , and S* = {*y*_{1}, …, *y*_{m}} *be the minimal gcd-closed set containing S. Then*

$$\mathrm{det}\left(f\left(S\right)\right)={\displaystyle \sum _{1\le {k}_{1}\text{<}\dots \text{<}{k}_{n}\le m}{\left(\mathrm{det}E{\left(S\right)}_{\left({k}_{1},\mathrm{...},{k}_{n}\right)}\right)}^{2}}{\displaystyle \prod _{i=1}^{n}{\alpha}_{\overline{S},f\left(y{k}_{i}\right)}},$$(5)*with E*(*S*)_{(k1, …, kn)} *being the* *n* × *n matrix whose columns are the k*_{1}th, … , *k*_{n}th columns of E(*S*).

*Proof*. Let *A* = *E*(*S*) · $\text{diag}\left(\sqrt{{\alpha}_{\overline{S},f}\left({y}_{1}\right)},\dots ,\sqrt{{\alpha}_{\overline{S},f}\left({y}_{m}\right)}\right).$ Then by Lemma 2.6, one has (*f*(*S*)) = *AA*^{T}. Using the Cauchy-Binet formula 8 we get

$$\mathrm{det}\left(f\left(S\right)\right)={\displaystyle \sum _{1\le {k}_{1}\le \mathrm{\dots}\le {k}_{n}\le m}\mathrm{det}{{\rm A}}_{\left({k}_{1},\mathrm{\dots},{k}_{n}\right)}}\cdot \mathrm{det}{{\rm A}}^{T}{}_{\left({k}_{1},\mathrm{\dots},{k}_{n}\right)}={\displaystyle \sum _{1\le {k}_{1}\le \mathrm{\dots}\le {k}_{n}\le m}{\left(\mathrm{det}{{\rm A}}_{\left({k}_{1},\mathrm{\dots},{k}_{n}\right)}\right)}^{2}},$$where *A*(*k*_{1}, …, *k*_{n}) is the *n* × *n* matrix whose columns are the *k*_{1}th, …, *k*_{n}th column of *A*. One can easily check that

$${{\rm A}}_{\left({k}_{1},\mathrm{\dots},{k}_{n}\right)}=E{\left(S\right)}_{\left({k}_{1},\mathrm{\dots},{k}_{n}\right)}\cdot \text{diag}\left(\sqrt{{\alpha}_{{}^{\overline{S},f}}\left(y{k}_{1}\right)},\mathrm{\dots},\sqrt{{\alpha}_{{}^{\overline{S},f}}\left(y{k}_{n}\right)}\right).$$It then follows that

$$\mathrm{det}\left({{\rm A}}_{\left({k}_{1},\mathrm{\dots},{k}_{n}\right)}\right)=\sqrt{{\displaystyle \prod}_{i=1}^{n}{\alpha}_{\overline{S},f}\left(y{k}_{i}\right)}\cdot \mathrm{det}\left(E{\left(S\right)}_{\left({k}_{1},\mathrm{\dots},{k}_{n}\right)}\right).$$So the desired formula (5) follows immediately. This finishes the proof of Lemma 2.7. □

In what follows, we write $S={\bigcup}_{i=1}^{h}{S}_{i}$ with *S*_{i} = {*x*_{i}_{1}, …, *x*_{ini} }(1 ≤ *i* ≤ *h*) being gcd closed and 1 < *x*_{i}_{1} < … < *x*_{ini} and gcd(lcm(*S*_{i}), lcm(*S*_{j})) = 1 for all integers *i* and *j* with 1 ≤ *i* ≠ *j* ≤ *h*. That is,

$$S=\text{\hspace{0.17em}\hspace{0.17em}}\left\{{x}_{11},\mathrm{\dots},{x}_{1{n}_{1}},\mathrm{\dots},{x}_{h}{}_{{}^{{}_{1}}},\mathrm{\dots},{x}_{h{n}_{h}}\right\}.$$(6)Let *S*: = *S* ∪{1} = {*x*_{11},…, *x*_{1n1},…, *x*_{h}_{1},…, *x*_{h}_{nh}, 1}. Clearly *S* is the minimal gcd-closed set containing *S*.

**Lemma 2.8**. *Let S be as in (6) and t be a given integer such that* 1 ≤ *t* ≤ *h*. *Let l*_{t} = *n*_{1} + … + *n*_{t}. *Let n*_{t} ≥ 2. *Then each of the following is true*.

