The question being taken from a logical point of view, I simplify the question (without providing context) but I can provide context if users ask to.
Let be three sets $A, B$ and $C$.
The question is how to simplify this logical statement:
($\forall a\in A, \exists b \in B$) and ($\forall b\in B, \exists c\in C$)
into:
$\forall a\in A, \exists c\in C$
To be more specific, we have: $r=lcm(\delta_1, ..., \delta_m)$ where the $\delta_i$ are all the possible orders of elements in the group $U_p$ ($p$ is an odd prime).
We write $r=q_1^{\alpha_1}... q_k^{\alpha_k}$
By number theory considerations we have that $\forall j\in [\![1,k]\!], \exists i\in [\![1,m]\!], \delta_i = a q_j ^{\alpha_j}$ (where $a\in\mathbb N).$
On the other hand, we have: $\forall i\in [\![1,m]\!], \exists x \in U_p$ such that $\delta_i$ is the order of $x$.
My question in this context is: how to prove that $\forall j \in [\![1,k]\!], \exists x\in U_p$ ?
I perfectly understand why we can simplify the first expression into the second one, but what I am asking here is: which formal rules of logic should we apply to the first expression in order to arrive to the second one.
Many thanks in advance.
Edit:
I realize it's not about "simplifying" but about inferring expression $2$ from expression $1$.
Note we respresent the statement $\forall a \in A, \exists b \in B$ in first-order logic (FOL) as $\forall x \exists y [x \in A \to y \in B]$ where $x,y$ are variables denoting arbitrary elements. In other words, we say "for every element $x$, there exists at least one element $y$ such that, if $x \in A$, then $y \in B$." Similarly, we respresent the statements $\forall b \in B, \exists c \in C$ and $\forall a \in A, \exists c \in C$ as $\forall y \exists z [y \in B \to z \in C]$ and $\forall x \exists z [x \in A \to z \in C]$, respectively. With this in mind, below I offer a proof of the argument
$$ \forall x \exists y [x \in A \to y \in B], \forall y \exists z [y \in B \to z \in C] \vdash \forall x \exists z [x \in A \to z \in C] $$
in a FOL natural deduction system using the following inference rules:
$ \begin{array}{ll} \text{$\forall$E} &\text{Universal Elimination} \\ \text{$\forall$I} &\text{Universal Introduction} \\ \text{$\exists$E} &\text{Existential Elimination} \\ \text{$\exists$I} &\text{Existential Introduction} \\ \text{MP} &\text{Modus Ponens} \\ \text{CP} &\text{Conditional Proof (Arrow Introduction)} \\ \text{Assumption for CP} &\text{Assumption for Conditional Proof} \\ \text{Assumption for TD} &\text{Assumption for Typical Disjunct} \\ \end{array} $
The proof is as follows:
$ \begin{array}{llll} \{1\} & 1. & \forall x \exists y [x \in A \to y \in B] & \text{premise} \\ \{2\} & 2. & \forall y \exists z [y \in B \to z \in C] & \text{premise} \\ \{1\} & 3. & \exists y [a \in A \to y \in B] & \text{1 $\forall$E} \\ \{4\} & 4. & a \in A \to b \in B & \text{Assumption of TD} \\ \{2\} & 5. & \exists z [b \in B \to z \in C] & \text{2 $\forall$E} \\ \{6\} & 6. & b \in B \to c \in C & \text{Assumption of TD} \\ \{7\} & 7. & a \in A & \text{Assumption for CP} \\ \{4,7\} & 8. & b \in B & \text{4,7 MP} \\ \{4,6,7\} & 9. & c \in C & \text{6,8 MP} \\ \{4,6\} & 10. & a \in A \to c \in C & \text{7,9 CP} \\ \{4,6\} & 11. & \exists z [a \in A \to z \in C] & \text{10 $\exists$I} \\ \{2,4\} & 12. & \exists z [a \in A \to z \in C] & \text{5,6,11 $\exists$E} \\ \{1,2\} & 13. & \exists z [a \in A \to z \in C] & \text{3,4,12 $\exists$E} \\ \{1,2\} & 14. & \forall x \exists z [x \in A \to z \in C] & \text{13 $\forall$I} & \square \\ \end{array} $
In short, we know from lines $1,2$ that for every $x \in A$ there is at least one $y \in B$ and for every $y \in B$ there is at least one $z \in C$, respectively. Let $a$ be any element in $A$ (line $3$). Then, according to line $1$ there must exist an element in $B$, which we will call $b$. In turn, according to line $2$ there must exist and element in $C$, which we will call $c$. Hence, if $a$ is an element $A$, then there must be some element $c \in C$ (line $10$). Now, notice the choice of $a,b,c$ was completely arbitrary, and I did not attribute any properties to these elements besides membership to their respective sets. For this reason, I can generalize my conclusions and state that for every $x \in A$ there is at least one $z \in C$ (line $14$).