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Propositional Logic Example Example
Example
Consider the following sentences:
1 If it rains, the aquaphobes won’t vote2 John will win only if the the aquaphobes and
vegetarians vote3 Either John or Peter will win, but not both4 If it rains, Peter will win.
How would these look in propositional logic?
NC State University 1 / 32First-Order Logic
Propositional Logic Example Example
Representation
• R (It rains); A (Aquaphobes Vote); J (John Wins); V(Vegetarians Vote); P (Peter Wins)
1 If it rains, the aquaphobes won’t vote:2 John will win only if the aquaphobes and vegetarians
vote:3 Either John or Peter will win, but not both:4 If it rains, Peter will win:
NC State University 2 / 32First-Order Logic
Propositional Logic Example Example
Representation
• R (It rains); A (Aquaphobes Vote); J (John Wins); V(Vegetarians Vote); P (Peter Wins)
1 If it rains, the aquaphobes won’t vote: R→ ¬A2 John will win only if the aquaphobes and vegetarians
vote:3 Either John or Peter will win, but not both:4 If it rains, Peter will win:
NC State University 2 / 32First-Order Logic
Propositional Logic Example Example
Representation
• R (It rains); A (Aquaphobes Vote); J (John Wins); V(Vegetarians Vote); P (Peter Wins)
1 If it rains, the aquaphobes won’t vote: R→ ¬A2 John will win only if the aquaphobes and vegetarians
vote: A ∧ V → J3 Either John or Peter will win, but not both:4 If it rains, Peter will win:
NC State University 2 / 32First-Order Logic
Propositional Logic Example Example
Representation
• R (It rains); A (Aquaphobes Vote); J (John Wins); V(Vegetarians Vote); P (Peter Wins)
1 If it rains, the aquaphobes won’t vote: R→ ¬A2 John will win only if the aquaphobes and vegetarians
vote: A ∧ V → J3 Either John or Peter will win, but not both:(J ∨ P) ∧ ¬(J ∧ P)
4 If it rains, Peter will win:
NC State University 2 / 32First-Order Logic
Propositional Logic Example Example
Representation
• R (It rains); A (Aquaphobes Vote); J (John Wins); V(Vegetarians Vote); P (Peter Wins)
1 If it rains, the aquaphobes won’t vote: R→ ¬A2 John will win only if the aquaphobes and vegetarians
vote: A ∧ V → J3 Either John or Peter will win, but not both:(J ∨ P) ∧ ¬(J ∧ P)
4 If it rains, Peter will win: R→ P
NC State University 2 / 32First-Order Logic
Propositional Logic Example Example
More Examples
Consider the following sentences:
• Tweety has feathers.• Donald has feathers.• John is the son of Jack.• If it rains, aquaphobes won’t vote.• If it is cloudy, some aquaphobes will vote.
Convert them into propositional logic.
NC State University 3 / 32First-Order Logic
FOL Motivation
Problems with Propositional Logic
• Propositional logic does not allow us to formulategeneral rules or represent relations and is declarative
• Meaning in propositional logic iscontext-independent
• Propositional logic has very limited expressive power• Reasons:
• No “domain” variables• No quantification
• Solution:• Introduce domain variables and quantification• FOL
NC State University 4 / 32First-Order Logic
FOL Motivation
First Order Logic and PropositionalLogic
Propositional logic assumes the world contains facts.
First-order logic assumes the world contains:
• Objects: people, houses, numbers, colors, baseballgames, wars, . . .
• Relations: red, round, siblingof, biggerthan, partof,comesbetween, . . .
• Functions: onemorethan, plus, . . .
NC State University 5 / 32First-Order Logic
FOL Motivation
First Order Logic and PropositionalLogic . . .
Ontological commitment
what a representation assumes about existence.Temporal logic assumes that facts hold at specific times.Higher-order logic takes relations and functions of FOL tobe objects themselves.
Epistemological commitment
what can be known about what exists.In PL and FOL, sentences are believed to be true or false(or to have an unknown value).
NC State University 6 / 32First-Order Logic
FOL Motivation
First Order Logic and PropositionalLogic . . .
Ontological commitment
what a representation assumes about existence.Temporal logic assumes that facts hold at specific times.Higher-order logic takes relations and functions of FOL tobe objects themselves.
