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$$\begin{array}{rlll} 1. & \sim H\lor \sim G & \text{Premise} & \\ 2. & H\& (G\lor H) & \text{Premise} & \text{DEDUCE $F\& H$} \end{array}$$

Using the rules of inference from standard propositional logic, we have to deduce F & H.

Well, if we can deduce F & H, we can also deduce F through simplification. That means that these two premises can imply any statement, F, ever.

It does not appear possible to deduce the conclusion.

I feel like the only thing we can deduce is "H & ~G." I even thought of what appears to be a counterexample. Let

H: I live at home.

G: I own a piece of gold.

F: I own a Ferrari.

Premise 1 translates to "It is not the case that I live at home or it is not the case that I own a piece of gold." which is true. Premise 2 translates to "I live at home and either I own a piece of gold or I live at home." which is also true.

However, the conclusion "F & H" translates to "I own a Ferrari and I live at home." which is false.

(original scan of problem here)

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1  
Your analysis seems correct to me. –  Trevor Wilson Sep 5 '13 at 2:05
4  
You're right, assuming that the problem is stated correctly. –  Brian M. Scott Sep 5 '13 at 2:07
    
You are right, but are these the only two assumptions given? There is nothing about $F$? –  Prahlad Vaidyanathan Sep 5 '13 at 2:09
2  
Perhaps you should be discussing this with your professor, not with us... –  GEdgar Sep 5 '13 at 2:26
1  
what do you mean by "the rules of inference from standard propositional logic" each book has its own "the rules of inference from standard propositional logic" –  Willemien Sep 5 '13 at 21:39

2 Answers 2

I suspect that Premise 1 is meant to be:
$ \lnot ( H \lor \lnot G ) $
which simplifies to:
$ \lnot H \land G $
and from which it follows, by and-ing to the simplification of Premise 2 as simply H:
$ H \land \lnot H \land G $
and then simply:
$ False $

From a falsehood, one can of course imply everything.

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We do an abbreviated truth-table analysis.

The sentence $H\land (G\lor H)$ is true precisely if $H$ is true.

To make $\lnot H \lor \lnot G$ true it is enough to make $G$ false.

Thus (1) and (2) are simultaneously satisfiable by making $G$ false and $H$ true. They cannot therefore lead to a contradiction.

Remark: But the instructor could be right: it may be a mistake, not a typo.

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If $H$ is false, then (2) is false.. –  Goos Sep 5 '13 at 2:14
    
Thanks! Had $G$ and $H$ transposed, $G$ comes before $H$. –  André Nicolas Sep 5 '13 at 2:20
    
Hey guys, it was a mistake. The professor found the mistake after I emailed a second time. This thread is no longer necessary. –  Joseph DiNatale Sep 5 '13 at 4:25
    
@user2521987: Good work!${}{}{}{}{}$ –  André Nicolas Sep 5 '13 at 4:34

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