First question I have solved I belive... show, s -> (q -> r) <-> (s ^ q) -> r using the defintion of implication and Boolean algebra.
s ->(~q V r) <-> ~ (s ^ q ) V r
~s V (~q V r ) <-> ~s V ~q V r
~s V ~q V r <-> ~s V ~q V r
L.H.S = R.H.S
Second question I'm having troubles with inference... given the premises:
(1) p -> (q -> r),
(2) p V ~ s,
(3) q Show that you can
conclude s -> r, using the rules of inference and the result in part above.
How should I go about the second question? I'm slightly confused when it comes to inference rules. Thanks.
For the result in part one, you need to go two steps further: You need to show that what you started with: $s \rightarrow (q\rightarrow r)$ is equivalent to $(s \land q)\rightarrow r.$ I'll start where you leave off:
$$\begin{align} (\lnot s \lor \lnot q \lor r) & \iff (\lnot(s \land q) \lor r)\tag{DeMorgan's} \\ \\ & \iff ((s \land q) \rightarrow r)\tag{implication}\end{align}$$
For the second question: Please double check your premises, as typed here, with the original problem statement. If you can use the result above, then you can use the result $p \rightarrow(q \rightarrow r) \equiv (p \land q) \rightarrow r$.
Now, with the added premise:
List the premises.
Start of subproof: Assume $s\equiv \lnot(\lnot s)$.
Infer that this implies $p$ by the second premise.
Conjoin ($\land$ "and") the just derived $p$ with premise three $q$ to get $(p \land q)$.
Use the first premise, along with the result from the first part (the statement equivalent to the first premise) to conclude $r$.
End of subproof.
Therefore $s \implies r$.
You can also prove the desired implication without the result from the first part of your question.