Epistemological commitment
what can be known about what exists.In PL and FOL, sentences are believed to be true or false(or to have an unknown value).NC State University 6 / 32
First-Order Logic
FOL WFFS
Parts of Formulas
Formulas have:
• Variables:• x,y,z• Can be instantiated
• Functions:• f,g,h• mappings
• Constants:• 3, John• Individuals
• Predicates:• P(x,y)• functions whose range
is {True,False}• Quantifiers:
• ∀ for all• ∃ there exists
NC State University 7 / 32First-Order Logic
FOL WFFS
FOL: WFFs
A term is
• A constant, or• A variable, or• f (t1, t2, . . . , tn) where f is an n-ary function symbol,
and ti are terms
NC State University 8 / 32First-Order Logic
FOL WFFS
FOL: WFFs
Atom
If P is an n-place predicate symbol, and t1, t2, . . . , tn areterms, then
P(t1, . . . , tn)
is an atom. It is sometimes referred to as an atomicformula.
Examples
• Reigns(Pope-Innocent-III, Europe, 1200)• Reigns(You, World, 2013)
NC State University 9 / 32First-Order Logic
FOL WFFS
FOL: WFFs. . .
A sentence in FOL is either an atomic sentence orcomplex sentence.
A complex sentence is defined as:Complex sentence:
Sentence, or¬ Sentence, orSentence ∧ Sentence, orSentence ∨ Sentence, orQuantifier Variable Sentence
NC State University 10 / 32First-Order Logic
FOL WFFS
Quantification
Universal
∀〈Variables〉〈Sentence〉Everyone at NCSU is smart: ∀xAt(x,NCSU)→ Smart(x)∀xP is true in a model m iff P is true where x is eachpossible object in the model.
Existential
∃〈Variables〉〈Sentence〉Someone at NCSU is smart: ∃xAt(x,NCSU) ∧ Smart(x)∃xP is true in a model m iff P is true where x is somepossible object in the model.
NC State University 11 / 32First-Order Logic
FOL WFFS
Quantification
Universal
∀〈Variables〉〈Sentence〉Everyone at NCSU is smart: ∀xAt(x,NCSU)→ Smart(x)∀xP is true in a model m iff P is true where x is eachpossible object in the model.
Existential
∃〈Variables〉〈Sentence〉Someone at NCSU is smart: ∃xAt(x,NCSU) ∧ Smart(x)∃xP is true in a model m iff P is true where x is somepossible object in the model.NC State University 11 / 32
First-Order Logic
FOL WFFS
Quantification
Question: What do the following mean?
• ∀xAt(x,NCSU) ∧ Smart(x)• ∃xAt(x,NCSU)→ Smart(x)
NC State University 12 / 32First-Order Logic
FOL WFFS
Quantification . . .
∃x∃y is the same as ∃y∃x. ∀x∀y is the same as ∀y∀x.
But... ∀x∃y is not the same as ∃x∀y.
Consider:∃x∀yLoves(x, y)∀y∃xLoves(x, y)∀x∃yLoves(x, y)
NC State University 13 / 32First-Order Logic
FOL WFFS
Quantification . . .
∃x∃y is the same as ∃y∃x. ∀x∀y is the same as ∀y∀x.
But... ∀x∃y is not the same as ∃x∀y.
Consider:∃x∀yLoves(x, y)“There is a person who loves everyone in the world”
∀y∃xLoves(x, y)∀x∃yLoves(x, y)
NC State University 13 / 32First-Order Logic
FOL WFFS
Quantification . . .
∃x∃y is the same as ∃y∃x. ∀x∀y is the same as ∀y∀x.
But... ∀x∃y is not the same as ∃x∀y.
Consider:∃x∀yLoves(x, y)“There is a person who loves everyone in the world”
∀y∃xLoves(x, y)“Everyone in the world is loved by at least one person”
∀x∃yLoves(x, y)
NC State University 13 / 32First-Order Logic
FOL WFFS
Quantification . . .
∃x∃y is the same as ∃y∃x. ∀x∀y is the same as ∀y∀x.
But... ∀x∃y is not the same as ∃x∀y.
Consider:∃x∀yLoves(x, y)“There is a person who loves everyone in the world”
∀y∃xLoves(x, y)“Everyone in the world is loved by at least one person”
∀x∃yLoves(x, y)“Everyone in the world loves at least one person”NC State University 13 / 32
First-Order Logic
FOL WFFS
Quantification . . .
