D. No-go result for $ I=0$ in $ a^{\mathrm {int}}$

The solution to the `homogeneous' equation (119) can be represented as

$\displaystyle \bar{a}_{0}^{\mathrm{int}}=\bar{a}_{0}^{\prime \mathrm{int}}+\bar{a} _{0}^{\prime \prime \mathrm{int}},$ (307)

$\displaystyle \gamma \bar{a}_{0}^{\prime \mathrm{int}}$ $\displaystyle =$ $\displaystyle 0,$ (308)
$\displaystyle \gamma \bar{a}_{0}^{\prime \prime \mathrm{int}}$ $\displaystyle =$ $\displaystyle \partial _{\mu }\bar{m}
_{0}^{\mu }$ (309)

and $ \bar{m}_{0}^{\mu }$ is a nonvanishing, local current.

According to the general result expressed by (75) in both antighost and pure ghost numbers equal to zero, equation (217) implies

$\displaystyle \bar{a}_{0}^{\prime \mathrm{int}}=\bar{a}_{0}^{\prime \mathrm{int}}\left( \left[ F_{\bar{A}}\right] \right) ,$ (310)

where $ F_{\bar{A}}$ are listed in (75). Solution (219) is assumed to provide a cross-coupling Lagrangian. Therefore, since $ R_{\mu \nu \rho \vert\alpha \beta }$ is the most general gauge-invariant quantity depending on the field $ t_{\mu \nu \vert\alpha }$, it follows that each interaction vertex from $ \bar{a}_{0}^{\prime \mathrm{int}}$ is required to be at least linear in $ R_{\mu \nu \rho \vert\alpha \beta }$ and to depend at least on a BF field. But $ R_{\mu \nu \rho \vert\alpha \beta }$ contains two spacetime derivatives, so the emerging interacting field equations would exhibit at least two spacetime derivatives acting on the BF field(s) from the interaction vertices. Nevertheless, this contradicts the general assumption on the preservation of the differential order of each field equation with respect to the free theory (see assumption ii) from the beginning of section 4), so we must set

$\displaystyle \bar{a}_{0}^{\prime \mathrm{int}}=0.$ (311)

Next, we solve equation (218). In view of this, we decompose $ \bar{a}
_{0}^{\prime \prime \mathrm{int}}$ with respect to the number of derivatives acting on the fields as

$\displaystyle \bar{a}_{0}^{\prime \prime \mathrm{int}}=\overset{(0)}{\pi }+\overset{(1)}{ \pi }+\overset{(2)}{\pi },$ (312)

where each $ \overset{(i)}{\pi }$ contains precisely $ i$ spacetime derivatives. Of course, each $ \overset{(i)}{\pi }$ is required to mix the BF and $ (2,1)$ field sectors in order to produce cross-interactions. In agreement with (221), equation (218) is equivalent to
$\displaystyle \gamma \overset{(0)}{\pi }$ $\displaystyle =$ $\displaystyle \partial _{\mu }\overset{(0)}{m}_{0}^{\mu },$ (313)
$\displaystyle \gamma \overset{(1)}{\pi }$ $\displaystyle =$ $\displaystyle \partial _{\mu }\overset{(1)}{m}_{0}^{\mu },$ (314)
$\displaystyle \gamma \overset{(2)}{\pi }$ $\displaystyle =$ $\displaystyle \partial _{\mu }\overset{(2)}{m}_{0}^{\mu }.$ (315)

Using definitions (45)-(47) and an integration by parts it is possible to show that
$\displaystyle \gamma \overset{(0)}{\pi }$ $\displaystyle =$ $\displaystyle \partial _{\mu }\overset{(0)}{m}_{0}^{\mu
}-\left( \partial _{\mu...
...}{\partial t_{\mu \lbrack \alpha
\vert\beta ]}}\right) A_{\alpha \beta } \notag$ (316)
    $\displaystyle +\left( \partial _{\lbrack \mu }\frac{\partial \overset{(0)}{\pi ...
...verset{(0)}{\pi }}{\partial B^{\nu
\rho ]}}\right) \eta ^{\mu \nu \rho } \notag$ (317)
    $\displaystyle -2\left( \partial _{\mu }\frac{\partial \overset{(0)}{\pi }}{\par...
...}{\partial K^{\nu \rho \lambda ]}}\right) \mathcal{G}
^{\mu \nu \rho \lambda }.$ (318)

