This section is devoted to studying the existence and stability of the equilibria of model (2). In this sense, we shall assume the following threshold parameter:
$$\begin{array}{}{\mathcal{R}}_{0}={\displaystyle \frac{{\beta}_{B}{\beta}_{T}}{{\lambda}_{B}{\mu}_{T}p}.}\end{array}$$

The value of this parameter means that each infected bovine produces
$\begin{array}{}\frac{{\beta}_{T}}{{\mu}_{T}p}\end{array}$
new infected ticks over its expected infectious period, and each infected tick produces
$\begin{array}{}\frac{{\beta}_{B}}{{\lambda}_{B}}\end{array}$
new infected bovines over its expected infectious period.

Contrary to the continues-time case, the (parametric) positive constraints obtained in Lemma 2.1 and Remark 2.2 become fundamental for the proofs of the results in our discrete-time case. Although such results are similar to the continuous case, the proofs need to be performed by using different techniques corresponding to discrete dynamical systems.

#### Proposition 3.1

*System (2) has a disease*-*free equilibrium*
$\begin{array}{}{E}_{1}^{\ast}\end{array}$
= (1,0,0)*for all the values of the (positive) parameters*, *while only if* 𝓡_{0} > 1, *there exists a unique endemic equilibrium*
$\begin{array}{}{E}_{2}^{\ast}=({S}_{{B}_{2}}^{\ast},{I}_{{B}_{2}}^{\ast},{I}_{{T}_{2}}^{\ast})\end{array}$
*in the region Ω*.

#### Proof

A fixed point
$\begin{array}{}{E}^{\ast}=({S}_{B}^{\ast},{I}_{B}^{\ast},{I}_{T}^{\ast})\end{array}$
of model (2) can be obtained by solving the equations below
$$\begin{array}{}\left\{\begin{array}{}\left({\mu}_{B}+{\alpha}_{B}\right)\left(1-{S}_{B}(t)-{I}_{B}(t)\right)+\left(1-{\beta}_{B}{I}_{T}(t)\right){S}_{B}(t)={S}_{B}(t),\\ {\beta}_{B}{S}_{B}(t){I}_{T}(t)+\left(1-{\lambda}_{B}\right){I}_{B}(t)={I}_{B}(t),\\ {\beta}_{T}\left(1-{I}_{T}(t)\right){I}_{B}(t)+(1-{\mu}_{T}p){I}_{T}(t)={I}_{T}(t).\end{array}\right.\end{array}$$(3)

This is equivalent to solve the following system
$$\begin{array}{}\left\{\begin{array}{}\left({\mu}_{B}+{\alpha}_{B}\right)\left(1-{S}_{B}(t)-{I}_{B}(t)\right)-{\beta}_{B}{I}_{T}(t){S}_{B}(t)=0,\\ {\beta}_{B}{S}_{B}(t){I}_{T}(t)-{\lambda}_{B}{I}_{B}(t)=0,\\ {\beta}_{T}\left(1-{I}_{T}(t)\right){I}_{B}(t)-{\mu}_{T}p{I}_{T}(t)=0.\end{array}\right.\end{array}$$(4)

If *I*_{B}(*t*) = 0 and *I*_{T}(*t*) = 0, from the first equation of system (4), one can easily check that *S*_{B}(*t*) = 1 independently of the values of the parameters. Therefore the disease-free equilibrium
$\begin{array}{}{E}_{1}^{\ast}\end{array}$ = (1, 0, 0) exists for any value of the parameters. This is epidemiologically meaningful since, when there is no infected individual, all the bovine population become susceptible.

