Transcript Resolution and Unification chapter 8

Resolution and Unification
Automated Reasoning, or theorem proving
based on chapter 8, Ben Coppin, Artificial Intelligence Illuminated
CSCE 4613
Artificial Inteligence
11.05.2013
Nikhil Thomas
Overview
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Resolution
Skolemization
Unification
Resolution Algorithm
Resolution for problem Solving Example
1. Some children will eat any food.
2. No children will eat food that is green.
3. All children like food made by Cadbury’s.
Prove:- No food made by Cadbury’s is green.
Resolution
• A proof method
• Has fixed number of steps
• Can be automated
Normal Forms
Well-Formed Formula (WFF)
In Conjunctive Normal Form (CNF) : set of phrases and-ed together
A1∧ A2∧ A3∧ . . . ∧ An
where each clause, Ai, is of the form
B1∨ B2∨ B3∨ . . . ∨ Bn
A ∧ (B ∨ C) ∧ (¬A ∨ ¬B ∨ ¬C ∨ D)
In Disjunctive Normal Form (DNF): set of phrases or-ed together
A ∨ (B ∧ C) ∨ (¬A ∧ ¬B ∧ ¬C ∧ D)
Conversion of WFF to CNF
Example
1. A ↔ B ≡ (A→B) ∧ (B→A)
2. A →B ≡ ¬A ∨ B
3. ¬(A ∧ B) ≡¬A ∨ ¬B
4. ¬(A ∨ B) ≡¬A ∧ ¬B
5. ¬¬A ≡A
6. A ∨ (B ∧ C) ≡(A ∨ B) ∧ (A ∨ C)
A↔ (B ∧ C)
WFF
(A→(B ∧ C)) ∧ ((B ∧ C)→A)
(¬A ∨ (B ∧ C)) ∧ (¬(B ∧ C) ∨ A)
(¬A ∨ (B ∧ C)) ∧ (¬B ∨ ¬C ∨ A)
(¬A ∨ B) ∧ (¬A ∨ C) ∧ (¬B ∨ ¬C ∨ A)
CNF
{(¬A, B), (¬A, C), (¬B, ¬C, A)}
Clause Form (set of clauses)
Conversion of WFF to CNF can be automated.
The Resolution Rule
A ∨ B ¬B ∨ C
A∨C
if A implies B and B implies C, then A implies C.
Resolution Rule applied to wff ’s in clause form,
{(A, B), (¬B, C)}
can be resolved to give
(A ∨ B) ∧ (¬A ∨ C) ∧ (¬B ∨ C)
{(A, B), (¬A, C), (¬B, C)}
{(B, C), (¬B, C)}
{(A, C)}
{C}
(A ∨ B) ∧ (¬A ∨ C) ∧ (¬B ∨ C) |= C
‘|=‘ : resolves to
Resolution Refutation
{(¬A, B), (¬A, ¬B, C), A, ¬C}
{(¬A, C), A, ¬C}
{C, ¬C}
⊥
= falsum
=> original clauses were inconsistent
{(¬A, B), (¬A, ¬B, C), A, ¬C} |= ⊥
Proof by Refutation
(a.k.a proof by contradiction)
• Proof by Refutation or Proof by Contradiction
 Negate conclusion
 Add it to set of premises
 Prove resultant set of expressions is unsatisfiable
 Implies argument is valid
If it rains and I don’t have an umbrella, then I will get wet.
It is raining, and I don’t have an umbrella.
Therefore, I will get wet.
{(¬A, B, C), A, ¬B, ¬C}
(A ∧ ¬B)→C
A ∧ ¬B
∴C
(A ∧ ¬B)→C
≡ ¬(A ∧ ¬B) ∨ C
≡ ¬A ∨ B ∨ C
{(B, C), ¬B, ¬C}
{C, ¬C}
⊥
I will get wet : true
Resolution Proof as a Tree
{(¬A, B), (¬B, C), (¬C, D), (¬D, E, F), A, ¬F)}
Did not resolve to ‘⊥’
(falsum)
∴ original conclusion was not valid.
Application of Resolution
Check whether solution exists for
combinatorial search problem
Eg: three-coloring problem for graphs
A graph representing a three-coloring problem
Resolution in Predicate Logic
Normal Forms for Predicate Logic
• Move ∀ and ∃ quantifiers to the beginning of the expression
• prenex normal form.