From (225) we observe that $ \overset{(0)}{\pi }$ is solution to (222) if and only if the following conditions are satisfied simultaneously

$\displaystyle \partial _{\mu }\frac{\partial \overset{(0)}{\pi }}{\partial t_{\...
...tial _{\lbrack \mu }\frac{ \partial \overset{(0)}{\pi }}{\partial H^{\nu ]}}=0,$ (319)
$\displaystyle \partial _{\mu }\frac{\partial \overset{(0)}{\pi }}{\partial V_{\...
...k \mu }\frac{ \partial \overset{(0)}{\pi }}{\partial K^{\nu \rho \lambda ]}}=0.$ (320)

Because $ \overset{(0)}{\pi }$ is derivative-free, the solutions to equations (226)-(227) read as

$\displaystyle \frac{\partial \overset{(0)}{\pi }}{\partial t_{\mu \nu \vert\alpha }}$ $\displaystyle =\tau ^{\mu \nu \vert\alpha },$ $\displaystyle \frac{\partial \overset{(0)}{\pi }}{\partial H^{\mu }}$ $\displaystyle =h_{\mu },$ $\displaystyle \frac{\partial \overset{(0)}{\pi }}{\partial V_{\mu }}$ $\displaystyle =v^{\mu },$ (321)
$\displaystyle \frac{\partial \overset{(0)}{\pi }}{\partial B^{\mu \nu }}$ $\displaystyle =b_{\mu \nu },$ $\displaystyle \frac{\partial \overset{(0)}{\pi }}{\partial \phi _{\mu \nu }}$ $\displaystyle =f_{\mu \nu },$ $\displaystyle \frac{\partial \overset{(0)}{\pi }}{\partial K^{\mu \nu \rho }}$ $\displaystyle =k_{\mu \nu \rho },$ (322)

where $ \tau ^{\mu \nu \vert\alpha }$, $ h_{\mu }$, $ v^{\mu }$, $ b_{\mu \nu }$, $ f_{\mu \nu }$, and $ k_{\mu \nu \rho }$ are some real, constant tensors. In addition, $ \tau ^{\mu \nu \vert\alpha }$ display the same mixed symmetry properties like the tensor field $ t^{\mu \nu \vert\alpha }$ and $ b_{\mu \nu }$, $ f_{\mu \nu }$, and $ k_{\mu \nu \rho }$ are completely antisymmetric. Because there are no such constant tensors in $ D=5$, we conclude that (226 )-(227) possess only the trivial solution, which further implies that

$\displaystyle \overset{(0)}{\pi }=0.$ (323)

Related to equation (223), we use again definitions (45)-( 47) and integrate twice by parts, obtaining

$\displaystyle \gamma \overset{(1)}{\pi }$ $\displaystyle =$ $\displaystyle \partial _{\mu }\overset{(1)}{m}_{0}^{\mu
}-\left( \partial _{\mu...
...{(1)}{\pi }}{\delta t_{\alpha \beta
\vert\mu }}\right) A_{\alpha \beta } \notag$ (324)
    $\displaystyle +\left( \partial _{\lbrack \mu }\frac{\delta \overset{(1)}{\pi }}...
...\overset{(1)}{\pi }}{\delta B^{\nu \rho ]}}\right) \eta ^{\mu
\nu \rho } \notag$ (325)
    $\displaystyle -2\left( \partial _{\mu }\frac{\delta \overset{(1)}{\pi }}{\delta...
...i }}{\delta K^{\nu \rho \lambda ]}}\right) \mathcal{G}^{\mu
\nu \rho \lambda }.$ (326)

Inspecting (231), we observe that $ \overset{(1)}{\pi }$ satisfies equation (223) if and only if the following relations take place simultaneously

$\displaystyle \partial _{\mu }\frac{\delta \overset{(1)}{\pi }}{\delta t_{\mu (...
... \partial _{\lbrack \mu }\frac{\delta \overset{(1)}{\pi }}{\delta H^{\nu ]}}=0,$ (327)
$\displaystyle \partial _{\mu }\frac{\delta \overset{(1)}{\pi }}{\delta V_{\mu }...
...lbrack \mu }\frac{\delta \overset{(1)}{\pi }}{\delta K^{\nu \rho \lambda ]}}=0.$ (328)