Quantifier duality: each can be expressed using the other.
∀xLikes(x, IceCream)
¬∃x¬Likes(x, IceCream)
And...
∃xLikes(x,Broccoli)¬∀x¬Likes(x,Broccoli)
NC State University 14 / 32First-Order Logic
FOL WFFS
Free vs Bound Variables
Definition
An occurrence of a variable in a formula is bound iff theoccurrence is in the scope of a quantifier employing thevariable; otherwise it is free.
Examples
• ∀x.P(x, y)• x is bound• y is free
Predicate calculus requires all variables to be bound.NC State University 15 / 32
First-Order Logic
FOL WFFS
More Examples
Consider the sample sentences again:
• Tweety has feathers.• Donald has feathers.• John is the son of Jack.• If it rains, aquaphobes won’t vote• If it is cloudy, some aquaphobes will vote
Convert them into Predicate Logic.
NC State University 16 / 32First-Order Logic
Predicate Logic
A few conventions for FOL
It is important to standardize a few thing to start with:
• Variables will always be lower case letters• Constants will always be upper case letters• Try not to use the same name for a function and a
predicate• predicateName/n represents a predicate that can
taken n arguments• A predicate may never be nested within another!
NC State University 17 / 32First-Order Logic
Predicate Logic
Example - Affine Spaces Theorem
Lexicon:Constants: noneVariables: x, yFunctions: nonePredicates:
affine(x) – x is an affine spaceperequiv(x,y) – spaces x and y are perceptuallyequivalenteuclidean(x) – x is a Euclidean spacelive(x) – we live in space xtrans(x,y) – space y is a transformation of space x
NC State University 18 / 32First-Order Logic
Predicate Logic
Example - Affine Spaces Theorem
Consider the sentences:
1 All affine spaces are perceptually equivalent.2 All Euclidean spaces are affine.3 Not all affine spaces are Euclidean.4 The space we live in is Euclidean.5 Not all transformations produce affine spaces.
Prove that:all affine transformations of the space we live in areperceptually equivalent to it.
NC State University 19 / 32First-Order Logic
Predicate Logic
Affine Spaces . . .
1 All affine spaces are perceptually equivalent.2 All Euclidean spaces are affine.3 Not all affine spaces are Euclidean.4 The space we live in is Euclidean.5 Not all transformations produce affine spaces.
NC State University 20 / 32First-Order Logic
Predicate Logic
Affine Spaces . . .
1 All affine spaces are perceptually equivalent.∀x(∀y(affine(x) ∧ affine(y)→ perequiv(x, y)))
2 All Euclidean spaces are affine.3 Not all affine spaces are Euclidean.4 The space we live in is Euclidean.5 Not all transformations produce affine spaces.
NC State University 20 / 32First-Order Logic
Predicate Logic
Affine Spaces . . .
1 All affine spaces are perceptually equivalent.∀x(∀y(affine(x) ∧ affine(y)→ perequiv(x, y)))
2 All Euclidean spaces are affine.∀x(euclidean(x)→ affine(x))
3 Not all affine spaces are Euclidean.4 The space we live in is Euclidean.5 Not all transformations produce affine spaces.
NC State University 20 / 32First-Order Logic
Predicate Logic
Affine Spaces . . .
1 All affine spaces are perceptually equivalent.∀x(∀y(affine(x) ∧ affine(y)→ perequiv(x, y)))
2 All Euclidean spaces are affine.∀x(euclidean(x)→ affine(x))
3 Not all affine spaces are Euclidean.∃x(affine(x) ∧ ¬euclidean(x))
4 The space we live in is Euclidean.5 Not all transformations produce affine spaces.
NC State University 20 / 32First-Order Logic
Predicate Logic
Affine Spaces . . .
1 All affine spaces are perceptually equivalent.∀x(∀y(affine(x) ∧ affine(y)→ perequiv(x, y)))
2 All Euclidean spaces are affine.∀x(euclidean(x)→ affine(x))
3 Not all affine spaces are Euclidean.∃x(affine(x) ∧ ¬euclidean(x))
4 The space we live in is Euclidean.∀x(live(x)→ euclidean(x))
5 Not all transformations produce affine spaces.
NC State University 20 / 32First-Order Logic
Predicate Logic
Affine Spaces . . .
