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Predicate Logic



Chapter 1  Predicate Logic (sections 1.3, 1.4)

1.1  Introduction


Exemple 1   ``Every man has a brain; Socrate is a man; therefore Socrate has a brain.''

Is propositional logic expressive enough for this sentence ?

What is missing ?

1.1.1  Quantifiers, Predicates, Validity


Exemple 2   ``Every man has a brain''

``For all man, this man has a brain''
New features: constants / variables / quantifier / predicates
Définition 1   Quantifiers: Universal quantifier: (for all x)

Existential quantifier: (there exists x)

Always place the quantifiers inside ( ).

1.1.2  Predicates:


Définition 2   Predicate: a statement that describes a property of a variable
Exemple 3   P(x) : x has a brain

``For all man, this man has a brain''

Translates into (for allx ) P(x)
As in propositional logic, we can build well formed formulas from predicates and logic connectives.
Exemple 4   (for allx ) B(x)-> R(x)': if x is blue, then x is not red.

1.1.3  Domain of interpretation:

If x is a car, the sentence above (for allx ) P(x) is false.

The truth of the sentence depends on where you take x from!
Définition 3   Domain of interpretation:

The collection of objects from which x may be selected
Exemple 5   (for allx ) (x>0)

DOI: all integers

DOI: all integers > 10

DOI: all humans

Exemple 6   For any thing, if it's a man then it has a brain

P(x): x is a man

Q(x): x has a brain

(for allx ) P(x)-> Q(x)
Exercice 1   Write a formula for the following sentence:

``There exists a man which is called Socrate''

1.1.4  N-Ary predicates:


Définition 4   Binary predicate: predicate which takes two variables

N-Ary predicate : predicate which takes several variables

1.1.5  Truth value

Can we define the Truth table of a wff ?
Problem 1   The domain of interpretation may be infinite.
To assess the truth value, we need to know:
  1. What is the domain of interpretation.
  2. What property of elements of the DOI does P(x) represent.
  3. An assignment for each constant.
Définition 5   This is called an interpretation.
Exemple 7   [, Exercise 1 p. 41]

What is the truth value of the following wffs for this interpretation:

DOI: integers; O(x): x is odd; L(x): x<10; G(x): x>9
  1. (there existsx ) O(x)

    There exists an odd number

  2. (for allx ) [L(x)-> O(x)]

    For any integer x, if x is strictly lower than 10, then x is odd;

    Any integer strictly lower than 10 is odd;

  3. (there existsx ) [L(x)/\ G(x)]

    There exists an integer x such that x is strictly lower than 10 and x is strictly larger than 9

  4. (for allx ) [L(x)\/ G(x)]

    For any integer, either x is strictly lower than 10 or x is strictly bigger that 9
Exercice 2   [, Exercise 3 p. 41]

Are the following wffs true for this interpretation:

DOI: states of the US.

Q(x,y): x is north of y; P(x): x starts with the letter M; a is "Mississippi".
  1. (for allx ) P(x)
  2. (for allx ) (for ally ) (for allz ) [ (Q(x,y)/\ Q(y,z)) -> Q(x,z) ]
  3. (there existsy ) (there existsx ) Q(x,y)
  4. (for allx ) (there existsy ) [P(y)/\ Q(x,y)]

    For any state x, there exists a state y such that y starts with M and x is north of y.

  5. (there existsy ) Q(a,y)

Exercice 3   Translate the following sentences into wffs, with:

DOI: world; D(x): x is a day; S(x): x is sunny; R(x): x is rainy;

M is "Monday"; T is "Tuesday".
  1. All days are sunny:
  2. Some days are not rainy:
  3. Every day that is sunny is not rainy:
  4. Some days are sunny and rainy:
  5. No day is both sunny and rainy:
  6. It is always a sunny day only if it is a rainy day:
  7. No day is sunny:
  8. Monday was sunny; therefore every day will be sunny:
  9. It rained both Monday and Tuesday:
  10. If some day is rainy, then every day will be sunny:

1.1.6  Dummy variables / free variables:


Définition 6   A dummy variable is a variable which is linked to a quantifier.