∀ :- Universal Quantifier (for all )
∃ :- Existential Quantifier (there exists one or for some)
1. A ↔B ≡ (A→B) ∧ (B→A)
2. A →B ≡ ¬A ∨ B
3. ¬(A ∧ B) ≡ ¬A ∨ ¬B
4. ¬(A ∨ B) ≡ ¬A ∧ ¬B
5. ¬¬A ≡ A
6. A ∨ (B ∧ C) ≡ (A ∨ B) ∧ (A ∨ C)
7. ¬(∀x)A(x) ≡ (∃x)¬A(x)
8. ¬(∃x)A(x) ≡ (∀x)¬A(x)
9. (∀x)A(x) ∧ B ≡ (∀x)(A(x) ∧ B)
10. (∀x)A(x) ∨ B ≡ (∀x)(A(x) ∨ B)
11. (∃x)A(x) ∧ B ≡ (∃x)(A(x) ∧ B)
12. (∃x)A(x) ∨ B ≡ (∃x)(A(x) ∨ B)
13. (∀x)A(x) ∧ (∀y)B(y) ≡ (∀x)(∀y)(A(x) ∧ B(y))
14. (∀x)A(x) ∧ (∃ y)B(y) ≡ (∀x)(∃y)(A(x) ∧ B(y))
15. (∃x)A(x) ∧ (∀y)B(y) ≡ (∃x)(∀y)(A(x) ∧ B(y))
16. (∃x)A(x) ∧ (∃y)B(y) ≡ (∃x)(∃y)(A(x) ∧ B(y))
Skolemization
Before resolution
• eliminate all the existential quantifiers, ∃, from the wff
• replacing existentially quantified variable by a constant
∃(x) P(x) convert to P(c)
: c is a constant that has not been used elsewhere in the wff
• P(c) is not logically equivalent to ∃(x) P(x)
• But useful for resolution to see whether a solution exists
• ‘c’ is called a Skolem constant
• slightly differently in cases where the ∃ follows a ∀ quantifier
(∀x)(∃y)(P(x,y))
(∀x)(P(x,f(x))
Replace ’y’ with a skolem function
Example of Skolemization
(∀x)(∃y)(∀z)(P(x) ∧ Q(y, z))
to
(∀w)(∀x)(∃y)(∀z)(P(x) ∧ Q(w, y, z)) to
(∀x)(∀z)(P(x) ∧ Q(f(x), z)
(∀w)(∀x)(∀z)(P(x) ∧ Q(w, f(w,x), z)
Unification
σ to indicate a substitution
• Sometimes clauses can be resolved by making substitutions
σ= {y/w, z/x}
{(P(w,x)), (¬P(y,z))} P with ¬P, are not currently identical
they have different arguments
A = P(w,x)
B = ¬P(y,z)
replace w with y and x with z
Aσ = P(y,z)
Bσ= ¬P(y,z)
{(P(y,z)), (¬P(y,z))} |= ⊥
substitution :
{y/w, z/x}
if a σ can be applied to a set of clauses {A, B, C,. . . }
such that
Aσ = Bσ= Cσ= . . .
Then
σ is called a unifier for the set {A, B, C, . . . }.
A constant, such as a, cannot be replaced by a variable.
A variable x cannot be replaced by a term that contains x.
Unification (contd)
composition
operator, o =
composition of those two substitutions
σ1 = {a1/x1, a2/x2, . . . , am/xm}
σ2 = {b1/y1, b2/y2, . . . , bn/yn}
σ1 o σ2 = {a1σ2/x1, a2σ2/x2, . . . , anσ2/xn, b1/y1, b2/y2, . . . , bn/yn}
If yi = xj then bi/yi can be eliminated.
If aiσ2 = xi then aiσ2/xi can be eliminated
most general unifier (mgu) : u1
for a set S = {A, B, C, . . . }
If any other unifier u2 can be expressed as the composition of
u1 with some other substitution
Example
σ1 = {a/x, x/y, f(a)/z}
σ2 = {y/x, f(z)/y}
σ1 o σ2 = {aσ2/x, xσ2/y, f(a)σ2/z, y/x, f(z)/y}
= {a/x, y/y, f(a)/z, y/x, f(z)/y}
= {a/x, f(a)/z, f(z)/y}
y/y is removed : aiσ2 = xi
y/x is removed : yi = xj
σ2 o σ1 = {yσ1/x, f(z)σ1/y, a/x, x/y, f(a)/z}
= {x/x, f(f(a))/y, a/x, x/y, f(a)/z}
= {f(f(a))/y, f(a)/z}
σ1 o σ2 ≠ σ2 o σ1.