Now, if we suppose that *I*_{B}(*t*) > 0, *I*_{T}(*t*) > 0 and consider the second and third equations of (4), we get
$$\begin{array}{}{S}_{B}(t)={\displaystyle \frac{{\lambda}_{B}{I}_{B}(t)}{{\beta}_{B}{I}_{T}(t)},\phantom{\rule{mediummathspace}{0ex}}{I}_{T}(t)=\frac{{\beta}_{T}{I}_{B}(t)}{{\beta}_{T}{I}_{B}(t)+{\mu}_{T}p}.}\end{array}$$(5)

Taking into account the above equalities, we can replace *S*_{B}(*t*) and *I*_{T}(*t*) into the first equation of (4) to obtain the following dependent function for the subpopulation of infected bovines:
$$\begin{array}{}F({I}_{B})=\left({\mu}_{B}+{\alpha}_{B}\right)\left(1-{\displaystyle \frac{1}{{\mathcal{R}}_{0}}}\right)\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}-{\displaystyle \frac{\left({\mu}_{B}+{\alpha}_{B}\right){\lambda}_{B}}{{\beta}_{B}}\left(1+\frac{{\beta}_{B}}{{\lambda}_{B}}+\frac{{\beta}_{B}}{{\mu}_{B}+{\alpha}_{B}}\right){I}_{B},}\end{array}$$(6)
where *F*(*I*_{B}) is obviously continuous and strictly decreasing in [0,1].

If 𝓡_{0} ≤ 1, then
$$\begin{array}{}F(0)=\left({\mu}_{B}+{\alpha}_{B}\right)\left(1-{\displaystyle \frac{1}{{\mathcal{R}}_{0}}}\right)\le 0\end{array}$$
and since *F* is strictly decreasing in [0, 1], there is no 0 <
$\begin{array}{}{I}_{B}^{\ast}<1\text{\hspace{0.17em}such that\hspace{0.17em}}F({I}_{B}^{\ast})=0.\end{array}$
Thus, the model (2) has only an equilibrium: the disease-free one.

Now, if 𝓡_{0} > 1, then we have
$$\begin{array}{}F(0)=\left({\mu}_{B}+{\alpha}_{B}\right)\left(1-{\displaystyle \frac{1}{{\mathcal{R}}_{0}}}\right)>0,\end{array}$$
while
$$\begin{array}{}F(1)=\left({\mu}_{B}+{\alpha}_{B}\right)\left(1-\frac{1}{{\mathcal{R}}_{0}}\right)\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}-\frac{\left({\mu}_{B}+{\alpha}_{B}\right){\lambda}_{B}}{{\beta}_{B}}\left(1+\frac{{\beta}_{B}}{{\lambda}_{B}}+\frac{{\beta}_{B}}{{\mu}_{B}+{\alpha}_{B}}\right)}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}<\left({\mu}_{B}+{\alpha}_{B}\right)-\left({\mu}_{B}+{\alpha}_{B}\right)\frac{{\lambda}_{B}}{{\beta}_{B}}\left(1+\frac{{\beta}_{B}}{{\lambda}_{B}}+\frac{{\beta}_{B}}{{\mu}_{B}+{\alpha}_{B}}\right)}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\left({\mu}_{B}+{\alpha}_{B}\right)\left[1-\left(1+\frac{{\lambda}_{B}}{{\beta}_{B}}+\frac{{\lambda}_{B}}{{\mu}_{B}+{\alpha}_{B}}\right)\right]}\\ {\displaystyle \phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}=\left({\mu}_{B}+{\alpha}_{B}\right)\left[-\left(\frac{{\lambda}_{B}}{{\beta}_{B}}+\frac{{\lambda}_{B}}{{\mu}_{B}+{\alpha}_{B}}\right)\right]<0}\end{array}$$
and, since *F* is continuous and strictly decreasing in [0,1], there exists a unique 0 <
$\begin{array}{}{I}_{B}^{\ast}\end{array}$
< 1 such that *F*(
$\begin{array}{}{I}_{B}^{\ast}\end{array}$) = 0. Therefore, model (2) has a unique endemic equilibrium
$\begin{array}{}{E}_{2}^{\ast}=({S}_{{B}_{2}}^{\ast},{I}_{{B}_{2}}^{\ast},{I}_{{T}_{2}}^{\ast})\end{array}$. In fact, one can check that
$$\begin{array}{}{\displaystyle {I}_{{B}_{2}}^{\ast}=\frac{\left({\mu}_{B}+{\alpha}_{B}\right)\left({\beta}_{B}{\beta}_{T}-{\lambda}_{B}{\mu}_{T}p\right)}{{\beta}_{T}{\alpha}_{B}\left({\beta}_{B}+{\lambda}_{B}\right)+{\mu}_{B}{\beta}_{T}{\lambda}_{B}+{\beta}_{T}{\beta}_{B}\left({\lambda}_{B}+{\mu}_{B}\right)}.}\end{array}$$

