Transversal and intersection of two foliations Let $\mathcal{F}_1$ and $\mathcal{F}_2$ be two foliations of a manifold. We say that $\mathcal{F}_1\pitchfork \mathcal{F}_2$ if $T_p L^{(1)}+T_pL^{(2)}=T_p M$ for any $p\in M$, where $L^{(1)}$ and $L^{(2)}$ are the leaves trough $p$.
Now if we have that $\mathcal{F_1}\pitchfork \mathcal{F}_2$, we define $\mathcal{F}_1\cap \mathcal{F}_2$ to be the foliation where the leaves are the connected components of $L^{(1)}\cap L^{(2)}$. Now we want to check that is a foliation with codimension the sum of the codimensions of $\mathcal{F}_1$ and $\mathcal{F}_2$.
I think the idea is to use the implicit function theorem but I am getting nowhere in concrete. That is take foliated charts for $\mathcal{F}_1$, $(x_1,...,x_k,y_1,...,y_{m-k})$, and foliated charts for $\mathcal{F}^2$, $(w_1,...,w_k',z_1,...,z_{m-k'})$ such that $p\cap L^{(1)}\cap L^{(2)}=\{p\in U: y_1(p)=ct,...,y_{m-k}(p)=ct,z_1(p)=ct,...,z_{m-k'}(p)=ct\}$ . Now from this I have tried and construct a new coordinate chart so that we get the result bu I got nowhere.
Any help or hints with this are aprecciated. Thanks in advance.
Attempt at solution :

First thing we note is that since $\mathcal{F}_1$ and $\mathcal{F}_2$ are foliations we will have that the leaves of this new foliation will cover $M$, are disjoint since we are taking the connected components, and will be path connected since we are taking the connected components and $M$ is locally path-connected. Now we need to find the foliated charts of $\mathcal{F}$ and check it's dimension. Let $p\in M$, note that since $L^{(1)}$ and $L^{(2)}$ are transversal we will have that $L^{(2)}\cap L^{(1)}$ is a submanifold of $L^{(1)}$.Using the local normal form we know that there exists open set of $p\in L^{(2)}, U,$ and coordinate chart $(V,\phi)=(V,x_1,...,x_k)$ for $L^{(1)}$ such that $U\cap L^{(1)}=\{p\in V : x^{k'}(p)=...=x^k(p)=0\}$, where $k'=dim L_1 -(dim L_1 + dim L_2 -dim M)= codim \mathcal{F}_2$. Now note that the way we gave the manifold structure to $L^{(1)}$ was that the topology was generated by the plaques $L^{(1)}\cap U'$ where $U'$  was a foliated chart for $p$, and the coordinate chart was the restriction of the foliated chart to the non-constant components. So we have the $(x_1,...,x_k)$ are associated with a chart of $M$, $(x^1,...,x^k,y^1,...,y^{d-k})$ such that the connected components of $L^{(1)}\cap U'$ are of the form $\{p\in U' : y^1(p)=ct,...,y^{d-k}(p)=ct\}$. Now we have $U$ is a plaque for $U''\cap L^{(2)}$. We also know that $p$ will be in a connected component of $L^{(1)}\cap L^{(2)}$, wich we now denote by $L$. If we consider the foliated chart to be $(W:=U'\cap U'',x^1,...,x^k,y^1,...,y^{d-k})$ we will have that the connected components of $W\cup L$ are of the form $\{p\in W: y^1(p)=ct,...,y^{d-k}(p)=ct,x^{k-k'}(p)=,...,x^k(p)=0\}$. To see the statement about the dimension we note that $codim \mathcal{F}_1+ dim L^{(1)}-(dim L^{(1)}+dim L^{(2)}-dim M)=codim\mathcal{F}_1 +codim \mathcal{F}_2 $.

