8/8/2019 CSE 402 HW 2 SOLN

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Cse402 Artificial

Intelligence

Assignment - 2

Submitted To Shruti Mam.

Submitted By Kshitij Goel.

10812548.

RB1805A03.

Ans.1. To check whether a particular argument is valid or not, we have to

construct a truth table incorporating all possible values existing and if the tautology

occurs then that argument is valid. In this example :

Let X: Paul lives in Dublin.

Let Y: Paul lives in Ireland.

X->Y says that if Paul lives in Dublin then Paul lives in Ireland which should be true

(already given)

X Y X->Y (X->Y)&Y ((X->Y)&Y)->X

T T T T T

F T T T T

F F T F T

It is clear from the above truth table that If Paul lives in Dublin, he lives in

Ireland. Paul lives in Ireland. Therefore Paul lives in Dublin.

Ans 2. Let is create a truth table comprising of all possibilities of situation :

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Box 1 Box 2 Box 3 Result

T F F 3

F T F 1

F F T 2

If Box 1 is true then definitely gold is not in it and in another two. Box 3 says

that gold is in Box 2 which is false. Therefore, gold has to be in Box 3.

If Box 2 is true then Box 1 says that gold is not present in it which is

absolutely false. Therefore, gold is in Box 1.

If Box 3 is true then it says that gold is in Box 2 which will be definitely

true.

Ans.3. Declarative representation It is the representation of knowledge which

exists but its use or its application is not known. When a program uses it, it will

declares the right application of the knowledge and put it into use.

Procedural representation It is one in which the control information that is

necessary to use the knowledge is considered to be embedded in the knowledge

itself and to use procedural representation we need to augment it with an

interpreter that follow the instruction given in the knowledge.

The real difference between the declarative and the procedure view of knowledge

lies in where control information resides.

Ans.4. (a) P Q ~P ~Q PVQ P&Q PVQ&~P ~Q&P&Q

PVQ&~PV~Q&P&Q

T T F F T T F F F

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T F F T T F F F F

F T T F T F T F F

F F T T F F F F F

Hence the above one is contradiction.

Ans.4. (b) PQ ~P ~Q P->~Q P->~Q->~p

T T F F F T

T F F T T F

F T T F T T

F F T T T T

Hence the above one is satisfiable.

Ans.4. (c) P Q ~P PVQ ~PVQ (PVQ)&(~PVQ)

(PVQ)&(~PVQ)&P

T T F T T T T

T F F T F F F

F T T T T T F

F F T F T F F

Hence the above one is satisfiable.

Ans.4. (d) PQ R ~Q P&Q P&QR P&QRV~Q

T T T F T T T

T T F F T F F

T F T T F T T

T F F T F T T

F T T F F T T

F T F F F T T

F F T T F T T

F F F T F T T

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Hence the above one is satisfiable.

Ans.4. (e) PQ P&Q PVQ ~(PVQ) (P&Q)~(PVQ)

T T T T F F

T F F T F T

F T F T F T

F F F F T T

Hence the above one is satisfiable.

Ans.5. Well formed formulae are given below:-

a. ~P V Q & R S U & Q

(((~P) V( Q & R)) S ) (U & Q)

b. P &~Q V P U ~ R

((P &(~Q) )V P) (U (~ R))

c. Q V P V ~R & S ~U & P R

((Q V P) V ((~R) & S)) ( ((~U )& P) R)

Ans. 6. 1. For all x, food(x) - > Likes(Mary, x)

Therefore, ~food(x) V Likes (Mary, x)

2. Food (Pizza).

3. Food (Apple)

4. for all x, for all y eats(x, y) - > food(y)

Therefore , ~eats(x, y) V food(y)

5. Eats (John, chicken)

Combining 4 & 5 we find get

6. food (Chicken).

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