# Are these arguments (rules of inference) the same?

The full question is:
If $m$, then $h$.
We have either $h$ or $w$.
Therefore, if not $h$, then I will not have $m$ and I have $w$.
Show that this above argument is logically valid.

Let $h,w,m$ be propositions. Argument 1 is
\begin{align}&m\rightarrow h \\ &h\vee w \end{align} $$\rule{2in}{.5pt}$$ $$\therefore\neg h \rightarrow (\neg m)\wedge w$$

Argument 2 is
\begin{align}&m\rightarrow h\\ &h\vee w \\ &\neg h \end{align} $$\rule{2in}{.5pt}$$$$\therefore (\neg m)\wedge w$$

Are these saying the same thing? I'm trying to be 100% correct about this. The first two compound propositions were premises, and the question said
"If not $h$, then I will not have $m$ and I have $w$".

So I'm not sure if I can say $\neg h$ is a premise (or assumption) from question, and this leads to $(\neg m) \wedge w$, meaning $\neg h \rightarrow (\neg m)\wedge w$ (this is for argument 2)

The solution says argument 1 is the solution.

\begin{align} \text{Given: }\qquad m &\rightarrow h \\ \therefore\quad \lnot\, h &\rightarrow \lnot\, m \\ \\ \text{Given: }\qquad h &\vee w \\ \therefore\quad \lnot\, h &\rightarrow w \\ \\ \text{taken together: }\quad\lnot\, h &\rightarrow \lnot\, m \land w \\ \end{align}

$\lnot\, h$ is not an outcome; the implications of $\lnot\, h$ are the outcome. However if you are given $\lnot\, h$ as a premise, you can then detach the implication from the statement and give those as an outcome. so:

\begin{align} \text{Given: }\quad\lnot\, h &\rightarrow \lnot\, m \land w \\ \text{and: }\qquad\lnot\, h \\ \therefore\quad \lnot\, m &\land w \\ \end{align}

• So argument 2 is the 100% correct one then? (we shouldn't ever have implication arrows in the conclusion/outcome?) – Natash1 Sep 16 '17 at 6:13
• Added the argument through to matching your argument 2 also – Joffan Sep 16 '17 at 6:21

Rather, I'd say your "argument 2" is the step before proving "argument 1".

You take the premises $\bbox[lemonchiffon]{m\to h}$ and $\bbox[lemonchiffon]{h\vee w}$, and make assumption $\bbox[lemonchiffon]{\neg h}$ to conclude $\bbox[lemonchiffon]{\neg m\wedge w}$, then you discharge the assumption to demonstrate what was to be proven: that if the premises are justified then, $\bbox[lemonchiffon]{\neg h\to (\neg m\vee w)}$.

Since $\bbox[lemonchiffon]{{m\to h, \neg h\vdash \neg m}}$ (by modus tollens) and $\bbox[lemonchiffon]{{h\vee w, \neg h\vdash w}}$ (by disjunctive syllogism), therefore $\bbox[lemonchiffon]{m\to h, h\vee w, \neg h~\vdash ~\neg m\wedge w}$ (by conjunctive introduction).   Since that, therefore $\bbox[lemonchiffon]{ m\to h, h\vee w~\vdash ~\neg h\to (\neg m\wedge w)}$ (by conditional introduction).   Thus demonstrating that the argument is valid.

$$\dfrac{m\to h\\ h\vee w}{\neg h\to (\neg m\wedge w)}$$