Use rules of inference to show Premises: 


*

*$p \land \lnot s$

*$q \to (r \to s)$


Conclusion:


*

*$(p \to q) \to \lnot r$


Use rules of inference to show the above argument is valid.
I only manage to get $(p \to q) \to (p \land \lnot r)$.


*

*$p \land \lnot s$ Assumption

*$q \to (r \to s)$ Assumption

*$\lnot q \lor (\lnot r \lor s)$ from 2, implication rule

*$\lnot s$ from 1, conjunctive simplification

*$\lnot q \lor \lnot r$ from 3,4 disjunctive syllogism

*$p$ from 1, conjunctive simplification

*$p \land (\lnot q \lor \lnot r)$ from 5,6 conjunction

*$(p \land \lnot q) \lor (p \land \lnot r)$ from 7, Distributive rule

*$(p \to q) \to (p \land \lnot r)$ from 8, implication rule

 A: You haven't specified what inference rules can be used, so here's a natural deduction argument of that demonstrates the consequence:


*

*$p \land \lnot s$ Given.

*$q \to (r \to s)$ Given.

*
*

*$p \to q$ Assumption.


*
*

*$p$ Conjunction elimination from 1.


*
*

*$q$ Conditional elimination (modus ponens) from 3 and 4.


*
*

*$r \to s$ Conditional elimination from 2 and 5.


*
*

*
*

*$r$ Assumption.



*
*

*
*

*$s$ Conditional elimination from 6 and 7.



*
*

*
*

*$\lnot s$ Conjunction elimination from 1.



*
*

*
*

*$\bot$ Contradiction introduction from 8 and 9.



*
*

*$\lnot r$ Negation introduction from 7–10.


*$(p \to q) \to \lnot r$ Conditional introduction from 3–11.

A: Joshua Taylor has given a correct proof using one common set of inference rules, but it appears, from the work you've shown, that your system of inference rules is a little different.  So let me try it with rules that look like what you used.  I'll start with the conclusion, $(p\to q)\to(p\land\neg r)$ that you arrived at.  The "implication rule" that you used to go from 2 to 3 should allow me to convert this to $(\neg(p\to q))\lor(p\land\neg r)$.  Next, a distributive law (admittedly dual to the one you used going from 7 to 8, so I hope both versions of distributivity are available to you) gives me $[(\neg(p\to q))\lor p]\land[(\neg(p\to q))\lor\neg r]$.  Then conjunctive simplification, which got you from 1 to 4, should produce $(\neg(p\to q))\lor\neg r$.  Finally, the implication rule, now in the other direction, as in your inference from 8 to 9, gives $(p\to q)\to\neg r$, as desired.
A: I'm going to solve it assuming that you have a very large and relatively standard set of inference rules.


*

*Assume $r$ for proof of a conditional. (Ultimately this will be the conditional $r\rightarrow \neg(p\rightarrow q)$)

*Infer $\neg s$ from the first premise.

*Infer $r \land \neg s$ from 1 and 2.

*Infer $\neg(\neg r \lor s)$ by De Morgan's and 3.

*Infer $\neg (r\rightarrow s)$ by equivalence of 4.

*Infer $\neg q$ from 5 and the second premise using Modus Tollens.

*Infer $p$ from the first premise.

*Infer $p\land \neg q$ from 6 and 7.

*Infer $\neg(\neg p \lor q)$ by De Morgan's of 8.

*Infer $\neg(p\rightarrow q)$ by equivalence of 9.

*Therefore, by lines 1 through 10, we have shown $r\rightarrow \neg(p\rightarrow q)$.

*By equivalence of contraposition with 11, infer $(p\rightarrow q)\rightarrow r$.