Substituting
$\begin{array}{}{I}_{{B}_{2}}^{\ast}\end{array}$
in the second equation of (5), we obtain
$$\begin{array}{}{\displaystyle {I}_{{T}_{2}}^{\ast}=\frac{\left({\mu}_{B}+{\alpha}_{B}\right)\left({\beta}_{B}{\beta}_{T}-{\lambda}_{B}{\mu}_{T}p\right)}{{\beta}_{T}{\beta}_{B}\left({\alpha}_{B}+{\mu}_{B}\right)+{\beta}_{B}{\mu}_{T}p\left({\alpha}_{B}+{\lambda}_{B}+{\mu}_{B}\right)}.}\end{array}$$

and, replacing
$\begin{array}{}{I}_{{B}_{2}}^{\ast}\text{\hspace{0.17em}and\hspace{0.17em}}{I}_{{T}_{2}}^{\ast}\end{array}$
in the first equation of (5), we have
$$\begin{array}{}{\displaystyle {S}_{{B}_{2}}^{\ast}=\frac{{\beta}_{T}{\lambda}_{B}\left({\alpha}_{B}+{\mu}_{B}\right)+{\lambda}_{B}{\mu}_{T}p\left({\alpha}_{B}+{\lambda}_{B}+{\mu}_{B}\right)}{{\beta}_{T}{\alpha}_{B}\left({\beta}_{B}+{\lambda}_{B}\right)+{\beta}_{T}{\lambda}_{B}{\mu}_{B}+{\beta}_{T}{\beta}_{B}\left({\lambda}_{B}+{\mu}_{B}\right)}.}\end{array}$$

($\begin{array}{}{S}_{{B}_{2}}^{\ast},{I}_{{B}_{2}}^{\ast},{I}_{{T}_{2}}^{\ast}\end{array}$)
is in the interior of *Ω*, as is demonstrated below. To do such a demonstration, we must verify that
$$\begin{array}{}0<{S}_{{B}_{2}}^{\ast}+{I}_{{B}_{2}}^{\ast}<1\phantom{\rule{1em}{0ex}}\text{\hspace{0.17em}and\hspace{0.17em}}\phantom{\rule{1em}{0ex}}0<{I}_{{T}_{2}}^{\ast}<1.\end{array}$$

First of all, observe that, since *R*_{0} > 1, we have that *λ*_{B}*μ*_{T}p − *β*_{T}*β*_{B} > 0. Besides, all the parameters involved in the expressions of
$\begin{array}{}{S}_{{B}_{2}}^{\ast},{I}_{{B}_{2}}^{\ast},{I}_{{T}_{2}}^{\ast}\end{array}$ are greater than zero. Hence, both conditions allow us to prove that
$\begin{array}{}{S}_{{B}_{2}}^{\ast},{I}_{{B}_{2}}^{\ast},{I}_{{T}_{2}}^{\ast}\end{array}$ > 0. In particular, we also have that
$\begin{array}{}{S}_{{B}_{2}}^{\ast}+{I}_{{B}_{2}}^{\ast}>0.\end{array}$

