# How to show that this logical argument is valid?

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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## 2 Answers

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. – Ross Millikan Oct 1 '12 at 4:28

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

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