May I ask for past material for the following subject, I will be taking these modules in 2012. Please kindly send to sedeya@gmail.com

(COS2621) Computer organization (Computer Science 221)

(COS2661) Formal logic 2 (Computer Science 261)

(INF3705) Advanced systems development (Information Systems 305)

(COS3701) Theory of computer science 3 (Computer Science 301)

(COS3711) Advanced programming (Computer Science 311)

(COS3721) Operating systems and architecture (Computer Science 321)

(COS3712) Computer graphics (Computer Science 340)

(COS3761) Formal logic 3 (Computer Science 361)

(COS3751) Techniques of artificial intelligence (Computer Science 351)

Thanks in advance.

Regards

Rutendo.]]>

some feedback on the exam paper. Would like your inputs

1a i) Only if is read as P --> Q = P only if Q

Thus q --> r and S

1a) ii) if he drank to much AND drives a BMW, then he does not drive too fast

1b) i)

FORALLx (P(x) and (not G(f(m),x)) and G(m,x) --> E(m,x)

1b) ii) THEREEXISTS an x, such that x is a party AND Mary goes to the part AND Mary's mother goes to the party AND mary's mother enjoys the party

1 c) i) K1p ^ K2p ^ K3p ^ K4p (ie. Each agent knows p)

1 c) ii) Agent 1 knows p AND

Agent 1 does NOT know that / if Agent 2 knows p or does not know P

in better english: Agent 1 knows p, but does not know if Agent 2 knows P or not.

Question 2:

Grive propositional Logic, we can dsign a truth table and assign T or F to each term and then finally get the result of the formula in 2^n steps. If predicate logic, we are unable to devise a truth table, because we might have infinite number of variables to consider via FORALLx. Propositional logic will terminate in 2^n, predicate logic might not terminate for all worlds (unless the world only has 4 instances of the variable).

Question 3: i and ii

x under a is bound - left branch of -->

x under B is bound - left branch of -->

y under B is bound - right branch of -->

3 iii)

if f(x) free for Y?

Y under right branch does not get replaced, because it is bound

2x Y's under left branch are free of Y quantifier, thus

substituting with F(x) would create problem, it would introduct x under THEREEXISTS x

THus it is NOT free.

Question 4

a)

1. p OR (q --> p) AND q --premise

2. p OR (q --> p) --or elim 1

3. q --or elim 1

BOX1

4. p --assumption

5. p --copy 4

BOX2

6. q -->p --assumption

7 q --copy 3

8. p -- --> elim 6

9. p OR elim, 2, 4-5, 6-8

4b)

1. ForALLx S(x) OR T(x) premise

2. ForALLx S(x) --> C(x) premise

3. ThereExistsx not C(x)

BOX1 x0

4. not C(x0) Assumption (actually by elim ThereExists of line 3, but the book says assumption for the reason)

5. S(x0) OR T(x0) -- forAll elim line 1

6. S(x0) --> C(x0) --for all elim line 2

BOX 2

7. S(x0) -- assumption

8. C(x0) -- --> elim line 6

9. CONTADICTION -- not elim line 4,8

10. T(x0) -- CONTRADICTION elim line 9

BOX 3

11. T(x0) assumption

12. T(x0) Copy line 11

13. T(x0) OR elim lines 5, 7-10, 11-12 still inside BOX1

14. ThereExists T(x) -- There exists INTRO, Lines 4 - 13

4(c) NOT QUITE SURE

still busy with the rest]]>

if BLOCK p then BLOCK BLOCK p (axiom 4)

if NOT BLOCK p then BLOCK NOT BLOCK p (axiom 5)

I assume, these work the other way around as well...

if BLOCK BLOCK p, then BLOCK p?

it was used like this in the last proof of assignment 3.

My logic behind the above assumption is the following

If I know something, then I know that I know it.

If I know that I know something, then I must know it.

you agree?]]>

we're given 3 valid formulas on page 314 . THey are always true in all worlds in all models.

Then on page 218 we are given formulas which are not valid, but should hold for certain interpretations of BLOCK and DIAMOND

now the decision on whether these hold, seem very very subjective.

for instance: It ought to be true

it says that if things ought to be true, then it should be permitted. hence saying if BLOCK p --> DIAMOND p (ie. if all should be permitted, then we should have at least one)

Are these interpretations fixed, should we know them based on the table on p 318. Or are they just given to help us understand the reasoning behind them?

many thanks]]>

However,

LS says that There exists x such that [not R(x)] OR [not Q(x)]. Meaning is not in the one OR its not in the other... (perhaps i got this wrong, but using de morgan I get, ~(R(x) ^ Q(x)) Meanings its never in both, just in one at a time and now im thinking perhaps its not in either one at all. meaning can be T&F, F&T or F&F

where as the RS says for all x we have either R(x) or Q(x),

this means that we can never have F&F. One of the two has got to be True.

THus I choose my model as such

A (of m) = 1 2 3 4 5 6

R (of m) = 1 3

Q (of m) = 2 4

In this model we have 1 and 3 only in R and 2 and 4 only in Q. 4 and 5 are nowehere. Thus LS satisfied.

the RH says all in A have got to be in R or in Q, which it is not, thus thie RH evaluates to False.