(i) *If* *x*_{t}_{ , nt} –1 *does not divide* x_{tnt}, *then* det(E_{lt} (*S*)) = det(E_{lt} –_{1}(*S* \ {*x*_{t},_{nt}_{ –1}})).

(ii) * If* *x*_{t nt}_{ –1} *divides* *x*_{tnt}, *then*

$$\mathrm{det}\left({E}_{{l}_{t}}\left(S\right)\right)=\mathrm{det}\left({E}_{{l}_{t}-1}\left(S\backslash \left\{{x}_{t,{n}_{{t}^{{}_{-1}}}}\right\}\right)\right)-\mathrm{det}\left({E}_{{l}_{t}-1}\left(S\backslash \left\{{x}_{t,{n}_{t}}\right\}\right)\right).$$*Proof*. Since *S* is as in (6), by the definition of *E*(*S*) we have

$$E\left(S\right)=\left(\begin{array}{ccccc}{E}_{1}& 0& \dots & 0& 1\\ 0& {E}_{2}& \dots & 0& 1\\ \dots & \dots & \dots & \dots & \dots \\ 0& 0& \dots & {E}_{h}& 1\end{array}\right),$$(7)where for 1 ≤ *l* ≤ *h*, one has

$${E}_{l}=\left(\begin{array}{cccccc}1& 0& 0& \mathrm{\dots}& 0& 0\\ 1& 1& 0& \mathrm{\dots}& 0& 0\\ 1& e{\prime}_{32}& 1& \mathrm{\dots}& 0& 0\\ \mathrm{\dots}& \mathrm{\dots}& \mathrm{\dots}& \mathrm{\dots}& \mathrm{\dots}& \mathrm{\dots}\\ 1& {e}_{{n}_{l}-1,2}^{\prime}& {e}_{nl-1,3}^{\prime}& \mathrm{\dots}& 1& 0\\ 1& {e}_{nl,2}^{\prime}& {e}_{nl,3}^{\prime}& \mathrm{\dots}& {e}_{nl,nl-1}^{\prime}& 1\end{array}\right)$$with ${{e}^{\prime}}_{ij}$ (1 ≤ *i*, *j* ≤ *n*_{t}) being defined as

$$e{\prime}_{ij}=\{\begin{array}{l}1,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{if}\text{\hspace{0.17em}\hspace{0.17em}}{x}_{lj}|{x}_{li},\\ 0,\text{\hspace{0.17em}\hspace{0.17em}}\mathrm{otherwise}.\end{array}$$But *E*_{lt}(*S*) is the *l*_{h} × *l*_{h} matrix obtained from *E*(*S*) by deleting its *l*_{t}th column. So one has

$${E}_{l}{}_{t}\left(S\right)=\left(\begin{array}{cccccc}{E}_{1}& \cdots & 0& \cdots & 0& 1\\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ 0& \cdots & {{E}^{\prime}}_{t}& \mathrm{\cdots}& 0& 1\\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ 0& \cdots & 0& \cdots & 0& 1\\ 0& \cdots & \cdots & \cdots & {E}_{h}& 1\end{array}\right)$$where

$$E{\prime}_{t}=\left(\begin{array}{ccccc}1& 0& 0& \mathrm{\dots}& 0\\ 1& 1& 0& \mathrm{\dots}& 0\\ 1& e{\prime}_{32}& 1& \mathrm{\dots}& 0\\ \mathrm{\dots}& \mathrm{\dots}& \mathrm{\dots}& \mathrm{\dots}& \mathrm{\dots}\\ 1& e{\prime}_{{n}_{t}-1,2}& e{\prime}_{{n}_{t}-1,3}& \mathrm{\dots}& 1\\ 1& e{\prime}_{{n}_{t}1,2}& e{\prime}_{{n}_{t},3}& \mathrm{\dots}& e{\prime}_{{n}_{t},{n}_{t}-1}\end{array}\right).$$(i). *x*_{t}, *n*_{t}_{–1}. ∤ *x*_{tnt}. Then one has that ${{e}^{\prime}}_{{n}_{t},{n}_{t}-1}=0.$ Thus the (*l*_{t} – 1)th column of *E*_{lt}(*S*) is $\underset{{l}_{t}-2}{\underbrace{(0,\dots ,0,}}1,\underset{{l}_{h}-{l}_{t}+1}{\underbrace{0,\dots ,0,{)}^{T}}}.$ Then using the Laplace expansion theorem, we obtain that