The solutions to equations (232)-(233) are expressed by

$\displaystyle \frac{\delta \overset{(1)}{\pi }}{\delta t_{\mu (\alpha \vert\bet...
...elta t_{\alpha \beta \vert\mu }}=\partial _{\nu }\tau ^{\alpha \beta \mu \nu },$ (329)
$\displaystyle \frac{\delta \overset{(1)}{\pi }}{\delta H^{\mu }}=\partial _{\mu...
...ta \overset{(1)}{\pi }}{\delta B^{\mu \nu }}=\partial _{\lbrack \mu }b_{\nu ]},$ (330)
$\displaystyle \frac{\delta \overset{(1)}{\pi }}{\delta \phi _{\mu \nu }}=\parti...
...t{(1)}{\pi }}{\delta K^{\mu \nu \rho }}=\partial _{\lbrack \mu }k_{\nu \rho ]},$ (331)

where the quantities $ s^{\mu \nu \alpha \beta }$, $ \tau ^{\alpha \beta \mu
\nu }$, $ h$, $ v^{\mu \nu }$, $ b_{\mu }$, $ f^{\mu \nu \rho }$, and $ k_{\mu
\nu }$ are some tensors depending at most on the undifferentiated fields $ \Phi ^{\alpha _{0}}$ from (2). In addition, they display the symmetry/antisymmetry properties

$\displaystyle s^{\mu \nu \alpha \beta }=-s^{\nu \mu \alpha \beta }=s^{\mu \nu \beta \alpha },$ (332)
$\displaystyle \tau ^{\alpha \beta \mu \nu }=-\tau ^{\beta \alpha \mu \nu }=-\tau ^{\alpha \beta \nu \mu },$ (333)
$\displaystyle \tau ^{\lbrack \alpha \beta \mu ]\nu }=0,$ (334)

and $ v^{\mu \nu }$, $ f^{\mu \nu \rho }$, and $ k_{\mu
\nu }$ are completely antisymmetric. Because both tensors $ s^{\mu \nu \alpha \beta }$ and $ \tau ^{\alpha \beta \mu
\nu }$ are derivative-free, their are related through

$\displaystyle s^{\mu \nu \alpha \beta }=\tau ^{\mu (\alpha \beta )\nu }.$ (335)

Using successively properties (237)-(239) and formula (240), it can be shown that $ \tau ^{\alpha \beta \mu
\nu }$ is completely antisymmetric. This last property together with (239) leads to

$\displaystyle \tau ^{\alpha \beta \mu \nu }=0,$    

which replaced in the latter equality from (234) produces

$\displaystyle \frac{\delta \overset{(1)}{\pi }}{\delta t_{\alpha \beta \vert\mu }}=0.$    

This means that the entire dependence of $ \overset{(1)}{\pi }$ on $ t_{\alpha
\beta \vert\mu }$ is trivial (reduces to a full divergence), and therefore $ \overset{(1)}{\pi }$ can at most describe self-interactions in the BF sector. Since there is no nontrivial solution to (223) that mixes the BF and $ (2,1)$ field sectors, we can safely take

$\displaystyle \overset{(1)}{\pi }=0.$ (336)

In the end of this section we analyze equation (224). Taking one more time into account definitions (45)-(47), it is easy to see that (224) implies that the EL derivatives of $ \overset{(2)}{\pi
}$ are subject to the equations

$\displaystyle \partial _{\mu }\frac{\delta \overset{(2)}{\pi }}{\delta t_{\mu (...
... _{\mu }\frac{\delta \overset{(2)}{\pi }}{\delta t_{\alpha \beta \vert\mu }}=0,$ (337)
$\displaystyle \partial _{\lbrack \mu }\frac{\delta \overset{(2)}{\pi }}{\delta ...
...tial _{\lbrack \mu }\frac{\delta \overset{(2)}{\pi }}{\delta B^{\nu \rho ]}}=0,$ (338)
$\displaystyle \partial _{\mu }\frac{\delta \overset{(2)}{\pi }}{\delta \phi _{\...
...lbrack \mu }\frac{\delta \overset{(2)}{\pi }}{\delta K^{\nu \rho \lambda ]}}=0.$ (339)

Because $ \overset{(2)}{\pi
}$ (and also its EL derivatives) contains two spacetime derivatives, the solution to both equations from (242) is of the type

