Lecture 2 [1in] Soundness and completeness of propositional logic

Lecture 2
Soundness and completeness
of propositional logic
August 31, 2005
Overview
I
Review of natural deduction.
I
Soundness and completeness.
I
Semantics of propositional formulas.
I
Soundness proof.
I
Completeness proof.
Propositional formulas
Grammar:
::= p j (:) j ( ^ ) j ( _ ) j ( ! )
Precedence rules:
Example:
: > _; ^ > !
p ^ (q _ : r ! q )
(p ^ ((q _ (:r )) ! q))
Natural deduction
1 ; : : : ; n
`
Formula can be proved from formulas 1 ; : : : ; .
n
Rules for reasoning about formulas.
I Introduction rules:
Connective in the conclusion.
Construct a complex formula.
I
..
.
!
Elimination rules:
Connective in a premise.
Extract information from proved formulas.
xxx !
!-e
!-i
Basic rules
Introduction
^
_
^
_
_
.
.
.
?
::
-i1
!
:
^
xxx
-i
_
_
!
^
Elimination
^
-e1
^
-e2
-i2
-i
^
_
.
.
.
xxx
xxx xxx
!
!
!
.
.
.
?
:
no rule
:
-i
_
.
.
.
:
? :
xxx ? ?
:: ::
-e
-e
-e
-e
-e
Proof
p ! q ; r ! a ` (p _ r ) ! (q _ s )
1:
p_r
assumption
2:
p
assumption
3:
p
4:
q
5:
q_s
!-e : 2 3
_-i1 : 4
6:
r
assumption
7:
r
8:
s
9:
q_s
10:
q_s
!q
premise
;
!s
premise
!-e : 6 7
_-i2 : 8
_-e : 2 9
;
Proof
p ! q ; r ! a ` (p _ r ) ! (q _ s )
1:
p_r
assumption
2:
p
assumption
3:
p
4:
q
5:
q_s
!-e : 2 3
_-i1 : 4
6:
r
assumption
7:
r
8:
s
9:
q_s
10:
q_s
!q
premise
;
!s
premise
!-e : 6 7
_-i2 : 8
_-e : 2 9
;
Reasoning vs. reality
Reasoning: argument based on observations and derivational rules.
Reality: reality.
How does reasoning relate to reality?
I Soundness: reasoning derives only true statements.
Trivial solution: reasoning that derives nothing.
I Completeness: reasoning derives all true statements.
Trivial solution: reasoning that derives everything.
Soundness, completeness, and auto mechanics
Reality: the fan belt is broken.
Mechanic's diagnosis:
Complete
Replace the
fan belt and the
engine
Replace the
car
Replace the
engine
Sound
Replace the
fan belt
I don’t know
Semantics
What is a true formula?
A formula is either true T or false F.
The meaning depends on the meaning of the subterms.
Examples:
I [ p ] = T and [ q ] = F implies [ p ^ q ] = F
I
[ p] = T implies [ :p] = F
I
[ p] = T and [ q] = F implies [ p ^ :q] = T
Truth tables
The value of a formula for all possible inputs.
Basic connectives:
> xxx ? xxx : xxx T
F
T
T
F
F
T
F
T
F
F
T
_
T
T
T
F
T
F
T
T
F
F
xxxxxx Other tables dene other functions.
T
T
F
F
T
F
T
F
T
F
T
F
^
T
F
F
F
!
T
F
T
T
Meaning of a formula
Begin with the meanings of atoms. Compute the value bottom-up.
Example: (p ! q) ^ (p ! r )
p q r p ! q p ! r (p ! q ) ^ ( p ! r )
T
T
T
T
F
F
F
F
I
I
T
T
F
F
T
T
F
F
T
F
T
F
T
F
T
F
Tautology: Always true.
Contradiction: always false.