1 All affine spaces are perceptually equivalent.∀x(∀y(affine(x) ∧ affine(y)→ perequiv(x, y)))
2 All Euclidean spaces are affine.∀x(euclidean(x)→ affine(x))
3 Not all affine spaces are Euclidean.∃x(affine(x) ∧ ¬euclidean(x))
4 The space we live in is Euclidean.∀x(live(x)→ euclidean(x))
5 Not all transformations produce affine spaces.∃x(∃y(trans(x, y) ∧ ¬affine(y)))
NC State University 20 / 32First-Order Logic
Predicate Logic
Affine Spaces . . .
Conclusion:all affine transformations of the space we livein are perceptually equivalent to it.
NC State University 21 / 32First-Order Logic
Predicate Logic
Affine Spaces . . .
Conclusion:all affine transformations of the space we livein are perceptually equivalent to it.∀x(∀y(live(x) ∧ trans(x, y) ∧ affine(y)→ perequiv(x, y)))
NC State University 21 / 32First-Order Logic
Theorem Proving with FOL
So, how do we prove a theorem?
In order to prove a theorem, we must:
• represent the given sentences in CNF• negate the conclusion – since we are proving by
contradiction• iteratively find a pair of clauses that can be resolved
till we reach a contradiction
NC State University 22 / 32First-Order Logic
Theorem Proving with FOL
So, how do we prove a theorem?
In order to prove a theorem, we must:
• represent the given sentences in CNF• negate the conclusion – since we are proving by
contradiction• iteratively find a pair of clauses that can be resolved
till we reach a contradiction
NC State University 22 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
CNF in predicate Logic
Consider the following clauses and identify which ofthese are in CNF (textbook, p. 346):
• ∀x p(x)→ q(x)• ¬∀x P(x) ∨ Q(x)• ∀x P(x)∨ ∃x Q(x)• ∀x∃y P(x) ∨ Q(y)
NC State University 23 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
CNF in predicate Logic
Consider the following clauses and identify which ofthese are in CNF (textbook, p. 346):
• ∀x p(x)→ q(x)Remove ‘→’: ∀x ¬p(x) ∨ q(x)
• ¬∀x P(x) ∨ Q(x)• ∀x P(x)∨ ∃x Q(x)• ∀x∃y P(x) ∨ Q(y)
NC State University 23 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
CNF in predicate Logic
Consider the following clauses and identify which ofthese are in CNF (textbook, p. 346):
• ∀x p(x)→ q(x)Remove ‘→’: ∀x ¬p(x) ∨ q(x)
• ¬∀x P(x) ∨ Q(x)Move the negation inwards: ∃x ¬P(x) ∧ ¬Q(x)
• ∀x P(x)∨ ∃x Q(x)• ∀x∃y P(x) ∨ Q(y)
NC State University 23 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
CNF in predicate Logic
Consider the following clauses and identify which ofthese are in CNF (textbook, p. 346):
• ∀x p(x)→ q(x)Remove ‘→’: ∀x ¬p(x) ∨ q(x)
• ¬∀x P(x) ∨ Q(x)Move the negation inwards: ∃x ¬P(x) ∧ ¬Q(x)
• ∀x P(x)∨ ∃x Q(x)Standardize the variables: ∀x1 P(x1)∨ ∃x2 Q(x2)
• ∀x∃y P(x) ∨ Q(y)
NC State University 23 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
CNF in predicate Logic
Consider the following clauses and identify which ofthese are in CNF (textbook, p. 346):
• ∀x p(x)→ q(x)Remove ‘→’: ∀x ¬p(x) ∨ q(x)
• ¬∀x P(x) ∨ Q(x)Move the negation inwards: ∃x ¬P(x) ∧ ¬Q(x)
• ∀x P(x)∨ ∃x Q(x)Standardize the variables: ∀x1 P(x1)∨ ∃x2 Q(x2)
• ∀x∃y P(x) ∨ Q(y)Interesting...