Unification Algorithm
To unify a set of clauses, S0:
Initialize: σ0 = {}
i=0
Loop: If Si has only one element, terminate and report that σi is
the mgu for S0.
If Si has more than one element, find the disagreement set
Di of Si (i.e., the substitutions that need to be made).
If Di contains a variable x and a term t where x is not contained
within t, then we say:
σi
1= σi o {t/x}
Si
1 = Si {t/x}
Increment i, and repeat the loop.
Unification Example
S0 = {P(a, x, y, z), P (x, y, z, w)}
(a is a constant, and x, y, z are variables)
We initialize σ0 = {} and i = 0.
D0 = {a,x}
σ1= σo o {a/x} = {a/x}
S1 = So {a/x} = {P(a, a, y, z), P (a, y, z, w)}
D1 = {a,y}
σ2 = {a/x} o {a/y} = {a/x, a/y}
S2 = S1 {a/x, a/y} = {P(a, a, a, z), P (a, a, z, w)}
D2 = {a, z}
σ3 = {a/x, a/y} o {a/z} = {a/x, a/y, a/z}
S3 = S2 {a/x, a/y, a/z} = {P(a, a, a, a), P(a, a, a, w)}
D3 = {a, w}
σ4 = {a/x, a/y, a/z} o {a/w} = {a/x, a/y, a/z, a/w}
S4 = S3 {a/x, a/y, a/z, a/w} = {P(a, a, a, a), P(a, a, a, a)}
= {P(a, a, a, a)}
S4 has just one element,
mgu, σ4 : {a/x, a/y, a/z, a/w}.
Resolution Algorithm
Example
• First, negate the conclusion and add it to the list of assumptions.
• Now convert the assumptions into prenex normal form.
• Next, skolemize the resulting expression.
• Now convert the expression into a set of clauses.
Now we resolve the clauses using the following method:
If our clauses are {A, B, C, . . . }
A has a literal LA and B has a literal LB
such that LA and ¬LB have a mgu σ, then
we can resolve A and B to give the
resolvent of these clauses,
which is: (Aσ - LAσ) U (Bσ - LBσ)
∪ set union operator,
- set difference operator.
{A(x,y), B(f(y)), C(x, y, z)}
{A(f(x), z), ¬B(z), C(f(a), b, z)}
{A(x,y), B(f(y)), C(x, y, z)}{f(y)/z} - B(f(y)){f(y)/z}
U (A(f(x), z), ¬B(z), C(f(a), b, z)} - ¬B(z){f(y)/z}
= {A(x,y), C(x, y, z)} U {A(f(x), z), C(f(a), b, z)}
= {A(x,y), C(x, y, z), A(f(x), z), C(f(a), b, z)}
Resolution for Problem Solving
premises:
1. Some children will eat any food.
2. No children will eat food that is green.
3. All children like food made by Cadbury’s.
Primises:
1. (∃x)(C(x) ∧ (∀y)(F(y)→L(x,y)))
2. (∀x)(C(x)→(∀y)((F(y) ∧ G(y))→¬L(x,y)))
3. (∀x)((F(x) ∧ M(c,x))→(∀y)(C(y)→L(y,x)))
Conclusion: No food made by Cadbury’s is green.
Conclusion: (∀x)((F(x) ∧ M(c,x))→¬G(x))
premises and the conclusion in predicate calculus
C(x) means “x is a child.”
F(x) means “x is food.”
L(x, y) means “x likes y.”
G(x) means “x is green.”
M(x, y) means “x makes y.”
c means “Cadbury’s.”
First, negate the conclusion and add it to the set of
premises
(∃x)(C(x) ∧ (∀y)(F(y)→L(x,y)))
∧ (∀x)(C(x)→(∀y)((F(y) ∧ G(y))→¬L(x,y)))
∧ (∀x)((F(x) ∧ M(c,x))→(∀y)(C(y)→L(y,x)))
∧ ¬ ((∀x)((F(x) ∧ M(c,x))→¬G(x)))