 A: Here is, I think, an answer. Let $\mathcal{D}_1$ be the tangent ditribution of the foliation $\mathcal{F}_1$ and $\mathcal{D}_2$ those of $\mathcal{F}_2$. Theses distribution are totally integrable, that is $[\mathcal{D}_i,\mathcal{D}_i]\subset \mathcal{D}_i$. Let $\mathcal{D} = \mathcal{D}_1\cap \mathcal{D}_2$ be the intersection distribution. It has constant rank as at any $p\in M$, $\dim\mathcal{D}_p = \dim\left({\mathcal{D}_1}_p\cap{\mathcal{D}_2}_p\right)=\dim{\mathcal{D}_1}_p+\dim{\mathcal{D}_2}_p-\dim\left({\mathcal{D}_1}_p+{\mathcal{D}_2}_p\right)$, and then $\mathrm{codim}\mathcal{D}_p = \mathrm{codim}{\mathcal{D}_1}_p + \mathrm{codim}{\mathcal{D}_2}_p$ (recall that $\mathcal{F}_1\pitchfork \mathcal{F}_2$)
Now, let us show that $\mathcal{D}$ is integrable, that is $[\mathcal{D},\mathcal{D}]\subset \mathcal{D}$. Let $X,Y$ be vector fields in $\mathcal{D}$. As $\mathcal{D}\subset \mathcal{D}_1$, $[X,Y]$ is in a vector field in $\mathcal{D}_1$. Symmetrically, it is a vector field in $\mathcal{D}_2$. Then, it is a vector field in $\mathcal{D}_1 \cap \mathcal{D}_2 = \mathcal{D}$, and $[\mathcal{D},\mathcal{D}]\subset \mathcal{D}$. By Frobenius theorem, there exists a foliation $\mathcal{F}$ those tangent bundle is $\mathcal{D}$. The integral submanifolds are the intersections of the leaves of the previous foliations.
Edit Here is another strategy. Fist, show that the intersection of two transverse submanifolds is a submanifold. Let $N$ and $N'$ be two trasverse submanifold with $\mathrm{codim} N= p, \mathrm{codim} N' =q$. Let $x \in N\cap N'$ be fixed. Take a chart $\varphi : U \to \mathbb{R}^n$ centered at $x$. Then $\varphi (U\cap N)$ and $\varphi(U\cap N')$ are submanifolds of $\varphi(U)$ of codimension $p$ and $q$. There exist $f = (f_1,\ldots,f_p): \mathbb{R}^n \to \mathbb{R}^p$ and $g=(g_1,\ldots,g_q): \mathbb{R}^n\to \mathbb{R}^q$ smooth submersions such that $\varphi(U\cap N) = f^{-1}(\{0\})$ and $\varphi(U\cap N')= g^{-1}(\{0\})$. Let $h = (f_1,\ldots,f_p,g_1,\ldots,g_q) : \mathbb{R}^n \to \mathbb{R}^{p+q}$. The transversality condition assures that $h$ is a sumbersion, and thus, $\varphi(U\cap\left(N\cap N'\right)) = h^{-1}(\{0\})$ is a submanifold of $U\cap \mathbb{R}^n$ of codimension $p+q$, and $U\cap (N \cap N') = \varphi^{-1}(h^{-1}(\{0\}) = (h\circ\varphi)^{-1}(\{0\})$ is a smooth submanifold of codimension $p+q$.
Let $\mathcal{F}^1$ and $\mathcal{F}^2$ be two transverse smooth foliations on $M^n$. If $L^1$ and $L^2$ are leaves of $\mathcal{F}^1$ and $\mathcal{F}^2$, and if $L^1\cap L^2 \neq 0$, then the result above shows that $L^1\cap L^2$ is a smooth submanifold of $M$ with codimension $\mathrm{codim}L_1 + \mathrm{codim}L_2$. As the dimension of the leaves of a foliation is constant, if two transverse leaves intersect, their intersection is a submanifold of dimension $n - \mathrm{codim}\mathcal{F}^1 -\mathrm{codim}\mathcal{F}^2$.
It is staightforward to show that $\{L^1\cap L_2 ~|~ L_1 \in \mathcal{F}^1,L_2\in \mathcal{F}^2,L_1\cap L_2\neq 0\}$ is a smooth foliation.
Remark: this shows that the distribution $D_1\cap D_2$ associated is smooth.