NC State University 23 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
CNF in predicate Logic
Consider the following clauses and identify which ofthese are in CNF (textbook, p. 346):
• ∀x p(x)→ q(x)Remove ‘→’: ∀x ¬p(x) ∨ q(x)
• ¬∀x P(x) ∨ Q(x)Move the negation inwards: ∃x ¬P(x) ∧ ¬Q(x)
• ∀x P(x)∨ ∃x Q(x)Standardize the variables: ∀x1 P(x1)∨ ∃x2 Q(x2)
• ∀x∃y P(x) ∨ Q(y)Interesting...
By the way, none of these are in CNF yet!!NC State University 23 / 32
First-Order Logic
Theorem Proving with FOL Conversion to CNF
Dealing with Quantifiers
We can handle ‘∃’ as follows:
• We eliminate the ‘∃’ by a process calledSkolemization
• We will use special terms called Skolem Constantsand Skolem Functions
• Consider:• ∀x∃y P(x) ∨ Q(y)
• ∃x∀y P(x) ∨ Q(y)
Once this is complete, we can drop the ‘∀’ quantifierwithout any change to the corresponding variables.
NC State University 24 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
Dealing with Quantifiers
We can handle ‘∃’ as follows:
• We eliminate the ‘∃’ by a process calledSkolemization
• We will use special terms called Skolem Constantsand Skolem Functions
• Consider:• ∀x∃y P(x) ∨ Q(y)
Replace y with Skolem function ‘f (x)’: ∀x P(x)∨Q(f (x))
• ∃x∀y P(x) ∨ Q(y)
Once this is complete, we can drop the ‘∀’ quantifierwithout any change to the corresponding variables.NC State University 24 / 32
First-Order Logic
Theorem Proving with FOL Conversion to CNF
Dealing with Quantifiers
We can handle ‘∃’ as follows:
• We eliminate the ‘∃’ by a process calledSkolemization
• We will use special terms called Skolem Constantsand Skolem Functions
• Consider:• ∀x∃y P(x) ∨ Q(y)
Replace y with Skolem function ‘f (x)’: ∀x P(x)∨Q(f (x))
• ∃x∀y P(x) ∨ Q(y)Replace x with Skolem constant ‘X1’: ∀y P(X1) ∨ Q(y)
Once this is complete, we can drop the ‘∀’ quantifierwithout any change to the corresponding variables.NC State University 24 / 32
First-Order Logic
Theorem Proving with FOL Conversion to CNF
Example
• ∀x∃y(man(x)→ woman(y) ∧ loves(x, y))• ∃x∀y∀z∃u∀v∃w P(x, y, z, u, v,w)
• ∃x→ A• ∃u→ f (y, z)• ∃w→ g(y, z, v)
NC State University 25 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
Example
• ∀x∃y(man(x)→ woman(y) ∧ loves(x, y))∀x(man(x)→ woman(f (x)) ∧ loves(x, f (x))
• ∃x∀y∀z∃u∀v∃w P(x, y, z, u, v,w)
• ∃x→ A• ∃u→ f (y, z)• ∃w→ g(y, z, v)
NC State University 25 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
Example
• ∀x∃y(man(x)→ woman(y) ∧ loves(x, y))∀x(man(x)→ woman(f (x)) ∧ loves(x, f (x))
• ∃x∀y∀z∃u∀v∃w P(x, y, z, u, v,w)• ∃x→ A
• ∃u→ f (y, z)• ∃w→ g(y, z, v)
NC State University 25 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
Example
• ∀x∃y(man(x)→ woman(y) ∧ loves(x, y))∀x(man(x)→ woman(f (x)) ∧ loves(x, f (x))
• ∃x∀y∀z∃u∀v∃w P(x, y, z, u, v,w)• ∃x→ A• ∃u→ f (y, z)
• ∃w→ g(y, z, v)
NC State University 25 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
Example
• ∀x∃y(man(x)→ woman(y) ∧ loves(x, y))∀x(man(x)→ woman(f (x)) ∧ loves(x, f (x))
• ∃x∀y∀z∃u∀v∃w P(x, y, z, u, v,w)• ∃x→ A• ∃u→ f (y, z)• ∃w→ g(y, z, v)
NC State University 25 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
Example
• ∀x∃y(man(x)→ woman(y) ∧ loves(x, y))∀x(man(x)→ woman(f (x)) ∧ loves(x, f (x))
• ∃x∀y∀z∃u∀v∃w P(x, y, z, u, v,w)• ∃x→ A• ∃u→ f (y, z)• ∃w→ g(y, z, v)
∀y∀z∀v P(A, y, z, f (y, z), v, g(y, z, v))
NC State University 25 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
Back to Affine Spaces . . .