1. Some children will eat any food.
2. No children will eat food that is green.
3. All children like food made by Cadbury’s.
(∃x)(C(x) ∧ (∀y)(F(y)→L(x,y)))
∧ (∀x)(C(x)→(∀y)((F(y) ∧ G(y))→¬L(x,y)))
∧ (∀x)((F(x) ∧ M(c,x))→(∀y)(C(y)→L(y,x)))
∧ ¬ ((∀x)((F(x) ∧ M(c,x))→¬G(x)))
Conclusion: No food made by Cadbury’s is green
(∃x)(C(x) ∧ (∀y)(F(y)→L(x,y)))
∧ (∀x)(C(x)→(∀y)((F(y) ∧ G(y))→¬L(x,y)))
∧ (∀x)((F(x) ∧ M(c,x))→(∀y)(C(y)→L(y,x)))
∧ ¬ ((∀x)((F(x) ∧ M(c,x))→¬G(x)))
convert this into a set of clauses
expression 1:
(∃x)(C(x) ∧ (∀y)(F(y)→L(x,y)))
‘→’ must be eliminated first:
(∃x)(C(x) ∧ (∀y)(¬F(y) ∨ L(x,y)))
bring the quantifiers to the front
(∃x)(∀y)(C(x) ∧ (¬F(y) ∨ L(x,y)))
skolemize, to eliminate the existential quantifier
(∀y)(C(a) ∧ (¬F(y) ∨ L(a,y)))
This can be expressed as the following clauses:
{C(a)}, {¬F(y), L(a,y)}
expression 2
(∀x)(C(x)→(∀y)((F(y) ∧ G(y))→¬L(x,y)))
→is eliminated first:
(∀x)(¬C(x) ∨ (∀y)(¬(F(y) ∧ G(y)) ∨ ¬L(x,y)))
Now DeMorgan’s law is applied:
(∀x)(¬C(x) ∨ (∀y)(¬F(y) ∨ ¬G(y) ∨ ¬L(x,y)))
Quantifiers are moved to the front:
(∀x)(∀y)(¬C(x) ∨ ¬F(y) ∨ ¬G(y) ∨ ¬L(x,y))
This can be written as the following single clause:
{¬C(x), ¬F(y), ¬(G(y), ¬L(x,y)}
1. Some children will eat any food.
2. No children will eat food that is green.
3. All children like food made by Cadbury’s.
Conclusion: No food made by Cadbury’s is green
(∃x)(C(x) ∧ (∀y)(F(y)→L(x,y)))
∧ (∀x)(C(x)→(∀y)((F(y) ∧ G(y))→¬L(x,y)))
∧ (∀x)((F(x) ∧ M(c,x))→(∀y)(C(y)→L(y,x)))
∧ ¬ ((∀x)((F(x) ∧ M(c,x))→¬G(x)))
expression 3:
(∀x)((F(x) ∧ M(c,x))→(∀y)(C(y)→L(y,x)))
We first eliminate →:
(∀x)(¬(F(x) ∧ M(c,x)) ∨ (∀y)(¬C(y) ∨ L(y,x)))
Next, we apply DeMorgan’s law:
(∀x)(¬F(x) ∨ ¬M(c,x) ∨ (∀y)(¬C(y) ∨ L(y,x)))
bring the quantifiers to the front of the expression:
(∀x)(∀y)(¬F(x) ∨ ¬M(c,x) ∨ ¬C(y) ∨ L(y,x))
This can be expressed as the following single clause:
{¬F(x), ¬M(c,x), ¬C(y), L(y,x)}
(∃x)(C(x) ∧ (∀y)(F(y)→L(x,y)))
∧ (∀x)(C(x)→(∀y)((F(y) ∧ G(y))→¬L(x,y)))
∧ (∀x)((F(x) ∧ M(c,x))→(∀y)(C(y)→L(y,x)))
∧ ¬ ((∀x)((F(x) ∧ M(c,x))→¬G(x)))
conclusion, which has been negated:
¬(∀x)((F(x) ∧ M(c,x))→¬G(x))
First, we eliminate →:
¬(∀x)(¬(F(x) ∧ M(c,x)) ∨ ¬G(x))
apply the quantifier equivalence to move the ¬ from the front
of the expression:
(∃x)¬(¬(F(x) ∧ M(c,x)) ∨ ¬G(x))
DeMorgan’s law can now be applied:
(∃x)(¬¬(F(x) ∧ M(c,x)) ∧ ¬¬G(x))
We can now remove ¬¬:
(∃x)(F(x) ∧ M(c,x) ∧ G(x))
This expression is now skolemized:
F(b) ∧ M(c,b) ∧ G(b))
This can be expressed as the following set of clauses:
