Learning rules of inference and would like to see what a "proper" way of setting out the work (not skipping steps and adding reasoning).
I found a solution by proof by contradiction but am wondering if it's valid to put in an assumption that you make, and not one that was given in the question.
Also, if there's another way to do this without proof by contradiction.

"Either Moriarty has escaped or the money is missing.
If Mr. Smith is not dead then Moriarty has not escaped.
Either the money is not missing or Mr. Smith is dead
Therefore, Mr. Smith is dead."

is a valid argument. Let

$m$ = \text{Moriarty has escaped},

$s$ = money is missing,

$ d$ = Mr. Smith is dead}.

$$\begin{align}&(1) m\vee s \qquad \text{premise}\\&(2)\neg d \rightarrow \neg m \qquad \text{premise}\\ &(3) \neg s \vee d \qquad \text{premise} \\ \\ &(4) \neg d \qquad \text{assumption} \\ &(5) \neg s \qquad \text{disjunctive syllogism (3),(4)}\\ &(6)\neg m \qquad \text{disjunctive syllogism (5),(1)}\\&(7) d \qquad\text{modus tollens (2),(6)}\\&(8) \neg d \wedge d \qquad \text{conjunctive addition (7),(4) }\\&(9) c \qquad \text{negation (8)}\\&(10) \neg d \rightarrow c \qquad \text{(what's the reasoning here???)} \\ &(11) d \qquad\text{proof by contradiction}\end{align} $$

where $c$ is the contradiction proposition.
Wondering if the steps are valid (and if putting in my own assumption is valid). Also, I'm not sure what the reasoning or name of reason for step (10) is.


Your proof by contradiction is essentially correct, except some local mistakes.

The disjunctive syllogism in step (6) allows you to conclude $ m $ and not $\lnot m $, indeed you need $ m $ (or equvalently $\lnot\lnot m $) to apply modus tollens in step (7).

Step (10) is an abstract nonsense and is useless. Actually, from step (8) you can directly derive $ d $ (your step 11): this is the essence of a proof by contradiction. If assuming $\lnot d $ you can deduce a contradiction (in this case the formula $\lnot d \land d $ in step 8), then your assumption $\lnot d $ is false and hence $ d $ is true.

  • $\begingroup$ Ah I see. I thought I had to make it clear that by assuming $\neg d$, this implies $c$ which is the contradiction. In these type of questions, is it always valid to make "an assumption" as introduced in step ($4$)? E.g. maybe I want to prove some conclusion is $p$, but I have to take cases $a$ and $\neg a$, to show both cases lead to conclude $p$. Would I do something like $$\begin{align}(\#) \quad &a \qquad \text{assumption} \\ \vdots\\ (\#\#) &p \\ (\# \# \#) &\neg a \qquad \text{assumption} \\ \vdots \\ p\end{align} $$ hence, $p$ (basically, is it valid to do it in "one" argument? $\endgroup$ – Natash1 Sep 16 '17 at 7:18
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    $\begingroup$ Yes, you're right. It is crucial that, whenever in a proof you make an additional assumption, at the end of the proof this assumption is discharged. This is exactly what you do with $ a $ and $\lnot a $ in your argument schema, which implicitly relies on the classical axiom of excluded middle $a \lor \lnot a $. $\endgroup$ – Taroccoesbrocco Sep 16 '17 at 15:17
  • $\begingroup$ Highly appreciate it. When you say "discharge" do you mean that I didn't refer to it again (after line $(\#\#)$)? So basically show that if $a$ holds, $p$ holds. Then never refer back (the assumption of $a$) to this again. Assume $\neg a$ holds, then $p$ holds. Then I can say $\therefore p$? $\endgroup$ – Natash1 Sep 17 '17 at 2:26
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    $\begingroup$ Yes, that's right. $\endgroup$ – Taroccoesbrocco Sep 17 '17 at 5:33

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