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Mathematics 15 Online
OpenStudy (anonymous):

There are three suspects for a murder: Adams, Brown, and Clark. Adams says "I didn't do it. The victim was an old acquaintance of Brown's. But Clark hated him." Brown states "I didn't do it. I didn't know the guy. Besides I was out of town all week." Clark says "I didn't do it. I saw both Adams and Brown in town around the victim that day; one of them must have done it." We know that exactly one of the suspects is guilty. Assume that the two innocent men are telling the truth, but that the guilty man might not be. Let the propositional variables have the following definitions:

OpenStudy (anonymous):

Let the propositional variables have the following definitions: A = Adams is innocent B = Brown is innocent C = Clark is innocent X = Brown knew the victim Y = Brown was out of town Z = Adams was out of town W = Clark hated the victim

OpenStudy (anonymous):

Formalize the description of each of the sentences below in propositional logic sentences using the variables above. Exactly one of the suspects is guilty: (A->~B)^(B->(~A^~C))^(C->~B)

OpenStudy (anonymous):

A word on notation: In all of the HTML pages (problems and exercises) we are using the following notation for propositional logic: ^ (caret) for and, v (lower case v) for or, ~ for not, -> (hyphen, >) for implication, and <-> (<, hyphen, >) for biconditionals. Propositional variables are upper case letters (and possibly numbers). The symbols true and false are written out in lower case.

OpenStudy (anonymous):

I FOUND MY LOGIC IS FALSE IN : Exactly one of the suspects is guilty: (A->~B)^(B->(~A^~C))^(C->~B) HOW SHOULD I CHANGE THIS

OpenStudy (anonymous):

I EVEN TRIED THIS ONE, TOO. BUT IT DIDN'T WORK : (~A -> B^C) ^ (~B-> A^C) ^ (~C -> A^B)

OpenStudy (anonymous):

SORRY, I FINALLY SOLVE THIS ONE (A^B^~C)v(B^~A^C)v(C^A^~B)

OpenStudy (anonymous):

Reference : 6.034(artificial intelligence) pset4

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