Semantic entailment
1 ; : : : j=
n
true when 1 ; : : : are true.
n
1. Compute truth tables of the 1 ; : : : .
n
2. Collect lines where 1 ; : : : are all true.
n
3. Evaluate in these cases.
Example
q ^ r j= (p ! q ) ^ (p ! r )
p q r q ^ r p ! q p ! r (p ! q ) ^ (p ! r )
T
T
T
T
F
F
F
F
T
T
F
F
T
T
F
F
T
F
T
F
T
F
T
F
T
F
F
F
T
F
F
F
Thus, q ^ r j= (p ! q) ^ (p ! r )
Example
q ^ r j= (p ! q ) ^ (p ! r )
p q r q ^ r p ! q p ! r (p ! q ) ^ (p ! r )
T
T
T
T
F
F
F
F
T
T
F
F
T
T
F
F
T
F
T
F
T
F
T
F
T
F
F
F
T
F
F
F
T
T
T
T
T
T
Thus, q ^ r j= (p ! q) ^ (p ! r )
Soundness and completeness
Relate provability to semantic entailment.
I Soundness:
1 ; : : : ; ` implies that 1 ; : : : ; j= .
n
I
n
Completeness:
1 ; : : : ; j= implies that 1 ; : : : ; ` .
n
n
Soundness
Let 1 ; : : : ; and be propositional logic formulas.
Then, if 1 ; : : : ; ` , then 1 ; : : : ; j= .
Theorem:
n
n
n
Proof idea: each proof step is justied by truth tables.
Proof: inductive denition
I
: premise is a proof of ` , where 2 .
I
Let:
prove ` prove ; ` For any rule R :
i
i
j
j
j
1
..
.
1
Then
: : : xxx 1
proves ` .
n
:::
n
m
1 ; : : : ; ; 1 ; : : : ; ; R
m
..
.
n
Base case
Given: ` Show: j= Proof of ` : : premise
Show: j= Evaluate for truth table lines where are all T.
is a premise, so 2 . Thus is T.
Induction case
Given: `
Show: j=
` must have a proof 1 ; : : : ; ; 1 ; : : : ; ; R , for R :
m
n
1
..
.
..
.
n
:::
1 : : : xxx 1
m
Induction hypothesis:
I proves ` implies j= .
I proves ; ` implies ; j= :
i
j
i
j
i
j
j
Proceed by cases on the possible rules.
j
n
Rule:
^
-i
Given: ` ^ , show: j= ^ .
Proof of ` ^ : 1 ; 2 ; ^-i
xxx
^
^-i
1 proves ` , 2 proves `
By induction: j= , j=
Evaluate ^ when [ ] = [ ] = T:
Thus, j= ^ .
T T
^
T
Rule:
^
-i
Given: ` ^ , show: j= ^ .
Proof of ` ^ : 1 ; 2 ; ^-i
xxx
^
^-i
1 proves ` , 2 proves `
By induction: j= , j=
Evaluate ^ when [ ] = [ ] = T:
Thus, j= ^ .
T T
^
T
Rule:
^
-i
Given: ` ^ , show: j= ^ .
Proof of ` ^ : 1 ; 2 ; ^-i
xxx
^
^-i
1 proves ` , 2 proves `
By induction: j= , j=
Evaluate ^ when [ ] = [ ] = T:
Thus, j= ^ .
T T
^
T
:
Rule: -e
Given: ` ?, show: j= ?.
Proof of ` ?: 1 ; 2 ; :-e
xxx :
? :-e
1 proves ` , 2 proves ` :
By induction: j= , j= :
Truth table for :
:
F
T
T
F
never all T. Thus, trivially: j= ?
NB: Truth table for ?:
?
F
:
Rule: -e
Given: ` ?, show: j= ?.
Proof of ` ?: 1 ; 2 ; :-e
xxx :
? :-e
1 proves ` , 2 proves ` :
By induction: j= , j= :
Truth table for :
:
F
T
T
F
never all T. Thus, trivially: j= ?
NB: Truth table for ?:
?
F
:
Rule: -e
Given: ` ?, show: j= ?.
Proof of ` ?: 1 ; 2 ; :-e
xxx :
? :-e
1 proves ` , 2 proves ` :
By induction: j= , j= :
Truth table for :
:
F
T
T
F
never all T. Thus, trivially: j= ?
NB: Truth table for ?:
?
F
Rule:
?
-e
Given: ` , show: j= .
Proof of ` : ; :-e
proves ` ?.
? ?-e
By induction: j= ?.
Truth table for ?:
?
F
never all T.
Thus, trivially: j= Rule:
?
-e
Given: ` , show: j= .
Proof of ` : ; :-e
proves ` ?.
