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I am asked to show the following argument is valid:

I know you need to use the rules of inference like modus ponens/converse fallacy but I'm confused because it doesn't look like any of the forms I've learned about?

$N\to B\lor S\\ S\to W\lor A \\ M\to N\land W \\ \text{therefore, }M\to B\lor A$

I don't want to use the truth table because it will be real long. If someone can get me started i would really appreciate the help. thx

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    After you ask a question here, if you get an acceptable answer, you should "accept" the answer by clicking the check mark ✓ next to it. This scores points for you and for the person who answered your question. If you don't do this, people are less likely to answer your later questions.2012-10-01

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No valid argument can prove this. Suppose $M, N, S$ and $W$ are true, and $A$ and $B$ are false. Then the three premisses are all true, but the conclusion is false.

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    As a way to find this, the only way to make the conclusion false is to have $M$ true and $A$ and $B$ false. To make the premises true you then (from the third) need $N$ and $W$ true. Then from the first you need $S$ to be true. Now check that the second is satisfied and you are done.2012-10-01
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You could turn your question into a binary decision diagram (BDD). If you assure that these BDDs are short and unique for logical equivalence classes, than you can read-off from the BDDs the following:

1) If the BDD has the form "true", then it is a tautology.

2) If the BDD has the form "false", then its negation is a tautology.

3) In all other cases the BDD nor its negation is a tautology.

In case 2) and 3) there is a model of the BDD that make it false. Let's give it a try, and let's convert your problem into a BDD via the program in ( * ) and ( ** ). We get the following result:

?- convert((n=>b v s)&(s=>w v a)&(m=>n & w)=>(m=>b v a),X). X = (a->true;b->true;m->(n->(s->(w->false;true);true);true);true) 

So it indeed differs from "true" and should be thus falsifiable. By extracting a CNF from the BDD we can also read off a counter model. Let's also do it for your problem:

?- cnf((a->true;b->true;m->(n->(s->(w->false;true);true);true);true),[],L). L = [not(w),not(s),not(n),not(m),b,a]  

So the only counter model is w=1, s=1, n=1, m=1, b=0, a=0.

Proof: If the BDD is not "true", then the CNF will at least contain one row. From this row we can directly construct a model that falsifies the row, for unmentioned propositional variables add either false or true, for mentioned propositional variables add a value that falsifies the literal in the row. Since by construction no propositional variable occurs twice in a row, the row can be falsified. Since the row is falsified the whole CNF is falsified.

( * ) Invert, Inter and Union on BDDs with Lexical Variable Order:
http://www.xlog.ch/jekejeke/principia/shannon.p

( ** ) Convert, DNF and CNF for BDDs:
http://www.xlog.ch/jekejeke/principia/nfs.p