Is my reasoning correct?]]>

Any thoughts on the final expectations, necessary bits, etc? I'm planning on spending the weekend going over this one, so I'll probably have more info to post then. Just thought I would start the discussion (I get the first post :D)...]]>

Complete = "If it's true, you can always prove it."]]>

1.Horn algorithm

2.Directed Acyclic Graph (DAG)

If you know of any other questions that won't be aske, why not add them to the above list?]]>

(iii) Used Theorem 1.37]]>

(ii) If he does not give her flowers nor chocolates, then she does not eat chocolates.

(b)

(i) For all x[S(x) --> there exists y(P(y) & F(x,y))[

(ii) If every x feeds every y, then x is a student and p is a parrot.

(c)

(i) (K1p & K2p) & (~K1K2p & ~K2K1p)

(ii) If agent 1 knows that agent 2 knows p, then agent 2 knows that agent 1 knows that agent 2 knows p]]>

(i) Does not hold - c is also accessible from itself but p is not true in c

(ii) Does not hold - p is not true in c

(iii) Holds

(iv) Holds]]>

Exists-x (~P(x) || Q(x)) ......... premise. (I'm pretty sure that's right. :D )

Open up an "Arb-Box"... x

.... deal with the cases here? Or take it to the next level?

Open up a box in which to get a contradiction of P(x

...

...

It ends in "bottom".

So after that inner box we have ~ P(x

....

This is a "Chi" to end the exists-+-"arb-that-exists" box, so we're allowed to take it out? ... This is NOT such a "Chi", is it? A little bell rings saying you can't use the arbitrary x in your Chi....

OK. Enough doubt! Carry on.

The idea Was, come out of the box with ~ P(x) && Q(x), which I'm almost certain is wrong.

Obviously were it not wrong, one would easily move from there to "Exists" to reach the conclusion sought.

There you go. I won't edit this. Mistakes are often far more interesting than correct answers.]]>

Let the expression "blah blah is odd" be P(n).

Base Case.

1

Induction Hypothesis.

Imagine for a moment that for k >= 1 P(k) just happens to be true.

In other words k

So by definition it has the general form 2m+1, where m is an integer.

So, equivalently, k

I'm pretty sure the "even number angle" is superfluous, but something in me just likes it, so there.

Now we examine the nature of P'(k + 1)

This is the assertion that "blah blah is even" or "da dee da is odd" (that this is true). So plug in k+1 and see what it does.

(k+1)

= k

= k

Now we notice that k

= 2m + 2k + 6 .... which is starting to look very much like an even number. We confirm that:

= 2(m+k+3)

So it follows from the induction hypothesis that P(k+1) is indeed TRUE.

So you could plug in a 1 in the place of k. Now you have P(k=1) is TRUE (base case), so then P(k+1=1+1=2) is TRUE. (Because we've just proven that this follows given the truth of the previous one). And so from P(2) we go to P(3). 3 to 4. 4,5. 5,6. .... infinity, if you please.

The short way of saying it is that "by the principle of mathematical induction blah blah blah".

Sometimes it's good to remind oneself of the mechanics of how it works at the very end.

OK. So now I offer you a skeleton of a 4b whose wheels fell off for me.]]>

3. ... ~p ........... T, 2.

4. ... "bottom" ............ ~e 1,3. .... end of assumption-box-2

5. ~ box ~p ..... ~i 2-4

6. box ~ box ~p ...... 5 , 5.

7. box diamond p ......... Def. diamond. ..... end of assumtion-box-1

8. p -> box diamond p .................... ->i 1-7

How do we introduce the "definition of diamond" construct? For now this looks an adequate way to justify it, but maybe there's some special notation in the book. Anyone know?]]>

3. dotted-box ... p -> q ... box e, 1.

4. .... p ... box e, 2. << I think maybe strictly they should be in the reverse of this order?>>

5. .... q ............ -> e, 3, 4. ..... end of dotted-box

6. box q ........ box i, 3-5, 5. .........................end of assumption box.

7. box p -> box q .................... ->i, 3-6]]>

I get a "contradictory" solution again.

1. Forall-x (~P(x) && Q(x)) ...... premise

2. box-1 ... The arbitrary x

3. ... ~P(x

4. ... box-2 .. P(x

5. ... .... "bottom" ......... ~e 3,4

6. ... .... Q(x

7. P(x

8. Forall-x (P(x) -> Q(x)) .......... forall-x i 2-7

(For some arbitrary x-value we get the implication. Then we must get it for all x-values).]]>

Does this look OK?

1. box1... ~p

2. .... box2 .... p (both those assumptions)

3. "bottom". ...................... ~e , 1, 2

4. p -> q ..........................."bottom" e, 3 .............. end of box 2.

5. p -> (p -> q) .................... -> i, 2-4 .................. end of box 1.

6. ~p -> (p -> (p -> q)) ........ -> i, 1-5]]>

pls can someone email me the tut letters and whatever solutions are available asap, my internet access is severely limited at the moment.

Thank you kindly!

34177531@mylife.unisa.ac.za]]>

I was wondering if someone on this forum would be able to help me. I did this subject a few years ago but seem to have misplaced my text book, the one we used was a unisa text. I am currently busy with my honours and I find myself needing some of the information that was in this text. Could someone please email me a pdf version of the text (If one is available) or perhaps point me to a textbook that contains the information

Thanks for the help]]>

Could someone please perhaps explain the logic behind it to me?]]>

http://www.cs.bham.ac.uk/research/projects/lics/tutor/index.html]]>