Convert the sentences into CNF:
1 All affine spaces are perceptually equivalent.∀x(∀y(affine(x) ∧ affine(y)→ perequiv(x, y)))
2 All Euclidean spaces are affine.∀x(euclidean(x)→ affine(x))
3 Not all affine spaces are Euclidean.∃x(affine(x) ∧ ¬euclidean(x))
NC State University 26 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
Back to Affine Spaces . . .
Convert the sentences into CNF:
1 All affine spaces are perceptually equivalent.∀x(∀y(affine(x) ∧ affine(y)→ perequiv(x, y)))¬affine(x) ∨ ¬affine(y) ∨ perequiv(x, y)
2 All Euclidean spaces are affine.∀x(euclidean(x)→ affine(x))
3 Not all affine spaces are Euclidean.∃x(affine(x) ∧ ¬euclidean(x))
NC State University 26 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
Back to Affine Spaces . . .
Convert the sentences into CNF:
1 All affine spaces are perceptually equivalent.∀x(∀y(affine(x) ∧ affine(y)→ perequiv(x, y)))¬affine(x) ∨ ¬affine(y) ∨ perequiv(x, y)
2 All Euclidean spaces are affine.∀x(euclidean(x)→ affine(x))¬euclidean(x) ∨ affine(x)
3 Not all affine spaces are Euclidean.∃x(affine(x) ∧ ¬euclidean(x))
NC State University 26 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
Back to Affine Spaces . . .
Convert the sentences into CNF:
1 All affine spaces are perceptually equivalent.∀x(∀y(affine(x) ∧ affine(y)→ perequiv(x, y)))¬affine(x) ∨ ¬affine(y) ∨ perequiv(x, y)
2 All Euclidean spaces are affine.∀x(euclidean(x)→ affine(x))¬euclidean(x) ∨ affine(x)
3 Not all affine spaces are Euclidean.∃x(affine(x) ∧ ¬euclidean(x))affine(A) ∧ ¬euclidean(A)
NC State University 26 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
Back to Affine Spaces . . .
Convert the sentences into CNF:
1 The space we live in is Euclidean.∀x(live(x)→ euclidean(x))
2 Not all transformations produce affine spaces.∃x(∃y(trans(x, y) ∧ ¬affine(y)))
NC State University 27 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
Back to Affine Spaces . . .
Convert the sentences into CNF:
1 The space we live in is Euclidean.∀x(live(x)→ euclidean(x))¬live(x) ∨ euclidean(x)
2 Not all transformations produce affine spaces.∃x(∃y(trans(x, y) ∧ ¬affine(y)))
NC State University 27 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
Back to Affine Spaces . . .
Convert the sentences into CNF:
1 The space we live in is Euclidean.∀x(live(x)→ euclidean(x))¬live(x) ∨ euclidean(x)
2 Not all transformations produce affine spaces.∃x(∃y(trans(x, y) ∧ ¬affine(y)))trans(B,C) ∧ ¬affine(C)
NC State University 27 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
Negated Conclusion for Affine Spaces
Conclusion:
All affine transformations of the space we live in areperceptually equivalent to it.