Secondly, observe that
$$\begin{array}{}{\displaystyle {S}_{{B}_{2}}^{\ast}+{I}_{{B}_{2}}^{\ast}=\frac{{\beta}_{T}{\lambda}_{B}\left({\alpha}_{B}+{\mu}_{B}\right)+{\lambda}_{B}^{2}{\mu}_{T}p+{\beta}_{B}{\beta}_{T}\left({\mu}_{B}+{\alpha}_{B}\right)}{{\beta}_{T}{\alpha}_{B}\left({\beta}_{B}+{\lambda}_{B}\right)+{\mu}_{B}{\beta}_{T}{\lambda}_{B}+{\beta}_{T}{\beta}_{B}\left({\lambda}_{B}+{\mu}_{B}\right)}<1,}\end{array}$$
if and only if
$$\begin{array}{}{\displaystyle {\beta}_{T}{\lambda}_{B}\left({\alpha}_{B}+{\mu}_{B}\right)+{\lambda}_{B}^{2}{\mu}_{T}p+{\beta}_{B}{\beta}_{T}\left({\mu}_{B}+{\alpha}_{B}\right)<{\beta}_{T}{\alpha}_{B}\left({\beta}_{B}+{\lambda}_{B}\right)}\\ +{\mu}_{B}{\beta}_{T}{\lambda}_{B}+{\beta}_{T}{\beta}_{B}\left({\lambda}_{B}+{\mu}_{B}\right),\end{array}$$
Canceling all the common terms, the inequality above becomes
$$\begin{array}{}{\lambda}_{B}^{2}{\mu}_{T}p<{\lambda}_{B}{\beta}_{T}{\beta}_{B}.\end{array}$$

But, this last inequality is equivalent to *λ*_{B}μ_{T}p − *β*_{T}β_{B} < 0, which holds since *R*_{0} > 1.

Finally, note that
$\begin{array}{}{I}_{{T}_{2}}^{\ast}\end{array}$
< 1 if and only if
$$\begin{array}{}\phantom{\rule{2em}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\phantom{\rule{thinmathspace}{0ex}}\left({\mu}_{B}+{\alpha}_{B}\right)\left({\beta}_{B}{\beta}_{T}-{\lambda}_{B}{\mu}_{T}p\right)<\\ {\beta}_{T}{\beta}_{B}\left({\alpha}_{B}+{\mu}_{B}\right)+{\beta}_{B}{\mu}_{T}p\left({\alpha}_{B}+{\lambda}_{B}+{\mu}_{B}\right)\end{array}$$

Again, canceling some common terms, the inequality above becomes the following equivalent one
$$\begin{array}{}\left({\mu}_{B}+{\alpha}_{B}\right)\left(-{\lambda}_{B}{\mu}_{T}p\right)<{\beta}_{B}{\mu}_{T}p\left({\alpha}_{B}+{\lambda}_{B}+{\mu}_{B}\right),\end{array}$$
which is true since its left hand side is less than zero, while its right hand side is greater than zero □.

Now, we are going to analyze the local stability of the disease-free fixed point. In order to do that, we consider the Jacobian matrix related to system (2) given by
$$\begin{array}{}{\displaystyle J\left({S}_{B}^{\ast},{I}_{B}^{\ast},{I}_{T}^{\ast}\right)}\\ =\left(\begin{array}{ccc}1-\left({\mu}_{B}+{\alpha}_{B}\right)-{\beta}_{B}{I}_{T}^{\ast}& -\left({\mu}_{B}+{\alpha}_{B}\right)& -{\beta}_{B}{S}_{B}^{\ast}\\ \\ {\beta}_{B}{I}_{T}^{\ast}& 1-{\lambda}_{B}& {\beta}_{B}{S}_{B}^{\ast}\\ \\ 0& {\beta}_{T}\left(1-{I}_{T}^{\ast}\right)& 1-{\mu}_{T}p-{\beta}_{T}{I}_{B}^{\ast}\\ \end{array}\right)\end{array}$$(7)

As in the continuous case, the demonstration takes into account the eigenvalues of this Jacobian matrix evaluated at the equilibria. Nevertheless, the method we apply here is different from the one employed for the continuous-time case in [6]. In this case, if all the eigenvalues of *J*(*E*^{*}) have magnitude less than one, then the equilibrium *E*^{*} is locally asymptotically stable, i.e., all solutions of system (2) sufficiently close to the equilibrium point approach it. Actually, we will prove that the conditions of the Jury-Criterion (see [14] or [24]) are satisfied by the equilibria.