? ?-e
By induction: j= ?.
Truth table for ?:
?
F
never all T.
Thus, trivially: j= Rule:
?
-e
Given: ` , show: j= .
Proof of ` : ; :-e
proves ` ?.
? ?-e
By induction: j= ?.
Truth table for ?:
?
F
never all T.
Thus, trivially: j= !
Rule -i
Given: ` ! , show: j= ! .
..
.
Proof of ` ! : ; !-i xxx
!-i
!
proves ; `
By induction: ; j= .
Show: when all true, so is ! .
T T
T F
!
T
F
xxx
F T
F F
!
T
T
Potential problem when [ ] = T and [ ] = F.
But, by induction, if all true and [ ] = T, then [ ] = T.
Thus, j= ! .
!
Rule -i
Given: ` ! , show: j= ! .
..
.
Proof of ` ! : ; !-i xxx
!-i
!
proves ; `
By induction: ; j= .
Show: when all true, so is ! .
T T
T F
!
T
F
xxx
F T
F F
!
T
T
Potential problem when [ ] = T and [ ] = F.
But, by induction, if all true and [ ] = T, then [ ] = T.
Thus, j= ! .
!
Rule -i
Given: ` ! , show: j= ! .
..
.
Proof of ` ! : ; !-i xxx
!-i
!
proves ; `
By induction: ; j= .
Show: when all true, so is ! .
T T
T F
!
T
F
xxx
F T
F F
!
T
T
Potential problem when [ ] = T and [ ] = F.
But, by induction, if all true and [ ] = T, then [ ] = T.
Thus, j= ! .
Completeness
Theorem: Let 1 ; : : : ; and
be propositional logic formulas.
Then, if 1 ; : : : ; j= , then 1 ; : : : ; ` .
n
n
n
Proof idea: construct a proof from a truth table.
Completeness proof structure
1. Eliminate premises:
1 ; 2 ; : : : ; j= implies
j= 1 ! (2 ! : : : ! ( ! )).
n
n
2. Show provability:
j= 1 ! (2 ! : : : ! ( ! )) implies
` 1 ! (2 ! : : : ! ( ! )).
n
n
3. Reintroduce premises:
` 1 ! (2 ! : : : ! ( ! )) implies
1 ; 2 ; : : : ; ` .
n
n
Eliminating premises
If 1 ; 2 ; : : : ; j= , then
j= 1 ! (2 ! : : : ! ( ! )).
Theorem:
n
n
Proof:
By induction on n.
Base case: n = 0.
Clearly j= implies j= .
Induction case:
We showed ; j= implies j= ! .
Thus, 1 ; 2 ; : : : ; j= implies 1 ; : : : n
n
1
j= ! .
By induction, j= 1 ! (2 ! : : : ! ( ! ))
n
n
Showing provability
Theorem:
I
Let be a formula such that p1 ; p2 ; : : : ; p are its only
propositional atoms.
I
Let ` be any line in 's truth table.
I
For any atom or formula , let ` () be if the truth table
entry in line ` for is T, and : if the truth table entry for is F.
n
Then, ` (p1 ); ` (p2 ); : : : ; ` (p ) ` ` () is provable.
n
Example
A truth table line:
(p ^ r ) ! ( q ^ s )
p q r s p ^ r q ^ s ( p ^ r ) ! (q ^ s )
T F T T
T
F
F
Constructed sequent:
p ; :q ; r ; s ` :((p ^ r ) ! (q ^ s ))
Another truth table line:
p q r s p ^ r q ^ s ( p ^ r ) ! (q ^ s )
F T F T
F
T
T
Constructed sequent:
:p; q; :r ; s ` (p ^ r ) ! (q ^ s )
Proof
Induction on the structure of .
Base case: p.
Truth table:
p p
T
F
T
F
` (p ) ` ` (p ) is p ` p , or :p ` :p :
I Proof of p ` p :
1: p premise
2: p 1
I Proof of :p ` :p :
1: :p premise
2: :p 1
Thus, ` (p) ` ` (p).