$$\mathrm{det}\left({E}_{{l}_{t}}\left(S\right)\right)=\mathrm{det}(\left(\begin{array}{cccccc}{E}_{1}& \dots & 0& \dots & 0& 1\\ \dots & \dots & \dots & \dots & \dots & \dots \\ 0& \dots & {E}_{t}^{\u2033}& \dots & 0& 1\\ \dots & \dots & \dots & \dots & \dots & \dots \\ 0& \dots & 0& \dots & 0& 1\\ 0& \dots & 0& \dots & {E}_{h}& 1\end{array}\right)),$$(8)with
$${{E}^{\u2033}}_{t}=\left(\begin{array}{ccccc}1& 0& 0& \cdots & 0\\ 1& 1& 0& \cdots & 0\\ 1& {{e}^{\prime}}_{32}& 1& \cdots & 0\\ \cdots & \cdots & \cdots & \cdots & \cdots \\ 1& {{e}^{\prime}}_{{n}_{t}-2,2}& {{e}^{\prime}}_{{n}_{t}-2,3}& \cdots & 1\\ 1& {{e}^{\prime}}_{{n}_{t},2}& {{e}^{\prime}}_{{n}_{t},3}& \cdots & {{e}^{\prime}}_{{n}_{t},{n}_{t}-2}\end{array}\right).$$

On the other hand, by the definition of *S*, one can easily deduce that *S* \ {*x*_{t}, _{nt –1}} consists of multiple coprime gcd-closed sets and *S* \ {*x*_{t}, _{nt –1}} is the minimal gcd-closed set containing the set *S* \ {*x*_{t}, _{nt –1}}. Hence by the definition of *E*_{lt}_{ – 1}(*S* \ {*x*_{t}, _{nt –1}}), one knows that the right-hand side of (8) is equal to det(*E*_{lt}_{ – 1}(*S* \ {*x*_{t}, _{nt –1}})). So the desired result follows. Part (i) is proved.

(ii). *x*_{t, nt}_{–1}. | *x*_{tnt}. Thus ${{e}^{\prime}}_{{n}_{t},{n}_{t}-1}=1.$ Clearly the (*l*_{t} – 1)th column of *E*_{lt}(*S*) is $\underset{{l}_{t}-2}{\underbrace{(0,\mathrm{\dots}\text{\hspace{0.17em}},0,}}1,1,\underset{{l}_{h}-{l}_{t}}{\underbrace{0,\mathrm{\dots}\text{\hspace{0.17em}},0,{)}^{T}}}.$ Applying the Laplace expansion theorem gives us that

$$\mathrm{det}\left({E}_{{l}_{t}}\left(S\right)\right)=\mathrm{det}(\left(\begin{array}{cccccc}{E}_{1}& \cdots & 0& \cdots & 0& 1\\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ 0& \cdots & {{E}^{\u2033}}_{t}& \cdots & 0& 1\\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ 0& \cdots & 0& \cdots & 0& 1\\ 0& \cdots & 0& \cdots & {E}_{h}& 1\end{array}\right))-\mathrm{det}(\left(\begin{array}{cccccc}{E}_{1}& \cdots & 0& \cdots & 0& 1\\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ 0& \cdots & {{E}^{\u2034}}_{t}& \cdots & 0& 1\\ \cdots & \cdots & \cdots & \cdots & \cdots & \cdots \\ 0& \cdots & 0& \cdots & 0& 1\\ 0& \cdots & 0& \cdots & {E}_{h}& 1\end{array}\right))$$(9)With

$${E}_{t}^{\u2034}=\left(\begin{array}{ccccc}1& 0& 0& \cdots & 0\\ 1& 1& 0& \cdots & 0\\ 1& {e}_{32}^{\prime}& 1& \cdots & 0\\ \cdots & \cdots & \cdots & \cdots & \cdots \\ 1& {e}_{{n}_{t}-2,2}^{\prime}& {e}_{{n}_{t}-2,3}^{\prime}& \cdots & 1\\ 1& {e}_{{n}_{t}-1,2}^{\prime}& {e}_{{n}_{t}-1,3}^{\prime}& \cdots & {e}_{{n}_{t}-1,{n}_{t}-2}^{\prime}\end{array}\right).$$Clearly *S* \ {*x*_{t}, _{nt}} consists of multiple coprime gcd-closed sets and *S* \ {*x*_{t}, _{nt}} is the minimal gcd-closed set containing *S* \ {*x*_{t}, _{nt}}. Thus by the definition of *E*_{lt}_{–1}(*S* \ {*x*_{t}, _{nt}}), we know that the right-hand side of (9) is equal to det(*E*_{lt}_{–1}(*S* \ {*x*_{t},_{nt–1}})) – det(*E*_{lt}_{–1}(*S* \ {*x*_{t}, _{nt}})) So part (ii) is true.