#### Theorem 3.2

*The disease*-*free equilibrium*
$\begin{array}{}{E}_{1}^{\ast}\end{array}$ *of (2) is locally asymptotically stable if* 𝓡_{0} < 1 *and unstable if* 𝓡_{0} > 1.

#### Proof

The characteristic polynomial at the disease-free equilibrium
$\begin{array}{}{E}_{1}^{\ast}\end{array}$ = (1,0,0) is
$$\begin{array}{}p(\gamma )=\left(1-{\mu}_{B}-{\alpha}_{B}-\gamma \right)\left({\gamma}^{2}+{a}_{1}\gamma +{a}_{2}\right),\end{array}$$
where *a*_{1} = *μ*_{T}p + *λ*_{B} − 2 and *a*_{2} = 1 + *μ*_{T}pλ_{B} − *β*_{T}β_{B} − *μ*_{T}p − *λ*_{B}.

The first root of the polynomial is *γ*_{1} = 1 −(*μ*_{B} + *α*_{B}). By Lemma (2.1), we have that |1 −(*μ*_{B} + *α*_{B})| = 1 −(*μ*_{B} + *α*_{B}) and since *μ*_{B},*α*_{B} are greater than zero,
$$\begin{array}{}|{\gamma}_{1}|=|1-({\mu}_{B}+{\alpha}_{B})|=1-({\mu}_{B}+{\alpha}_{B})<1.\end{array}$$

Now, to analyze if the other two roots satisfy |*γ*_{2,3}| < 1, we need to check the Jury conditions.

The characteristic polynomial of the Jacobian matrix *J*(
$\begin{array}{}{E}_{1}^{\ast}\end{array}$) can be rewritten as
$$\begin{array}{}p(\gamma )=\left(1-{\mu}_{B}-{\alpha}_{B}-\gamma \right)\left({\gamma}^{2}-tr(\hat{J})\gamma +det(\hat{J})\right),\end{array}$$
where *Ĵ* is the submatrix 2 × 2 of *J*(
$\begin{array}{}{E}_{1}^{\ast}\end{array}$) given by,
$$\begin{array}{}{\displaystyle \hat{J}\left(1,0,0\right)=\left(\begin{array}{cc}1-{\lambda}_{B}& {\beta}_{B}\\ \\ {\beta}_{T}& 1-{\mu}_{T}p\\ \end{array}\right).}\end{array}$$(8)

In this context, Tr(*Ĵ*) = (1 − *λ*_{B}) + (1 − *μ*_{T}p) and det(*Ĵ*) = (1 − *λ*_{B})(1 − *μ*_{T}p) − *β*_{T}β_{B}. The (simplified) Jury criterion (see [24]) states that the eigenvalues of *Ĵ* have magnitude less than one if and only if |Tr(*Ĵ*)| < det(*Ĵ*) + 1 < 2.

Note that, the inequality det(*Ĵ*) + 1 < 2 is equivalent to det(*Ĵ*) < 1. This is the one we are going to prove next. Since 1 − *λ*_{B} ≤ 1 and 1 − *μ*_{T}p ≤ 1, we also have that (1 − *λ*_{B})(1 − *μ*_{T}p) ≤ 1 and, taking into account that *β*_{T},*β*_{B} > 0, we have that
$$\begin{array}{}det(\hat{J})=(1-{\lambda}_{B})(1-{\mu}_{T}p)-{\beta}_{T}{\beta}_{B}<1\end{array}$$
is always satisfied.

On the other hand, observe that in this case Tr(*Ĵ*) ≥ 0, since 1 − *λ*_{B} ≥ 0 and 1 − *μ*_{T}p ≥ 0. Hence, |Tr(*Ĵ*)| = Tr(*Ĵ*) and we must demonstrate that
$$\begin{array}{}(1-{\lambda}_{B})+(1-{\mu}_{T}p)<1+(1-{\lambda}_{B})(1-{\mu}_{T}p)-{\beta}_{T}{\beta}_{B}\end{array}$$
which is equivalent to
$$\begin{array}{}2-{\lambda}_{B}-{\mu}_{T}p<2-{\lambda}_{B}-{\mu}_{T}p+{\lambda}_{B}{\mu}_{T}p-{\beta}_{T}{\beta}_{B}.\end{array}$$

After removing all the common terms in both sides, the inequality Tr(*Ĵ*) < 1 + det(*Ĵ*) becomes
$$\begin{array}{}0<{\mu}_{T}p{\lambda}_{B}-{\beta}_{T}{\beta}_{B}\end{array}$$
which is true if and only if 𝓡_{0} < 1 □.