$\displaystyle \frac{\delta \overset{(2)}{\pi }}{\delta t_{\mu \nu \vert\alpha }}=\partial _{\rho }\partial _{\beta }\bar{\tau}^{\mu \nu \rho \vert\alpha \beta },$ (340)

where $ \bar{\tau}^{\mu \nu \rho \vert\alpha \beta }$ depends only on the undifferentiated fields $ \Phi ^{\alpha _{0}}$ and exhibits the mixed symmetry $ \left( 3,2\right) $. This means that $ \bar{\tau}^{\mu \nu \rho \vert\alpha \beta }$ is simultaneously antisymmetric in its first three and respectively last two indices and satisfies the identity $ \bar{\tau}^{[\mu
\nu \rho \vert\alpha ]\beta }=0$. The solutions to the remaining equations, (243) and (247), can be represented as

$\displaystyle \frac{\delta \overset{(2)}{\pi }}{\delta H^{\mu }}=\partial _{\mu...
...erset{(2)}{\pi }}{\delta B^{\mu \nu }}=\partial _{\lbrack \mu }\bar{b}_{\nu ]},$ (341)
$\displaystyle \frac{\delta \overset{(2)}{\pi }}{\delta \phi _{\mu \nu }}=\parti...
...{\pi }}{\delta K^{\mu \nu \rho }}=\partial _{\lbrack \mu }\bar{k}_{\nu \rho ]},$ (342)

where the functions $ \bar{v}^{\mu \nu }$, $ \bar{f}^{\mu \nu \rho }$, and $ \bar{k}_{\mu \nu }$ are completely antisymmetric and contain a single spacetime derivative.

Let $ N$ be a derivation in the algebra of the fields $ t_{\mu \nu \vert\alpha }$, $ H^{\mu }$, $ V_{\mu }$, $ B^{\mu \nu }$, $ \phi _{\mu \nu }$, $ K^{\mu \nu \rho }$, and of their derivatives, which counts the powers of these fields and of their derivatives

$\displaystyle N$ $\displaystyle =$ $\displaystyle \sum\limits_{n\geq 0}\left( \left( \partial _{\mu _{1}\cdots \mu
...rtial \left(
\partial _{\mu _{1}\cdots \mu _{n}}H^{\mu }\right) }\right. \notag$ (343)
    $\displaystyle +\left( \partial _{\mu _{1}\cdots \mu _{n}}V_{\mu }\right) \frac{...
...\partial \left( \partial _{\mu _{1}\cdots \mu _{n}}B^{\mu \nu
}\right) } \notag$ (344)
    $\displaystyle \left. +\left( \partial _{\mu _{1}\cdots \mu _{n}}\phi _{\mu \nu ...
...l \left( \partial _{\mu _{1}\cdots \mu
_{n}}K^{\mu \nu \rho }\right) }\right) .$ (345)

We emphasize that $ N$ does not `see' either the scalar field $ \varphi $ or its spacetime derivatives. It is easy to check that for every nonintegrated density $ \Psi $ we have
$\displaystyle N\Psi$ $\displaystyle =$ $\displaystyle \frac{\delta \Psi }{\delta t_{\mu \nu \vert\alpha }}t_{\mu \nu \v...
V_{\mu }}V_{\mu }+\frac{\delta \Psi }{\delta B^{\mu \nu }}B^{\mu \nu }
\notag$ (346)
    $\displaystyle +\frac{\delta \Psi }{\delta \phi _{\mu \nu }}\phi _{\mu \nu }+\fr...
\Psi }{\delta K^{\mu \nu \rho }}K^{\mu \nu \rho }+\partial _{\mu }s^{\mu }.$ (347)

If $ \Psi ^{(n)}$ is a homogeneous polynomial of degree $ n$ in the fields $ t_{\mu \nu \vert\alpha }$, $ H^{\mu }$, $ V_{\mu }$, $ B^{\mu \nu }$, $ \phi _{\mu \nu }$, $ K^{\mu \nu \rho }$ and their derivatives (such a polynomial may depend also on $ \varphi $ and its spacetime derivatives, but the homogeneity does not take them into consideration since $ \Psi $ is allowed to be a series in $ \varphi $), then