This concludes the proof of Lemma 2.8. □

In ending this section, we show the following relation between ${S}_{f}^{(1)}(T)$ and ${S}_{f}^{\left(2\right)}\left(T\right)$ which is also needed in the proof of Theorem 1.3.

**Lemma 2.9**. *Let f be an arithmetic function and T be a set of distinct positive integers. If f*(*x*) ≠ 0 *for any x* ∈ *T and f*(*x*) = 1 *then one has that* ${S}_{\frac{1}{f}}^{\left(1\right)}\left(T\right){\prod}_{x\in T}\text{\hspace{0.17em}}f{\left(x\right)}^{2}={S}_{f}^{\left(2\right)}\left(T\right).$

*Proof*. Since *f*(*x*) ≠ 0 for any *x* ∈ *T* and *f*(*x*) = 1, it follows that

$$\begin{array}{l}{S}_{\frac{1}{f}}^{\left(1\right)}\left(T\right){\displaystyle \prod}_{x\in T}f{\left(x\right)}^{2}\\ =\left({\displaystyle \sum _{y\in T}{\displaystyle \prod}_{x\in T\backslash \left\{y\right\}}}\left(\frac{1}{f\left(x\right)}-1\right)+{\displaystyle \prod}_{y\in T}\left(\frac{1}{f\left(y\right)}-1\right)\right){\displaystyle \prod}_{x\in T}f{\left(x\right)}^{2}\\ =\left({\displaystyle \sum _{y\in T}{\displaystyle \prod}_{x\in T\backslash \left\{y\right\}}}\frac{1-f\left(x\right)}{f\left(x\right)}-+{\displaystyle \prod}_{y\in T}\frac{1-f\left(y\right)}{f\left(y\right)}\right){\displaystyle \prod}_{x\in T}f{\left(x\right)}^{2}\\ =\left({\displaystyle \sum _{y\in T}f\left(y\right){\displaystyle \prod}_{x\in T\backslash \left\{y\right\}}}\left(1-f\left(x\right)\right)+{\displaystyle \prod}_{y\in T}\left(1-f\left(y\right)\right)\right){\displaystyle \prod}_{x\in T}f\left(x\right)\\ ={\left(-1\right)}^{\left|T\right|+1}\left({\displaystyle \sum _{y\in T}\left(\left(f\left(y\right)-1\right)+1\right){\displaystyle \prod}_{x\in T\backslash \left\{y\right\}}}\left(f\left(x\right)-1\right)+{\left(-1\right)}^{\left|T\right|}{\displaystyle \prod}_{y\in T}\left(f\left(y\right)-1\right)\right){\displaystyle \prod}_{x\in T}f\left(x\right)\\ ={\left(-1\right)}^{\left|T\right|+1}({\displaystyle \sum _{y\in T}{\displaystyle \prod}_{x\in T}\left(f\left(x\right)-1\right)+{\displaystyle \sum _{y\in T}{\displaystyle \prod}_{x\in T\backslash \left\{y\right\}}}}\left(f\left(x\right)-1\right)-{\displaystyle \prod}_{y\in T}\left(f\left(y\right)-1\right){\displaystyle \prod}_{x\in T}f\left(x\right)\\ ={\left(-1\right)}^{\left|T\right|+1}\left(\left({\displaystyle \prod}_{x\in T}\left(f\left(x\right)-1\right)\right)\right){\displaystyle \sum _{y\in T}1}+{\displaystyle \sum _{y\in T}{\displaystyle \prod}_{x\in T\backslash \left\{y\right\}}}\left(f\left(x\right)-1\right)-{\displaystyle \prod}_{y\in T}\left(f\left(y\right)-1\right){\displaystyle \prod}_{x\in T}f\left(x\right)\\ ={\left(-1\right)}^{\left|T\right|+1}({\displaystyle \sum _{y\in T}{\displaystyle \prod}_{x\in T\backslash \left\{y\right\}}\left(f\left(x\right)-1\right)+}\left(\left|T\right|-1\right){\displaystyle \prod}_{y\in T}\left(f\left(y\right)-1\right){\displaystyle \prod}_{x\in T}f\left(x\right)\\ ={S}_{f}^{\left(2\right)}\left(T\right)\end{array}$$as desired. So Lemma 2.9 is proved. □

## Comments (0)