Introduction to Theorem Proving

1
Introduction to Theorem Proving

• Suppose p1, p2, - - - -, pn and C are
  propositions for which (p1 /\ p2 /\ - - - - -/\
  pn) -gt C is a tautology. Then
• p1 /\ p2 /\ - - - /\ pn is called the premise
• C is called the conclusion
• of a syllogism.
• - So, a syllogism is a deductive reasoning
  scheme composed of premises and a conclusion.
• We also say that p1,p2, - - - -, C is a
  Theorem. Then we can represent the theorem as
  follows.
• p1, p2, - - - , pn gtC

Note that I switched -gt to mean implies and gt
to mean a theorem because I could not find the
right symbol on my keyboard.
2
Checking for Theorem

• There are 2 standard ways to check for theorems
  using the Truth-Table
• Forward chaining
• Check to see that whenever p1, p2, - - -, pn are
  all True, then C is also True
• Backward chaining
• Check to see that whenever C is False, there is
  at least one of the p1, p2, - - - , pn is also
  False

3
Strange Example 1
Lets try a little strange set of premises and
conclusion. (Note that premises and conclusions
are all propositions and the truth value of a
compound proposition depends on that of its
components.) let p1 a printer is a device
p2 if Ram is in class, then a printer is a
device C Ram is in class
P2 (C-gtp1)
C
p1
Forward chaining p1 and p2 are T in rows 1 and
2 but C is F in row 2.
T
T
T
T
T
F
Also Backward chaining C is F in rows 2 and 4,
but in row 2 both p1 and p2 are T.
F
T
F
F
F
T
So, p1, p2, C is not a theorem or
p1, p2 /gt C
4
Rewording Example 1
Lets reword this to a more normal look let
p1 Ram is in class p2 if Ram is in
class, then a printer is a device C a
printer is a device
Forward chaining p1 and p2 are T in row 1 and
C is also T in row 1. There is no other row with
p1 and p2 both having T value.
P2 (p1 -gt C)
C
p1
T
T
T
F
T
F
Also Backward chaining C is F in rows 2 and 4.
In row 2, p2 is also F, and in row 4, p1 is also
F.
F
T
T
F
F
T
So, p1, p2, C is a theorem or p1,
p2 gt C
This is an example of modus ponens where we
have p1, p1-gtC gt C.
5
Theorem Proving by Manipulating Syntax

• We have used the Truth table for demonstrating
  tautologies, proof with forward chaining, and
  backward chaining.
• Can we demonstrate an expression or set of
  statements to be a theorem by manipulating the
  syntax ?
• An expression or a set of statements is proven to
  be a theorem if we show that the expression is
  derived from a set of premises.
• A procedure often used in derivation is
• Start with a premise which may be a given axiom
  or a previously proven theorem
• Utilize a logical sequence of tautologies or
  previously proven theorems to arrive at the
  expression which needs to be proven.
• Another procedure often used is the reverse of
  above
• Start with the expression that needs to be proven
  as theorem
• Utilize a sequence of tautologies or previously
  proven theorems to arrive at a known theorem or
  axiom.
• The term utilize a sequence of - - - means to
  replace any of the expression by any statement
  that is logically equivalent. (the most obvious
  is to use the law of inference such as modus
  ponens)

6
List of Some Tautologies

• p lt-gt p
• p /\ q lt - gt q /\ p
• p \/ q lt - gt q \/ p
• (p /\ q) /\ r lt - gt p /\ (q /\ r)
  (associative law)
• (p \/ q) \/ r lt - gt p \/ (q \/ r)
  (associative law)
• p /\ (q \/ r) lt - gt (p /\ q) \/ (p/\ r)
  (distributive law)
• p \/ (q /\ r) lt - gt (p \/ q) /\ (p \/ r)
  (distributive law)
• (p /\ q) lt - gt p \/ q (De Morgan)
• (p \/ q) lt - gt p /\ q (De Morgan)
• p \/ p lt - gt p
• p /\ p lt - gt p
• p /\ q -gt q
• p /\ q -gt p
• p \/ False - gt p
• p /\ True -gt p
• p -gt p \/ q
• q -gt p v q