In the next theorem, we prove the global stability of the disease-free equilibrium when 𝓡_{0} ≤ 1, using the LaSalle Invariance Principle for discrete-time systems given in [25].

#### Theorem 3.3

*The disease*-*free equilibrium*
$\begin{array}{}{E}_{1}^{\ast}\end{array}$ *of system (2) is globally asymptotically stable if* 𝓡_{0} ≤ 1.

#### Proof

First of all, we relocate our disease-free equilibrium to the origin of coordinates. That is, we perform the change of coordinates
$$\begin{array}{}{X}_{B}(t)=1-{S}_{B}(t)\end{array}$$
in the system (2) and it becomes
$$\begin{array}{}\left\{\begin{array}{}{X}_{B}(t+1)=\left[1-({\mu}_{B}+{\alpha}_{B})-{\beta}_{B}{I}_{T}(t)\right]{X}_{B}(t)\\ +({\mu}_{B}+{\alpha}_{B}){I}_{B}(t)+{\beta}_{B}{I}_{T}(t),\\ {I}_{B}(t+1)={\beta}_{B}\left(1-{X}_{B}(t)\right){I}_{T}(t)+\left(1-{\lambda}_{B}\right){I}_{B}(t),\\ {I}_{T}(t+1)={\beta}_{T}\left(1-{I}_{T}(t)\right){I}_{B}(t)+(1-{\mu}_{T}p){I}_{T}(t).\end{array}\right.\end{array}$$(9)

We consider the following Lyapunov function *V* : *Ω* → ℝ_{+} defined by
$$\begin{array}{}V\left({X}_{B}(t),{I}_{B}(t),{I}_{T}(t)\right)={\beta}_{T}{I}_{B}(t)+{\lambda}_{B}{I}_{T}(t),\end{array}$$
where *V* ∈ *C*(*Ω*,ℝ*R*_{+}), such that *V*(0, 0, 0) = 0 and *V*(*X*_{B}(*t*),*I*_{B}(*t*),*I*_{T}(*t*)) ≥ 0 in *Ω* − (0, 0, 0).

Then, the difference
$$\begin{array}{}\mathit{\Delta}V\left({X}_{B},{I}_{B},{I}_{T}\right)=V\left({X}_{B}(t+1),{I}_{B}(t+1),{I}_{T}(t+1)\right)\\ \phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{2em}{0ex}}\phantom{\rule{1em}{0ex}}-V\left({X}_{B}(t),{I}_{B}(t),{I}_{T}(t)\right)\end{array}$$
is given by
$$\begin{array}{}{\beta}_{T}\left[{\beta}_{B}\left(1-{X}_{B}(t)\right){I}_{T}(t)+\left(1-{\lambda}_{B}\right){I}_{B}(t)\right]\\ +{\lambda}_{B}\left[{\beta}_{T}\left(1-{I}_{T}(t)\right){I}_{B}(t)+\left(1-{\mu}_{T}p\right){I}_{T}(t)\right]\\ -{\beta}_{T}{I}_{B}(t)-{\lambda}_{B}{I}_{T}(t).\end{array}$$

Simplifying and grouping terms, we have
$$\begin{array}{}\mathit{\Delta}V\left({X}_{B}(t),{I}_{B}(t),{I}_{T}(t)\right)=\\ \left[\left({\beta}_{T}{\beta}_{B}-{\lambda}_{B}{\mu}_{T}p\right)-{\beta}_{T}{\beta}_{B}{X}_{B}(t)-{\lambda}_{B}{\beta}_{T}{I}_{B}(t)\right]{I}_{T}(t).\end{array}$$(10)

At this point, we shall prove that, if 𝓡_{0} ≤ 1, this last expression is non-positive for every (*X*_{B}(*t*),*I*_{B}(*t*),*I*_{T}(*t*)) in *Ω*.