$\displaystyle N\Psi ^{(n)}=n\Psi ^{(n)}.$    

Based on results (245)-(247), we can write
$\displaystyle N\overset{(2)}{\pi }$ $\displaystyle =$ $\displaystyle -\tfrac{1}{3}\bar{\tau}^{\mu \nu \rho \vert\alpha \beta
}R_{\mu \...
...ial _{\lbrack \mu }V_{\nu ]}+2\bar{b}_{\mu }\partial
_{\nu }B^{\mu \nu } \notag$ (348)
    $\displaystyle -\tfrac{1}{3}\bar{f}^{\mu \nu \rho }\partial _{\lbrack \mu }\phi ...
...3\bar{k}_{\mu \nu }\partial _{\rho }K^{\mu \nu \rho }+\partial _{\mu
}m^{\mu }.$ (349)

We decompose $ \overset{(2)}{\pi
}$ along the degree $ n$ as

$\displaystyle \overset{(2)}{\pi }=\sum\limits_{n\geq 2}\overset{(2)}{\pi }^{(n)},$ (350)

where $ N\overset{(2)}{\pi }^{(n)}=n\overset{(2)}{\pi }^{(n)}$ ($ n\geq 2$ in ( 251) because $ \overset{(2)}{\pi
}$, and hence every $ \overset{(2)}{
\pi }^{(n)}$, is assumed to describe cross-interactions between the BF model and the tensor field with the mixed symmetry $ (2,1)$), and find that

$\displaystyle N\overset{(2)}{\pi }=\sum\limits_{n\geq 2}n\overset{(2)}{\pi }^{(n)}.$ (351)

Comparing (252) with (250), it follows that decomposition ( 251) induces a similar one with respect to each function $ \bar{\tau}^{\mu \nu \rho \vert\alpha \beta }$, $ \bar{h}$, $ \bar{v}^{\mu \nu }$, $ \bar{b}
_{\mu }$, $ \bar{f}^{\mu \nu \rho }$, and $ \bar{k}_{\mu \nu }$

$\displaystyle \bar{\tau}^{\mu \nu \rho \vert\alpha \beta }$ $\displaystyle =\sum\limits_{n\geq 2}\bar{\tau} _{(n-1)}^{\mu \nu \rho \vert\alpha \beta },$ $\displaystyle \bar{h}$ $\displaystyle =\sum\limits_{n\geq 2} \bar{h}_{(n-1)},$ $\displaystyle \bar{v}^{\mu \nu }$ $\displaystyle =\sum\limits_{n\geq 2}\bar{v} _{(n-1)}^{\mu \nu },$ (352)
$\displaystyle \bar{b}^{\mu }$ $\displaystyle =\sum\limits_{n\geq 2}\bar{b}_{(n-1)}^{\mu },$ $\displaystyle \bar{f}^{\mu \nu \rho }$ $\displaystyle =\sum\limits_{n\geq 2}\bar{f}_{(n-1)}^{\mu \nu \rho },$ $\displaystyle \bar{k} ^{\mu \nu }$ $\displaystyle =\sum\limits_{n\geq 2}\bar{k}_{(n-1)}^{\mu \nu }.$ (353)

Inserting (253) and (254) in (250) and comparing the resulting expression with (252), we get
$\displaystyle \overset{(2)}{\pi }^{(n)}$ $\displaystyle =$ $\displaystyle -\tfrac{1}{3n}\bar{\tau}_{(n-1)}^{\mu \nu \rho
\vert\alpha \beta ...
...\tfrac{1}{2n}\bar{v}_{(n-1)}^{\mu \nu
}\partial _{\lbrack \mu }V_{\nu ]} \notag$ (354)
    $\displaystyle +\tfrac{2}{n}\bar{b}_{(n-1)}^{\mu }\partial ^{\nu }B_{\mu \nu }-\...
...)}^{\mu \nu }\partial ^{\rho }K_{\mu \nu \rho
}+\partial _{\mu }m_{(n)}^{\mu }.$ (355)

Replacing the last result, (255), into (251), we further obtain
$\displaystyle \overset{(2)}{\pi }$ $\displaystyle =$ $\displaystyle -\tfrac{1}{3}\hat{\tau}^{\mu \nu \rho \vert\alpha \beta
}R_{\mu \...
...ial _{\lbrack \mu }V_{\nu ]}+2\hat{b}_{\mu }\partial
_{\nu }B^{\mu \nu } \notag$ (356)
    $\displaystyle -\tfrac{1}{3}\hat{f}^{\mu \nu \rho }\partial _{\lbrack \mu }\phi ...
...k}_{\mu \nu }\partial _{\rho }K^{\mu \nu \rho }+\partial _{\mu
}\hat{m}^{\mu },$ (357)