Note that these are almost reversed
7
List of Some Theorems

• 1. p, q gt p/\ q
• 2. p, p- gtq gt q (modus ponens)
• 3. p, p \/ q gt q
• 4. q, p-gtq gt p (modus tolens)
• 5. p \/ q, p-gt r, q -gt r gt r
• 6. p gt q, q -gt r gt p -gt r
• 7. p, p -gtq, q -gt r gt r
• 8. p \/ ( q /\ q) ? p
• 9. p /\ ( q \/ q) ? p
• 10. p -gt q ? p \/ q
• 11. p -gt q ? q -gt p
• 12 (p -gt q) ? p /\ q
• 13. p lt-gt q ? (p -gt q) /\ ( q -gt p)
• 14. plt-gtq ? (p /\ q) \/ (p /\ q)
• 15. p -gt ( q -gt r) ? (p /\q) -gtr

Note the symbolisms gt means is a theorem ?
means is the same or equals
8
Some Examples of proving theorems
withmanipulating symbols

• p, p v q gt q
• p ? ( p v q) -gt q
• p ? ( p v q) v q
• p v (p v q) v q
• p v q v (p v q)
• (p v q ) v (p v q)
• True
• thus p ? ( p v q) -gt q is a tautology

• p, q gt ( p ? q)
• p ?q -gt (p ? q)
• (p ? q) V (p ? q)
• True
• thus p ?q -gt (p ? q) is a tautology

9
More example of proving with manipulating
symbols
14. Plt-gtQ ? (P?Q) V (P ? Q)
Plt-gtQ (P-gtQ) ? (Q-gtP) (P VQ) ? (Q
V P) P ? (Q V P) V Q ? (Q v P)
(P ? Q) V(P ? P) V (Q ? Q) V (Q ? P)
(P ? Q) V (F) V ( F)
V (Q ? P) (P ? Q) V
(Q ? P) (Q ? P ) V (P ? Q)

• P V (Q ? Q) ? P
• P v ( Q ? Q)
• P v (F)
• P

10
Conjunctive Normal Form Disjunctive Normal Form
Conjunctive Normal Form is a conjunction (AND) of
expressions where each expression is a
disjunction (Or) of literals. note a
literal is a proposition such as P, Q, or
P. Example of Conjunctive Normal Form (A V B
V C) ? (B V C)
Disjunctive Normal Form is a disjunction (OR) of
expressions where each expression is a
conjunction (AND) of literals. note a
literal is a proposition such as P, Q, or
P. Example of Disjunctive Normal Form (A ? B
? C) V (B ? C)
11
Can Any Logical Expression be Expressed in either
Disjunctive Normal Form or Conjunctive Normal
Form?
Lets look at an example with truth table
B
Exp
A
Exp is true on 1, 3rd, 4th line (A ? B) or (A
?B) or (A ?B) are true.
T
T
T
T
F
F
T
T
F
T
F
F
We can express Exp with the following
Disjunctive Normal Form
Exp (A ? B) V (A ?B) V (A ?B)
12
Simplify the sample Disjunctive Normal Form

• (A ? B) V (A ?B) V (A ?B)
• A V(A ?B) ? B V (A ?B) V (A ? B)
• (AVA) ? (AVB) ? (BVA) ?(BVB) V (A
  ?B)
• T ? (AVB) ? (BVA) ? (B)
  V (A ?B)
• (A V B) ? ( B)
  V (A ?B)
• (A ? B) V (B ? B)
  V (A ?B)
• (A ? B) V (B)
  V (A ?B)
• (B)
  V (A ?B)
• (B V A) ? (B V B)
• (B V A) ? T
• ( B V A )
• which is ( A V B ) or A -gt B or the
  Exp expression of the truth table in the
  previous slide

We used the following absorption rules in the
above derivation (see page 46 of your text
book) P V (P
? Q ) lt-gt P
P ? (P V Q ) lt-gt P
Check the absorption rules out with truth table
to satisfy yourself.
13
Can we turn the Disjunctive Normal Form of Exp to
Conjunctive Normal Form?
Recall that from the Truth Table, Exp (A ? B)
V (A ?B) V (A ?B)
To turn this into Conjunctive Normal Form, lets
look at using DeMorgan Law. So lets look at
Exp.
Exp (A ? B) V (A ? B) V (A ? B)
(A ? B) ? (A ? B) ?
(A ? B) (A V B ) ? ( A
V B) ? (A V B) (A V B
) ? ( A v (B ? B) ) (A V B
) ? ( A V False ) (A v B
) ? (A) (A ? A) V ( A
? B) F V ( A ?
B) ( A ? B)
But ---- we want Exp, which is (Exp)
So, Exp (A ? B) Exp A V
B or A -gt B This is the same as the
truth table expression, Exp.
Conjunctive Normal Form Form
14
Recapture the Steps

• From the Truth Table Expression, Exp, pick all
  the rows where the Exp has the value True.
• Express the each of those True rows as a
  conjunction of the literals, based on the truth
  values of the literals for that row.
• The disjunction of the conjunctions formulate the
  Disjunctive Normal Form of that Exp.
• To obtain the Conjunctive Normal Form we need to
  apply the DeMorgan Law to convert Exp to Exp.