As by hypotheses 𝓡_{0} ≤ 1, equivalently, we have that
$$\begin{array}{}\left({\beta}_{T}{\beta}_{B}-{\lambda}_{B}{\mu}_{T}p\right)\le 0.\end{array}$$

Additionally, as all the parameters are greater than zero and *X*_{B}(*t*), *I*_{B}(*t*), *I*_{T}(*t*) are non-negative in *Ω*, we can deduce that
$$\begin{array}{}\mathit{\Delta}V\left({X}_{B}(t),{I}_{B}(t),{I}_{T}(t)\right)\le 0,\phantom{\rule{1em}{0ex}}\phantom{\rule{1em}{0ex}}\text{for\hspace{0.17em}}{\mathcal{R}}_{0}\le 1.\end{array}$$

Now, we need to obtain the maximal positively invariant set *G*^{*} contained in the subset *G* ⊂ *Ω* given by
$$\begin{array}{}G=\{\left({X}_{B},{I}_{B},{I}_{T}\right)\in \mathit{\Omega}:\phantom{\rule{1em}{0ex}}\mathit{\Delta}V\left({X}_{B},{I}_{B},{I}_{T}\right)=0\}.\end{array}$$

We shall distinguish two cases, depending on the values of the threshold parameter *R*_{0}:

𝓡_{0} < 1: In this case, expression (10) equals zero if and only if *I*_{T}(*t*) = 0. The system (9) in such points becomes
$$\begin{array}{}\left\{\begin{array}{}{X}_{B}(t+1)=\left[1-({\mu}_{B}+{\alpha}_{B})\right]{X}_{B}(t)+({\mu}_{B}+{\alpha}_{B}){I}_{B}(t),\\ {I}_{B}(t+1)=(1-{\lambda}_{B}){I}_{B}(t),\\ {I}_{T}(t+1)={\beta}_{T}{I}_{B}(t).\end{array}\right.\end{array}$$(11)

Observe that, for every initial state of the form (*X*_{B}(*t*), *I*_{B}(*t*),0), with *I*_{B}(*t*) > 0, the following state in its orbit verifies that *I*_{T}(*t* + 1) = *β*_{T}I_{B}(*t*) > 0. Thus, no orbit of a point of the form (*X*_{B}(*t*), *I*_{B}(*t*),0), with *I*_{B}(*t*) > 0 is contained in such a set of points *G* ⊂ *Ω* for which *Δ* *V*(*X*_{B},*I*_{B},*I*_{T}) = 0. Nevertheless, if we avoid this problem considering only the points of *G* for which *I*_{B}(*t*) = 0, one can easily check that, for any point in such a subset, the iteration of the system (9) reduces to
$$\begin{array}{}\left\{\begin{array}{}{X}_{B}(t+1)=\left[1-({\mu}_{B}+{\alpha}_{B})\right]{X}_{B}(t),\\ {I}_{B}(t+1)=0,\\ {I}_{T}(t+1)=0.\end{array}\right.\end{array}$$(12)

That is, the orbit of any initial state in the subset of *G* given by *I*_{B}(*t*) = 0, *I*_{T}(*t*) = 0 remains in such a subset, i.e., the largest positively invariant set contained in *G* is
$$\begin{array}{}{G}^{\ast}=\{\left({X}_{B},{I}_{B},{I}_{T}\right)\in \mathit{\Omega}:{I}_{B}={I}_{T}=0\}.\end{array}$$

Moreover, note that, since 0 ≤ [1 −(*μ*_{B} + *α*_{B})] < 1, the (disease-free) equilibrium (0, 0, 0) is *G*^{*}–globally asymptotically stable. At this point, since all the orbits of the system remain in *Ω*, all of them are bounded. Therefore, applying the LaSalle Invariance Principle for discrete dynamical systems given in Theorem 3.3 of [25], we can conclude that the disease-free equilibrium is globally asymptotically stable in *Ω*.

