In developing the argument where the premises $p$ and $r$ lead to conclusion $t$, we can use a rule of inference on $p$ such that $p\implies q$ and one on $r$ such that $r\implies s$, after which we can apply another rule of inference on $q$ and $s$ such that $q \wedge s\implies t$, thus showing that $p\wedge r\implies t$.
For example, I was reading the following example:
- $\exists x(A(x)\wedge \neg B(x))$ [premise $p$]
- $A(c)\wedge \neg B(c)$ [existential instantiation from $p$]
- $A(c)$ [simplification from the above; this is statement $q$]
- $\forall x(A(x)\rightarrow C(x))$ [premise $r$]
- $A(c)\rightarrow C(c)$ [universal instantiation from $r$; this is $s$]
- $C(c)$ [modus ponens on $q$ and $s$; this is $t$]
where steps 1 to 3 establish $p\implies q$, steps 4 to 5 establish $r\implies s$, and step 6 is $q\wedge s \implies t$. Intuitively, I can understand why the example works. This feels like a double case of hypothetical syllogism (two chains of implications converging at the modus ponens step), but anyhow I can't justify rigorously through the use of symbols. I've tried $$(p\rightarrow q)\wedge (r \rightarrow s) \wedge (q\wedge s \rightarrow t)\Leftrightarrow(p\wedge r \rightarrow t)$$
but I can't seem to get the manipulations to work. Any guidance will be much appreciated.