Can we infer p → (r∧s) from (p∧t) → (r∨s)? Is it possible to write p → (r∧s) from (p∧t) → (r∨s) using rules of inference or any other implications in discrete mathematics?


*

*(p∧t) → (r∨s)

*q → (u∧t)

*u → p

*¬s
Need to prove q → r. How do I start with this problem? There's nothing much common between the given premises.
 A: $q$ implies $p\land t$ and hence $r\land s$ (never mind $r$).
A: 
Is it possible to write p → (r∧s) from (p∧t) → (r∧s) using rules of inference or any other implications in discrete mathematics?

No.  But fortunately you do not need to do any such thing.
Hint: You do not need to use all four premises in a proof.   If $\alpha, \beta,\gamma\vdash \eta$ then $\alpha,\beta,\gamma,\delta\vdash \eta$


*

*Accept the premises, $(p\land t)\to (r\land s),u\to p, q\to(u\land t),\neg s$.


*

*Assume $q$. 


*

*Now $q$ and premise $q\to (u\land t)$ entails $u\land t$.

*Of course $u\land t$ entails $u$ and also entails $t$. 

*$\vdots$

*$p$ and $t$ entails $p\land t$

*$\vdots$

*$r\land s$ entails $r$.


*Obtaining $q\to r$, by discharging the assumption


*Therefore premises $(p\land t)\to (r\land s),u\to p, q\to(u\land t),
\neg s$ entails $q\to r$.

A: In answer to the first question in your post (and the title of your post): 
No, you cannot infer $p \rightarrow (r \land s)$ from $(p \land t) \rightarrow (r \land s)$
Just intuitively: the given statement says that you can know that $r$ and $s$ are both true if you know that both $p$ and $t$ are true ... so if you only know that $p$ is true, can you infer $r$ and $s$? Clearly not.
As a concrete example: we can say that someone is a bachelor if this person is male and unmarried: $(m \land u) \rightarrow b$. So, if I tell you that someone is male, is that enough to infer that that person is a bachelor? (i.e. Can we infer $m \rightarrow b$?) Again, clearly not!
So notice that in the alleged inference, we weaken the antecedent of the conditional we have, but as a result, the resulting conditional becomes a stronger statement, and thus indeed cannot be inferred. Thinking that you can commits the logical fallacy of weakening the antecedent.
Now, if you had $(p \lor t) \rightarrow (r \land s)$ as a given, then you can infer $p \rightarrow (r \land s)$. Indeed, $(p \lor t) \rightarrow (r \land s)$ is equivalent to $(p \rightarrow (r \land s)) \land (t \rightarrow (r \land s))$, and thus by strengthening the antecedent, we weaken the conditional as a whole, and are therefore dealing with a valid inference called strengthening the antecedent