The name of the variable is irrelevant.

Same thing as a local variable in a program.
Exemple 8   (there existsx ) Q(x) is the same wff as (there existsz ) Q(z)

(there existsx ) (for ally ) Q(x,y) is the same wff as (there existsz ) (for allt ) Q(z,t) or (there existsy ) (for allx ) Q(y,x)
Définition 7   A variable is free if it is not linked to a quantifier.

Same thing as a global variable in a program.

1.1.7  Validity


Résumé 2   Where are we?
  1. We can build all the wff of predicate logic.
  2. Given a wff P, and an interpretation, we can decide the truth value of P.
Définition 8   Argument: P1/\ P2/\···/\ Pk -> Q
Exercice 4   [, Exercise 16 p. 35]

Give interpretations to prove that the following wffs are not valid:
  1. (there existsx ) A(x)/\(there existsx ) B(x) -> (there existsx ) [ A(x)/\ B(x) ]
  2. (for allx ) (there existsy ) P(x,y) -> (there existsx ) (for ally ) P(x,y)
  3. (for allx ) [P(x)-> Q(x)] -> (for allx ) [P(x)\/ Q(x)]
  4. (for allx ) [ A(x)' ] <-> [(for allx ) A(x) ]'

1.2  Predicate logic

There is an infinity of interpretations, so there is no algorithm to check the validity of a predicate.

We will have to use REASON:

1.2.1  Universal instantiation:


Exemple 9   We want to be able to prove the following argument:

Every human is mortal; Socrate is a man; Therefore Socrate is mortal.

H(x): x is a human; M(x): x is a mortal; s: Socrate

(for allx ) [H(x) -> M(x)]/\ H(s) -> M(s)
Définition 9   Rule of universal instantiation (ui):

From: (for allx ) P(x)

Can derive: P(s)

Note: s can be any constant.
We decide that this rule is valid, because it is intuitively valid.

Proof sequence for: (for allx ) [H(x)-> M(x)]/\ H(s) -> M(s)
1. (for allx ) [H(x)-> M(x)] (hyp)
2. H(s) (hyp)
3. H(s)-> M(s) (ui 1)
4. M(s) (mp 2, 3)

Exercice 5   Prove (for allx ) [ P(x)-> R(x) ]/\[ R(y)' ] -> [ P(y)' ]

1.2.2  Universal generalization:


Exemple 10   ``Every human is a living being. Every living being is mortal.Therefore every human is mortal.''

H(x): is a human

L(x): is a living being

M(x) is mortal

(for allx ) [H(x)-> L(x)] /\ (for allx ) [L(x)-> M(x)] -> (for allx ) [H(x)-> M(x)]
This is clearly something we want to be able to prove, but we cannot use hypothetical syllogism directly!
Définition 10   Rule of universal generalization (ug)

From: P(s)

Can derive: (for allx ) P(x)

s must be an arbitrary element of the domain.
Exemple 11   (for allx ) [H(x)-> L(x)] /\ (for allx ) [L(x)-> M(x)] -> (for allx ) [H(x)-> M(x)]
1. (for allx ) H(x)-> L(x) (hyp)
2. (for allx ) L(x)-> M(x) (hyp)
3. H(s)-> M(s) (ui 1)
4. L(s)-> M(s) (ui 2)
5. H(s)-> M(s) (hs 3,4)
6. (for allx ) [H(x)-> M(x)] (ug 5)

Exemple 12   s: Socrate; H(x): x is a man; M(x): x is mortal:
1. M(s) (hyp)
2. H(s)'\/ M(s) (add 1)
3. H(s)-> M(s) (imp 2)
4. (for allx ) [H(x)-> M(x)] (ug 3)

Socrate is a mortal, therefore every man is a mortal.
This proof sequence is incorrect: you cannot apply ug at step 4.

