Mock Exam Q7

Hi, please comment on my solution,
the image is quite large you may have to click on it so that it can download in a seperate browser window/tab only then does it enlarge it and make it more readable

Here is the url, it does not seem to allow you to click on it

Proposed Solution - Link to Larger Image

Here is my proposed answers for 7b to 7d, i cannot figure out the answer for 7e

Once again here is the link as well because it is so small below
Click Here - Link to Larger Image

Hi I wrote the below mail to the lecturer about 7f (only notice now that there is a numbering problem so actualy e)

Mock exam question 7 f
As there are no sentence supplied for this question. Will it be correct to state that the sentences must be converted to CNF and that the Resolution algorithm must be applied to proof that alpha implies beta. Should a description be given of the algorithm.

His answer:
CNF and the resolution algorithm should work. Sometime a simple truth table also does the job. No, you do not need to describe the algorithm unless it is explicitly asked.

Ok, when I draw a truth table it is only true if:
alpha = true
beta = false
now what now? what conclusion should one derive from this?



thanks

oh, and by the way what did you think of all my other proposed answers?
7a) to 7d)?

I have not done "a" yet.

7c is valid see p210 Valid sentences are also known as tautologies and P V notP is valid or a tautology same if you draw a truth table for p -> p.

What worries me about e is that for alpha to entail beta, (alpha^not beta) must be unsatisfiable. On p211 it states that a sentence is satisfiable if it is true in some model which is the case in the truth table that you drew up. They way I see the question is that I was not supplied with "real" sentences for alpha and beta as per p215-216 as a result I can not prove it so kept my answer general.

Henk

Aaa, ok my mistake!!!

7.c) is VALID since it does lead to a tautology ! ! !

if it was P and notP then it is a whole different story though.

100% correct.

Hey Ian

I might be wrong, but I got a slightly different answer for (a):
(-P \/ -Q \/ R) /\ (-P \/ Q) /\ P /\ -R

Additionally, what happen when we apply resolution laws to this?
I get Q/\-Q, which is invalid.

*** NEWSFLASH ***

SanozOr you are ... CORRECT ! ! ! !

Please find below the correction, at first I did not think that I was wrong so I did not try and simplify the sentence and just went for it and came to the same result as you proposed in the above post!!! so shocked in getting it wrong myself initially here is the correction ! ! !

Link to correction of proposed answer ...

What did you use to draw those pretty symbols ?

you guys rock.

f:

From the truth table you have drawn, we can conclude that A only entails B when (A & ! is NOT true, as A entails B means that wherever A is true, B should be true, and this is not satisfied by A&!B.
If we draw the column for !(A&!, we see that the scenario where A is true and B is false is excluded so A will entail B.