$\displaystyle \hat{\tau}^{\mu \nu \rho \vert\alpha \beta }$ $\displaystyle =\sum\limits_{n\geq 2}\tfrac{1}{n} \bar{\tau}_{(n-1)}^{\mu \nu \rho \vert\alpha \beta },$ $\displaystyle \hat{h}$ $\displaystyle =\sum\limits_{n\geq 2}\tfrac{1}{n}\bar{h}_{(n-1)},$ $\displaystyle \hat{v}^{\mu \nu }$ $\displaystyle =\sum\limits_{n\geq 2}\tfrac{1}{n}\bar{v}_{(n-1)}^{\mu \nu },$ (358)
$\displaystyle \hat{b}^{\mu }$ $\displaystyle =\sum\limits_{n\geq 2}\tfrac{1}{n}\bar{b}_{(n-1)}^{\mu },$ $\displaystyle \hat{f}^{\mu \nu \rho }$ $\displaystyle =\sum\limits_{n\geq 2}\tfrac{1}{n}\bar{f} _{(n-1)}^{\mu \nu \rho },$ $\displaystyle \hat{k}^{\mu \nu }$ $\displaystyle =\sum\limits_{n\geq 2}\tfrac{ 1}{n}\bar{k}_{(n-1)}^{\mu \nu }.$ (359)

So far, we showed that the solution to (224) can be put in the form ( 256). By means of definitions (36)-(37), we can bring (256) to the expression
$\displaystyle \overset{(2)}{\pi }$ $\displaystyle =$ $\displaystyle -\tfrac{1}{3}\hat{\tau}^{\mu \nu \rho \vert\alpha \beta
}R_{\mu \nu \rho \vert\alpha \beta }+\partial _{\mu }\hat{m}^{\mu } \notag$ (360)
    $\displaystyle +\delta \left( -\varphi ^{\ast }\hat{h}-B_{\mu \nu }^{\ast }\hat{...
...\ast }\hat{f}^{\mu
\nu \rho }-3\phi ^{\ast \mu \nu }\hat{k}_{\mu \nu }\right) .$ (361)

The $ \delta $-exact modulo $ d$ terms in the right-hand side of (259) produce purely trivial interactions, which can be eliminated via field redefinitions. This is due to the isomorphism $ H^{i}\left( s\vert d\right) \simeq
H^{i}\left( \gamma \vert d,H_{0}\left( \delta \right) \right) $ in all positive values of the ghost number and respectively of the pure ghost number [42], which at $ i=0$ allows one to state that any solution of equation ( 224) that is $ \delta $-exact modulo $ d$ is in fact a trivial cocycle from $ H^{0}\left( s\vert d\right) $. In conclusion, the only nontrivial solution to (224) can be written as

$\displaystyle \overset{(2)}{\pi }=-\tfrac{1}{3}\hat{\tau}^{\mu \nu \rho \vert\alpha \beta }R_{\mu \nu \rho \vert\alpha \beta },$ (362)

where $ \hat{\tau}^{\mu \nu \rho \vert\alpha \beta }$ displays the mixed symmetry $ \left( 3,2\right) $, is derivative-free, and is required to depend at least on one field from the BF sector. But $ R_{\mu \nu \rho \vert\alpha \beta }$ already contains two spacetime derivatives, so such a $ \overset{(2)}{\pi
}$ disagrees with the hypothesis on the differential order of the interacting field equations (see also the discussion following formula (219)), which means that we must set

$\displaystyle \overset{(2)}{\pi }=0.$ (363)

Substituting results (230), (241), and (261) into decomposition (221), we obtain

$\displaystyle \bar{a}_{0}^{\prime \prime \mathrm{int}}=0,$ (364)

which combined with (220) proves that indeed there is no nontrivial solution to the `homogeneous' equation (119) that complies with all the working hypotheses

$\displaystyle \bar{a}_{0}^{\mathrm{int}}=0.$ (365)

Ashkbiz Danehkar