Indeed s is Socrate, and not an arbitrary element of the domain.
Exercice 6   Prove the following arguments:
  1. (for allx ) [P(x)/\ Q(x)] -> (for allx ) P(x)/\(for allx ) Q(x)
  2. (for allx ) P(x)/\(for allx ) Q(x) -> (for allx ) (P(x)/\ Q(x))
  3. (for allx ) [P(x)/\(for ally ) Q(x,y)] -> (for allx ) (for ally ) Q(x,y)

1.2.3  Existential instantiation:


Exemple 13   DOI: contents of the fridge.

There exists a fruit; therefore, I can take a fruit.

F(x): x is a fruit

(there existsx ) F(x) -> F(s)
Définition 11   Rule of existential instantiation (ei):

From: (there existsx ) P(x)

Can derive: P(s)

s must be a newly created variable
Exemple 14   (for allx ) [P(x)-> Q(x)] /\(there existsy ) P(y) -> Q(s)
1. (for allx ) [P(x)-> Q(x)] (hyp)
2. (there existsy ) P(y) (hyp)
3. P(s) (ei 2)
4. P(s)-> Q(s) (ui 1)
5. Q(s) (mp 3,4)

1.2.4  Existential generalization:


Exemple 15   DOI: contents of the fridge

I see a fruit, therefore there exists a fruit.

F(s) -> (there existsx ) F(x)
Définition 12   Rule of existential generalization (eg)

From: F(s)

Can derive:(there existsx ) F(x)
Exemple 16   (for allx ) [P(x)-> Q(x)] /\ (there existsy ) P(y) -> (there existsy ) Q(y)
1. (for allx ) [P(x)-> Q(x)] (hyp)
2. (there existsy ) P(y) (hyp)
3. P(s) (ei 2)
4. P(s)-> Q(s) (ui 1)
5. Q(s) (mp 3,4)
6. (there existsy ) Q(y) (eg 5)


Exemple 17   (there existsx ) P(x)/\(there existsx ) Q(x) -> (there existsx ) [P(x)/\ Q(x)]
1. (there existsx ) P(x) (hyp)
2. (there existsx ) Q(x) (hyp)
3. P(s) (ei 1)
4. Q(s) (ei 2)
5. P(s)/\ Q(s) (add 3,4)
6. (there existsx ) [P(x)/\ Q(x)] (eg 5)


We have seen that the previous argument is incorrect:

So, where is the flaw in the following proof ?

For example, s cannot have been created by existential instantiation
Exercice 7   Prove or disprove the following arguments
  1. (there existsx ) (for ally ) Q(x,y) -> (for ally ) (there existsx ) Q(x,y)
  2. (for allx ) (there existsy ) Q(x,y) -> (there existsy ) (for allx ) Q(x,y)

1.2.5  Deduction Method and Temporary Hypothesis:


Exemple 18   P(s)->(for ally ) Q(x,y) -> (for ally ) [P(s)-> Q(x,y)]

1.3  Conclusion

Goal: formalization of arguments and proofs.
Propositional logic:
 

propositions A, B, ...

connectives, well formed formula

truth table

argument valid iff wff is a tautology

proofs: Ex 24, 37

Predicate logic
 

variables, predicates

domain of interpretation

connectives, wff

interpretation

truth table: all possible interpretations (infinite)

argument valid iff wff is true for all possible interpretations

proofs:
Those two logic are correct and complete.

They are still not powerful enough to represent all real life argument.

For this we would need of more powerful logics (2nd order), that allows for quantifying other sets.

Problem: it's often difficult, if not impossible to prove that those logics are complete and correct!

We won't need to go into those details.

We have seen enough low-level logic to help us write safely less formal proofs.
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Predicate Logic / CSM MACS-358 / Nicolas M. Thiéry
Last modified: Fri Jan 19 13:12:15 2007