{{F(b)}, {M(c,b)}, {G(b)}}
1. Some children will eat any food.
2. No children will eat food that is green.
3. All children like food made by Cadbury’s.
(∃x)(C(x) ∧ (∀y)(F(y)→L(x,y)))
∧ (∀x)(C(x)→(∀y)((F(y) ∧ G(y))→¬L(x,y)))
∧ (∀x)((F(x) ∧ M(c,x))→(∀y)(C(y)→L(y,x)))
∧ ¬ ((∀x)((F(x) ∧ M(c,x))→¬G(x)))
Clauses
1. {C(a)}
2. {¬F(y), L(a,y)}
3. {¬C(x), ¬F(y), ¬(G(y), ¬L(x,y)}
4. {¬F(x), ¬M(c,x), ¬C(y), L(y,x)}
5. {F(b)}
6. {M(c,b)}
7. {G(b)}
Conclusion: No food made by Cadbury’s is green
apply resolution :
First, unify lines 1 and 3 using {a/x} and resolve these two,
to give
8. {¬F(y), ¬(G(y), ¬L(a,y)}
the unifier {b/y} can be applied to lines 2 and 5, which are
then
resolved to give
9. {L(a,b)}
Now we apply {b/y} to resolve line 5 with line 8 to give
10. {¬(G(b), ¬L(a,b)}
Now lines 9 and 10 can be resolved to give
11. {¬(G(b)}
Finally, line 7 can be resolved with line 11 to give
12.
⊥ Falsum
 set of clauses derived from the premises 1, 2, and 3, and
the negation of the conclusion, are unsatisfiable.
 conclusion does indeed follow from the premises,and so
the argument is a valid one
Horn Clauses
• a clause that has, at most, one positive literal
A ∨ ¬B ∨ ¬C ∨ ¬D ∨ ¬E . . .
• can take three forms
 fact: clause with no negative literals
 rule relation: one positive literal, and one or more negative literals
 goal, or a headless clause: no positive literal
• A program in PROLOG consists of a set of Horn clauses
• PROLOG applies unification and resolution to attempt to derive conclusions
from a set ofrules that are defined in the language
• the programmer
 define rules and facts
 set up a goal
 PROLOG attempt to prove the goal, using unification and resolution.
Herbrand Universes
• Herbrand universe (for a ser of Clauses S)
 the set of constants that are contained within S, and the set of functions in S applied to those constants
 Constants and functions are known as ground terms.
S is {{A(x), B(y, a), C(z)}, {D(x, a, b), ¬E(y, c, b)}}
HS = {a, b, c}
S is {{A(x), B(y, a), C(z)}, {D(x, a, b), ¬E(y, f(x, y))}}
HS = {a, b, c, f(a, a), f(a, b), f (b, a), f(b, b), f(f(a), a), f(f(a), f(a)) . . . }
S is {{A(x), B(y, z), C(z)}, {D(x, y, z), ¬E(y, x)}}
HS = {a}

ground instance of a clause in S is a version of that clause in which any variables it contains have been replaced
by ground terms from HS.
S is {{A(x), B(y, a), C(z)}, {D(x, a, b), ¬E(y, c, b)}}
{A(a), B(b, a), C(a)}
•
The Herbrand Base HS(S)
 set of ground atoms that can be obtained by replacing variables in S by members of HS
 A ground atom is a formula that contains no variables, only ground terms
S is {{A(x), B(y, a), C(z)}, {D(x, a, b), ¬E(y, c, b)}}
HS = {a, b, c}
HS(S) = {{A(a), B(a, a), C(a), D(a, a, b), ¬E(a, c, b),A(b), B(b, a), C(b), D(b, a, b), ¬E(b, c, b),. . . }
Summary
• Automated Reasoning and Theorem Proving
• Proof by Refutation or Proof by Contradiction
 Negate conclusion
 Prove resultant set of expressions is unsatisfiable
 Implies argument is valid
• Resolution
 Well Formed Formula to CNF
 Prenex Normal Form
 Skolemization
 Unification
 Resolution
Never theorize before you have data. Invariably, you end up twisting facts to suit
theories, instead of theories to suit facts.
- Sherlock Holmes
http://www.imdb.com/title/tt0988045/trivia?tab=qt&ref_=tt_trv_qu
https://sherlockxxi.wikispaces.com/Funny+things+and+curiosities+about+Sherlock+Holmes
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