15
Introduction to Resolution Theorem

• A Popular Theorem used in A.I. is the Resolution
  Theorem ( a simplified version)
• p \/ r, q \/ r
  gt p \/ q
• Lets look at the first term , p \/ r.
• Start with the known theorem, p-gtr is equal to
  p \/ r, thus we have (p) -gt r ? (p) \/ r
• Using the tautology, p lt-gt p, we can convert
  the above to the (p) -gt r ? p \/ r. So we can
  replace the p \/ r with (p) -gtr in the original
  expression.
• Next, q \/ r is the same as r \/ q from one of
  the known tautologies. And r \/ q can be
  replaced with r -gt q from the same earlier
  theorem.
• Now we have (p) -gt r, r -gt q gt p v q
• Finally replace p \/ q with p -gt q and we have
  
• (p) -gt r, r -gt q gt p -gt q which
  is the same as one of
• the earlier known theorem in the form
  of p -gt q, q -gt r gt p -gt r
• Thus p \/ r, q \/ r gt p \/ q.

16
General Resolution Theorem

• The General form of Resolution theorem states
  that there may be any number of disjuncts,
  including the case of just one, in either of the
  two expressions. The only requirements is that
  one expression must have a disjunct that is the
  negation of a disjunct in the other.
• E1 \/ E2 \/ - - - \/ Ek - - - \/ En, E1 \/ E2
  \/ - - - \/ Ek - - -\/ En gt E1 \/ E2 - - - \/
  En
• Note that the Ek as a disjunct that is
  dropped.
• Special Case 1 E1, E1 \/ E2 gt E2
• this is the same as E1, E1 -gt E2 gt E2
  (modus ponens)
• Special Case 2 E1 \/ E2, E2 gt E1
• This is the same as E1 -gt E2, E2 gt E1
  (modus tolens)

Note that we are dealing with Conjunctive Normal
Form
17
Refutation Principle or Proof by Contradiction

• AI and many other fields use a method called
  refutation principle together with the Resolution
  Theorem.
• Show that the negation of the conclusion is
  inconsistent with the premise(s).
• Example
• (prem) local database \/ remote database
• prove local database v remote database,
  (local database) gt remote database
• Proof
• 1. Assume remote database (negation of
  conclusion)
• 2. local database \/ remote database
  (premise)
• 3. local database
  (premise from the prove statement)
• 4. local database \/ remote database, local
  database (combine 2 and 3 above)
• 5. remote database
  (4 above and resolution theorem)
• 6. remote database, remote base (1 and 5
  above ------ contradiction)
• 7. empty (or False)
  ( 6 above and resolution theorem)

18
General Approach to Utilizing Refutation Principle

• 1. Express the premises and conclusion in
  disjunctive form.
• 2. Assume the negation of the conclusion and show
  inconsistency via utilizing the Resolution
  Theorem.
• Show that the negation of the conclusion leads
  to inconsistency (False), thus the conclusion
  must be True

19
Resolution-Refutation Tree with Modus Ponens
- The familiar modus ponens A -gt B
A --------- B - Express the premise in
disjunctive form along with negation of
conclusion A V B A V False
Assume B
A V B
A V False
B
B V False
False
The Resolution-Refutation tree shows that B
assumption is False thus B must be True
20
Example of a Resolution-Refutation Tree used in
A.I.
Prove A-gtB, B-gtC, C-gtD gt A-gtD

• Assume ( A -gt D)
• A\/B, B\/C, C\/D gt (A \/ D)
• We have 3 disjunctive terms and 1 conjunctive
  term
• A\/B, B\/C, C\/D, A /\ D
• A\/B, B\/C, C\/D, A /\ D
• A\/B, B\/C, C\/D, A, D
• Place these on a reverse tree
• Process using Resolution Theorem

A\/B
B\/C
C\/D
A\/C
A
A\/D
D
D
Empty (False)