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Discussion #19 Resolution with Predicate CalculusPowerPoint Presentation

Discussion #19 Resolution with Predicate Calculus

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Discussion #19 Resolution with Predicate Calculus

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- Horn Clauses
- Dropping Quantifiers
- Unification
- Resolution

- For resolution, the premises are in CNF.
- CNF: (… … …) (… … …) … (… … …)
- Each (… … …) is a clause.

- A horn clause is a disjunction of literals that contains at most one positive (non-negative) literal.
- Horn clause: P Q R
- Horn clause: P
- Horn clause: P
- Not a horn clause: P Q R too many positive literals

- Note that, in general, a horn clause is an implication with a conjunction of all positive literals on the lhs and one positive (non-negated) literal on the rhs.
- Horn clause: P Q R Q R P (Q R) P
(Q R) P

- Degenerate cases:
- Horn clause: P F P T P T P
- Horn clause: P P F P F
- Horn clause: P1 P2 … Pn (P1 P2 … Pn) F
P1 P2 … Pn F

- Not a horn clause: P Q R R P Q R (P Q)

- Horn clause: P Q R Q R P (Q R) P

- One positive literal and one or more negative literals.
- e.g. P Q R (Q R) P
- These are the Datalog rules.

- One positive literal and no negative literals.
- i.e. P T P
- “We see P by itself as being true.”
- These are the Datalog facts.

- No positive literal.
- P P F or P1 P2 … Pn P1 P2 … Pn F
- “When the consequent is false, we wonder what’s going on.”
- These are the Datalog queries either one query or several.

- For horn clauses in Datalog, we put all clauses in Prenex Normal Form.
- Further, to be a valid horn clause, all variables must be universally quantified.
- Sample horn clauses with variables and quantifiers:
xyz(P(x,y) Q(x,z) R(z))

x(R(x))

- Now we discard the universal quantifiers but don’t forget them. One consequence is that variables are local to their own horn clause (i.e. the x’s in P and Q are the same x, but the x’s in P and R are different).
P(x,y) Q(x,z) R(z)

R(x)

- We then convert the horn clauses to implications with all positive literals.
P(x,y) Q(x,z) R(z)

R(x) F

- Finally, we modify the syntax so that it corresponds to Datalog.
R(z) :- P(x,y),Q(x,z)

R(x)?

- Because every variable is universally quantified, we can instantiate it to any (and every) constant in the domain. (We are in context of a derivation, so all statements are true either by assumption, premise, or by a rule of inference.)
R(a) :- P(x,y),Q(x,a)Sza

R(a)?Sxa

- So, we’re asking if R(a) is true, but the rule says that R(a) is true if
P(x,y) Q(x,a) is true. Formally, we are doing resolution

P(x,y) Q(x,a) R(a)

R(a)

P(x,y) Q(x,a)

- We can resolve only when the complementary literals are instantiated identically (unified).

R(a) P(x,y) Q(x,a)

R(a)

P(x,y) Q(x,a)

IfR(x,y) P(x,z) Q(z,y)

P(a,b)

Q(b,j)

Q(b,k)

ThenR(a,j)

1. P(x,z) Q(z,y) R(x,y)

2. P(a,b)

3. Q(b,j)

4. Q(b,k)

5. R(a,j) conclusion

unification6. P(a,z) Q(z,j) R(a,j)UI, Sxa Syj

7. P(a,z) Q(z,j)res 5,6

unification8. P(a,b) Q(b,j)Szb

9. Q(b,j) res 2,8

10. (succeed!)res 9,3

Domain = {a,b,j,k}

Ifaunt(x,y) sister(x,z) parent(z,y)

sister(ann,bob)

parent(bob,jay)

parent(bob,kay)

Thenaunt(ann,jay)

1. sister(x,z) parent(z,y) aunt(x,y)

2. sister(ann,bob)

3. parent(bob,jay)

4. parent(bob,kay)

5. aunt(ann,jay)conclusion

6. sister(ann,z) parent(z,jay) aunt(ann,jay) UI, Sxann Syjay

7. sister(ann,z) parent(z,jay)res 5,6

8. sister(ann,bob) parent(bob,jay)Szbob

9. sister(ann,bob) res 3,8

10. (succeed!)res 2,9

Domain = {ann,bob,jay,kay}

Facts:

sister('ann','bob').

parent('bob','jay').

parent('bob','kay').

Rules:

aunt(x,y) :- sister(x,z), parent(z,y).

Queries:

aunt('ann','jay')?

Observe: for aunt('ann','jay') we have exactly the previous proof!

Output: yes.

What do we do for aunt('ann',x)?

- Try all proofs with all substitutions from the domain if a substituted value v for x works, output v.
- What’s the domain? Usual choice is all constants in the facts = {'ann', 'bob', 'jay', 'kay'}
Try first to prove aunt('ann','ann'). (Assume support strategy)

1.aunt('ann','ann')conclusion

2. aunt('ann','ann') :- sister('ann',z), parent(z,'ann'). Sxann

Now observe that the resolution of 1 & 2 is

sister('ann',z) parent(z,'ann')(sp)a (sp)a

And that this is only F if both sister('ann',z) and parent(z,'ann') are T. Thus, we can essentially start over and prove these using resolution one at a time as sub-proofs.

