Mathematically correct?

Mathematically correct?

${P}_m := \{ \Phi_{1m}, \Phi_{2m}, \ldots , \Phi_{nm} \}$ where
$$|\Phi_{im} \rangle = \wp_{1m} | \phi_1 \rangle + \wp_{2m} | \phi_{2m}
\rangle + \cdots + \wp_{km} | \phi_{km} \rangle$$ such that $\wp_{im}$ is
a probability amplitude for $\Phi_{im} \in \mathbb{P}_m$ being in it's
$i^{th}$ state, $\phi_{im} $ Probability amplitudes are defined in such a
way that $|\wp_{1m}|^2 + |\wp_{2m}|^2 + \cdots + |\wp_{km}|^2 = 1$ and let
$\mathbb{P}_m | i \rangle$ be the set $\mathbb{P}_m$ with each of its
elements in the $i^{th}$ state. We define the sets to not be pairwise
disjoint in the following way:
Let $N$ be an index set such that $\forall m \in N$ let $\mathbb{P}_m$ be
a set. Therefore we have $$ \mathbb{P}_m \mid \bigcap_{m \in N}
\mathbb{P}_m | i \rangle\neq O$$ in other words $\exists \phi_{i1} ,
\phi_{i2} , \ldots , \phi_{in} \in \mathbb{P}_1, \mathbb{P}_2, \ldots,
\mathbb{P}_n \mid \phi_{i1} \equiv \phi_{i2} \equiv \cdots \equiv
\phi_{in} $
The Mathematics of Probabilistic Sets
A Probabilistic Set can be thought of as a set of normalised wave
functions $\Phi_{im}$ that have at least one state in common and therefore
have a finite probability of intersection.
Union of Probabilistic Sets
The union of probabilistic sets can be defined quite trivially as
$$\bigcup_{m \in N} \mathbb{P}_m : = \{ \Phi_{11}, \Phi_{12}, \ldots,
\Phi_{1N}, \Phi_{21}, \Phi_{22}, \ldots, \Phi_{2N}, \ldots ,\Phi_{1n},
\Phi_{2n}, \ldots, \Phi_{nN} \}$$ Intersection of Probabilistic Sets It
should be fairly clear that two probabilistic sets $\mathbb{P}_m$ and
$\mathbb{P}_K$ do not have a well defined intersection. Instead we may
discuss the \textit{probability} of two such sets intersecting. \ The
probability of intersection of the probabilistic sets shall be denoted
$$\mathcal{P} \Big( \bigcap_{m \in N} \mathbb{P}_m | i \rangle \Big)$$ We
can define $\Phi_{im} (\phi_{im})$ to be the wave function for $\Phi_{im}$
where $\phi_{im} \equiv [\alpha_i, \beta_i]$. So we define a state
$\phi_{im}$ to be the corresponding interval $[\alpha_i,\beta_i]$ on the
wave function $\Phi_{im} (\phi_{im})$. It not follows that the probability
of finding the element $\Phi_{im}$ in the state $\phi_{im}$ is given by $$
P_{\phi_{im} \in \Phi_{im}} = \int_{\alpha_i}^{\beta_i} d \phi_{im} \Big|
\Phi_{im} (\phi_{im}) \Big|^2$$ If only one such $\phi_{im}$ per set
exists satisfying $\phi_{i1} , \phi_{i2} , \ldots , \phi_{in} \in
\mathbb{P}_1, \mathbb{P}_2, \ldots, \mathbb{P}_n \mid \phi_{i1} \equiv
\phi_{i2} \equiv \cdots \equiv \phi_{in} $, then the probability of the
intersection of sets is given by $$ \mathcal{P} \Big( \bigcap_{m \in N}
\mathbb{P}_m | i \rangle \Big) = \prod_{m \in N} \Big(
\int_{\alpha_i}^{\beta_i} d \phi_{im} \Big| \Phi_{im} (\phi_{im}) \Big|^2
\Big)_m$$ However if two such states are equivalent per set, call this
second equivalent state $\phi_{jm}$ the probability becomes $$ \mathcal{P}
\Big( \bigcap_{m \in N} \mathbb{P}_m | i \rangle \Big) = \prod_{m \in N}
\Big( \int_{\alpha_i}^{\beta_i} d \phi_{im} \Big| \Phi_{im} (\phi_{im})
\Big|^2 + \int_{\alpha_j}^{\beta_j} d \phi_{jm} \Big| \Phi_{jm}
(\phi_{jm}) \Big|^2 \Big)_m$$ Now we extend this to the obvious case of:
what if such elements share $\mathfrak{R}$ equivalent states? Well
intuitively the probability of such an intersection occurring approaches
$1$ as $\mathfrak{R} \to n$. Trivially following from the previous two
equations, the probability of intersection is given by $$ \mathcal{P}
\Big( \bigcap_{m \in N} \mathbb{P}_m | i \rangle \Big) = \prod_{m \in N}
\Big( \sum_{\mathfrak{r} \leq \mathfrak{R}} \Big(
\int_{\alpha_i}^{\beta_i} d \phi_{im} \Big| \Phi_{im} (\phi_{im}) \Big|^2
\Big)_\mathfrak{r} \Big)_m$$
Is everything I'm doing thus far correct? Is anything I'm doing thus far
correct? I was just brainstorming ideas and this is what I came up with so
I'd like to know if I'm right. Thanks