∀x(∀y(live(x) ∧ trans(x, y) ∧ affine(y)→ perequiv(x, y)))
NC State University 28 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
Negated Conclusion for Affine Spaces
Negated Conclusion:
¬∀x(∀y(live(x) ∧ trans(x, y) ∧ affine(y)→perequiv(x, y)))
⇒ ¬∀x(∀y(¬(live(x) ∧ trans(x, y) ∧ affine(y)) ∨perequiv(x, y)))
⇒ ∃x(∃y(live(x) ∧ trans(x, y) ∧ affine(y) ∧¬perequiv(x, y)))
⇒ live(D) ∧ trans(D,E) ∧ affine(E) ∧ ¬perequiv(D,E)
NC State University 29 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
Negated Conclusion for Affine Spaces
Negated Conclusion:
¬∀x(∀y(live(x) ∧ trans(x, y) ∧ affine(y)→perequiv(x, y)))
⇒ ¬∀x(∀y(¬(live(x) ∧ trans(x, y) ∧ affine(y)) ∨perequiv(x, y)))
⇒ ∃x(∃y(live(x) ∧ trans(x, y) ∧ affine(y) ∧¬perequiv(x, y)))
⇒ live(D) ∧ trans(D,E) ∧ affine(E) ∧ ¬perequiv(D,E)
NC State University 29 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
Negated Conclusion for Affine Spaces
Negated Conclusion:
¬∀x(∀y(live(x) ∧ trans(x, y) ∧ affine(y)→perequiv(x, y)))
⇒ ¬∀x(∀y(¬(live(x) ∧ trans(x, y) ∧ affine(y)) ∨perequiv(x, y)))
⇒ ∃x(∃y(live(x) ∧ trans(x, y) ∧ affine(y) ∧¬perequiv(x, y)))
⇒ live(D) ∧ trans(D,E) ∧ affine(E) ∧ ¬perequiv(D,E)
NC State University 29 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
Negated Conclusion for Affine Spaces
Negated Conclusion:
¬∀x(∀y(live(x) ∧ trans(x, y) ∧ affine(y)→perequiv(x, y)))
⇒ ¬∀x(∀y(¬(live(x) ∧ trans(x, y) ∧ affine(y)) ∨perequiv(x, y)))
⇒ ∃x(∃y(live(x) ∧ trans(x, y) ∧ affine(y) ∧¬perequiv(x, y)))
⇒ live(D) ∧ trans(D,E) ∧ affine(E) ∧ ¬perequiv(D,E)
NC State University 29 / 32First-Order Logic
Theorem Proving with FOL Conversion to CNF
Negated Conclusion for Affine Spaces
Negated Conclusion:
¬∀x(∀y(live(x) ∧ trans(x, y) ∧ affine(y)→perequiv(x, y)))
⇒ ¬∀x(∀y(¬(live(x) ∧ trans(x, y) ∧ affine(y)) ∨perequiv(x, y)))
⇒ ∃x(∃y(live(x) ∧ trans(x, y) ∧ affine(y) ∧¬perequiv(x, y)))
⇒ live(D) ∧ trans(D,E) ∧ affine(E) ∧ ¬perequiv(D,E)
NC State University 29 / 32First-Order Logic
Theorem Proving with FOL Resolution
Resolution
Consider:
¬euclidean(x2) ∨ affine(x2)
¬affine(c)
These have a pair of complementary predicates. But, canwe resolve them? If so, how?
Now assume:
¬euclidean(x2) ∨ affine(f (x2))
¬affine(c)
Can we resolve them? If so, how?
The idea is to make the 2 predicates ‘similar’ so that theycan be resolved. So, we try to find a set of substitutionsthat unify the predicates.
NC State University 30 / 32First-Order Logic
Theorem Proving with FOL Resolution
Substitutions
Definition
A substitution is a set of pairs {t1/v1, t2/v2, . . . , tn/vn},where the ti are terms, the vi are variables, and no two viare the same.
Substitutions always replace variables with terms. Legalsubstitutions are
• constant-for-var• function-for-var• var-for-var
The vars substituted for may occur in either literal.
Example:
p(x,A, f (g(y)), z) and ¬p(C, s, f (t),C) unify under thesubstitution:
{C/x,A/s, g(y)/t,C/z}
NC State University 31 / 32First-Order Logic
Theorem Proving with FOL Resolution
Substitutions
Definition
A substitution is a set of pairs {t1/v1, t2/v2, . . . , tn/vn},where the ti are terms, the vi are variables, and no two viare the same.
Substitutions always replace variables with terms. Legalsubstitutions are
• constant-for-var• function-for-var• var-for-var
The vars substituted for may occur in either literal.
Example:
p(x,A, f (g(y)), z) and ¬p(C, s, f (t),C) unify under thesubstitution: {C/x,A/s, g(y)/t,C/z}
NC State University 31 / 32First-Order Logic
Theorem Proving with FOL Resolution
Unification
Definition
The process of finding a substitution that makes twoliterals complementary is called unification.
Two literals for which a unifying substitution exists arecalled unifiable.Importance of Unification
• It is the basis for FOL resolution.• It is the main way rule-based systems determine
which rules apply in a situation.• It is the way variables are treated in logic
programming.NC State University 32 / 32First-Order Logic