𝓡_{0} = 1: In this case, expression (10) equals zero if and only if *I*_{T}(*t*) = 0 or *X*_{B}(*t*) = *I*_{B}(*t*) = 0. Therefore, the set *G* ⊂*Ω* for which *Δ* *V*(*X*_{B},*I*_{B},*I*_{T}) = 0 is the following:
$$\begin{array}{}G=\{\left({X}_{B},{I}_{B},{I}_{T}\right)\in \mathit{\Omega}:{I}_{T}=0\}\cup \{\left({X}_{B},{I}_{B},{I}_{T}\right)\\ \phantom{\rule{1em}{0ex}}\in \mathit{\Omega}:{X}_{B}{I}_{B}=0\}.\end{array}$$

Observe that for any initial point of the form (0, 0, *I*_{T}(*t*)) with *I*_{T}(*t*) > 0, the system (9) becomes
$$\begin{array}{}\left\{\begin{array}{}{X}_{B}(t+1)={\beta}_{B}{I}_{T}(t),\\ {I}_{B}(t+1)={\beta}_{B}{I}_{T}(t),\\ {I}_{T}(t+1)=(1-{\mu}_{T}p){I}_{T}(t).\end{array}\right.\end{array}$$(13)

This proves that no orbit of an initial state in the subset {(*X*_{B},*I*_{B},*I*_{T}) ∈ *Ω*: *X*_{B}= *I*_{B} = 0} with *I*_{T}(*t*) > 0 remains in such a subset of *G*. In fact, only the orbit originated by (0, 0, 0) remains in this subset.

At this point, proceeding as in the previous case, the largest positively invariant set contained in *G* is
$$\begin{array}{}{G}^{\ast}=\{\left({X}_{B},{I}_{B},{I}_{T}\right)\in \mathit{\Omega}:{I}_{B}={I}_{T}=0\}.\end{array}$$

Moreover, note that, since 0 ≤ [1 −(*μ*_{B} + *α*_{B})] < 1, the (disease-free) equilibrium (0, 0, 0) is *G*^{*}-globally asymptotically stable, and, since all the orbits of the system remain in *Ω*, all of them are bounded.

Therefore, applying the LaSalle Invariance Principle for discrete dynamical systems again, we can conclude that the disease-free equilibrium is globally asymptotically stable in *Ω*, also in this case.

□

This last result is epidemiologically significant, because it indicates that if a small number of infective individuals is introduced in a susceptible population, then the disease vanishes.

For 𝓡_{0} > 1, the endemic equilibrium
$\begin{array}{}{E}_{2}^{\ast}\end{array}$ is locally asymptotically stable, as shown by numerical simulations. Actually, it can be seen that
$\begin{array}{}{E}_{2}^{\ast}\end{array}$ is globally asymptotically stable in *Ω* − {(1, 0, 0)}. That is, every initial condition (*S*_{B}(0),*I*_{B}(0),*I*_{T}(0)) ∈ *Ω* produces a trajectory (*S*_{B}(*t*),*I*_{B}(*t*),*I*_{T}(*t*)) ∈ *Ω* which converges to the unique interior fixed point
$\begin{array}{}{E}_{2}^{\ast}\end{array}$. However,
$\begin{array}{}{E}_{2}^{\ast}\end{array}$ is not globally asymptotically stable in *Ω*, because the disease-free point is also in *Ω*.

Besides, when 𝓡_{0} < 1, the system has also the two equilibria,
$\begin{array}{}{E}_{1}^{\ast}\end{array}$ asymptotically stable in *Ω* and
$\begin{array}{}{E}_{2}^{\ast}\end{array}$ unstable, being
$\begin{array}{}{E}_{1}^{\ast}\end{array}$ in the outside of *Ω*.

Moreover, when 𝓡_{0} = 1,
$\begin{array}{}{E}_{1}^{\ast}\end{array}$ is the unique point of the system being a non-hyperbolic fixed point which is asymptotically stable in *Ω*.

Such issues allow us to infer the following corollary.

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