3a. sister('ann',z)conclusion

3b. sister('ann','ann')Szann

3c. (fail!) can’t resolve, no fact: sister('ann','ann')

1. aunt('ann',x)conclusion

2. aunt('ann','ann')Sxann

3. aunt('ann','ann') :- sister('ann',z), parent(z,'ann'). SxannSyann

So, we try the next element of the domain for z, namely, bob.

3a. sister('ann',z)conclusion

3b. sister('ann','bob')Szbob

3c. (succeed!) resolution with fact: sister('ann','bob')

Great! Let’s try the 2nd half with bob (the z’s must be the same).

3d. parent(z,'ann')conclusion

3e. parent('bob','ann') Szbob

3f. (fail!)again, no fact: parent('bob','ann')

In fact, z='jay', z='kay' all fail on both sister and parent. So, we backtrack to step 2 and try the next element of the domain for x, namely, 'bob'.

Continuing…

2. aunt('ann','bob')conclusion w/Sxbob

3. aunt('ann','bob') :- sister('ann',z), parent(z,'bob'). SxannSybob

3a. sister('ann',z)conclusion

3b. sister('ann','ann')Szann

3c. (fail!) no fact: sister('ann','ann')

So, we try the next element of the domain for z, namely, bob.

3a. sister('ann',z)conclusion

3b. sister('ann','bob')Szbob

3c. (succeed!) res with fact: sister('ann','bob')

3d. parent(z,'bob')conclusion

3e. parent('bob','bob') Szbob

3f. (fail!)no fact: parent('bob','bob')

Further, z='jay' fails, z='kay' fails, so, we backtrack again.

Continuing with next element in domain for x, namely 'jay' …

2. aunt('ann','jay')Sxjay

3. aunt('ann','jay') :- sister('ann',z), parent(z,'jay'). SxannSyjay

3a. sister('ann',z)conclusion

3b. sister('ann','ann')Szann

3c. (fail!) no fact: sister('ann','ann')

3a. sister('ann',z)conclusion

3b. sister('ann','bob')Szbob

3c. (succeed!) res with fact: sister('ann','bob')

3d. parent(z,'jay')conclusion

3e. parent('bob','jay') Szbob

3f. (succeed!) res with fact: parent('bob','jay')

So, both sub-proofs are satisfied and we output “x='jay'”.

Are there any more?

From what point do we continue?

We can continue from as if we had failed, choosing the next element in domain for z, namely 'jay' …

3a. sister('ann',z)conclusion

3b. sister('ann','jay')Szjay

3c. (fail!) no fact: sister('ann','jay')

3a. sister('ann',z)conclusion

3b. sister('ann','kay')Szkay

3c. (fail!) no fact: sister('ann','kay')

So, we again backtrack to step 2 and the next element for x …

(Alternatively, since we have succeeded, we can jump back to the query and try the next element of the domain at that level.)

Try to resolve using last element of domain for x, namely, 'kay' …

2. aunt('ann','kay')Sxkay

3. aunt('ann','kay') :- sister('ann',z), parent(z,'kay'). SxannSykay

3a. sister('ann',z)conclusion

3b. sister('ann','ann')Szann

3c. (fail!) no fact: sister('ann','ann')

3a. sister('ann',z)conclusion

3b. sister('ann','bob')Szbob

3c. (succeed!) res with fact: sister('ann','bob')

3d. parent(z,'kay')conclusion

3e. parent('bob','kay') Szbob

3f. (succeed!) res with fact: parent('bob','kay')

So, we output “x='kay'”.

Continuing, as if we had failed, we try z = 'jay' and z = 'kay', which both fail. (Alternatively, since we succeeded, we can jump back to the query and try the next element of the domain at that level.) We’ve tried all possible substitutions, so we’re done.

Sx=y

Sz

'ann'

Sx'ann'

'ann'

'ann'

Fail (sister)

'bob'

Fail (parent)

'jay'

Fail (sister)

'kay'

Fail (sister)

'bob'

'ann'

Fail (sister)

'bob'

Fail (parent)

'jay'

Fail (sister)

'kay'

Fail (sister)

'jay'

'ann'

Fail (sister)

Facts:

sister('ann','bob').

parent('bob','jay').

parent('bob','kay').

Rules:

aunt(x,y) :- sister(x,z), parent(z,y).

Queries:

aunt('ann',x)?

'bob'

Succeeds!

'jay'

Fail (sister)

'kay'

Fail (sister)

'kay'

'ann'

Fail (sister)

'bob'

Succeeds!

'jay'

Fail (sister)

'kay'

Fail (sister)

Thank goodness computers can do this faster and better than we can.