Negation
Possible truth table lines:
:
:
F
T
T
F
If [ ] = F and [ : ] = T:
xxxShow ` (p1 ); ` (p2 ); : : : ; ` (p ) ` :
n
By induction, since [ ] = F:
` (p1 ); ` (p2 ); : : : ; ` (p ) ` :
n
Negation, continued
If [ ] = T and [ : ] = F:
xxxShow ` (p1 ); ` (p2 ); : : : ; ` (p ) ` ::
n
By induction, since [ ] = T:
` (p1 ); ` (p2 ); : : : ; ` (p ) `
n
To prove ` (p1 ); ` (p2 ); : : : ; ` (p ) ` :: :
First prove ` (p1 ); ` (p2 ); : : : ; ` (p ) ` .
Use :-i.
n
n
Implication
1 ! 2
Proof strategy:
I Consider possible truth values of 1 ! 2 .
I Use induction to nd proofs of ` (1 ) and ` (2 ).
I Construct a proof of ` (1 ! 2 ).
1 2 1 ! 2
T
T
F
F
Four lines, so four cases.
T
F
T
F
T
F
T
T
Use of the induction hypothesis
1 2 1 ! 2
T
T
F
F
T
F
T
F
T
F
T
T
Consider 1 , 2 :
I Propositional atoms of 1 : a1 ; : : : ; a .
I Propositional atoms of 2 : b1 ; : : : ; b .
Atoms of 1 ! 2 = fa1 ; : : : ; a g [ fb1 ; : : : ; b g.
x
y
x
y
Thus, a truth table line for 1 ! 2 also gives meaning to 1 and
2 .
Using the induction hypothesis
For a truth table line ` for 1 ! 2 ,
by induction:
` (a1 ); : : : ; ` (a ) ` ` (1 )
` (b1 ); : : : ; ` (b ) ` ` (2 )
Our goal:
x
y
` (a1 ); : : : ; ` (a ); ` (b1 ); : : : ; ` (b ) ` ` (1 ! 2 )
x
y
Proof structure:
I Proof of ` (a1 ); : : : ; ` (a ) ` ` (1 ).
I Proof of ` (b1 ); : : : ; ` (b ) ` ` (2 ).
I ` (1 ) ^ ` (2 ) : ^-i
I Derive ` (1 ! 2 ) from ` (1 ) ^ ` (2 ).
x
y
Cases
I
1 is T, 2 is T, and is T:
Show 1 ^ 2 implies 1 ! 2 .
I
1 is T, 2 is F, and is F:
Show 1 ^ :2 implies :(1 ! 2 ).
I
1 is F, 2 is T, and is T:
Show :1 ^ 2 implies 1 ! 2 .
I
1 is F, 2 is F, and is T:
Show :1 ^ :2 implies 1 ! 2 .
Conjunction and disjunction similar.
Combining sequents
We have a \complete" collection of sequents:
` (p1 ); ` (p2 ); : : : ; ` (p ) ` ` ()
n
Note: j= implies 8`:` () = .
We want to prove:
`
Proof idea:
Use the Law of the Excluded Middle.
Example (p
^ q) ! q
Constructed sequents:
p ; q ` (p ^ q ) ! q
p ; :q ` (p ^ q ) ! q
:p; q ` (p ^ q) ! q
:p; :q ` (p ^ q) ! q
Consider p; q ` (p ^ q) ! q and p; :q ` (p ^ q) ! q.
Proof of p ` (p ^ q) ! q:
q _ :q
1:
LEM
2:
3
4:
5
6:
q
proof of
:q
assumption
p ; q ` (p ^ q ) ! q
assumption
p ; :q ` (p ^ q ) ! q
(p ^ q ) ! q
_-e : 1 5
proof of
Proof idea
Goal: Reduce ` (p1 ); ` (p2 ); : : : ; ` (p ) ` to ` .
n
Proof by induction on n.
Find pairs of sequents that only dier in ` (p ).
n
Use LEM to prove a single sequent without ` (p ) in the premise.
n
Reintroducing premises
Theorem:
If ` 1 ! (2 ! : : : ! ( ! )), then
` .
1 ; 2 ; : : : ; n
n
Proof idea:
Convert ` 1 ! (2 ! : : : ! ( ! )) to
1 ` 2 ! : : : ! ( ! ), and proceed by induction.
n
n
Problem: Theorem not general enough.
Restatement:
If ` 1 ! (2 ! : : : ! ( ! )), then ; 1 ; 2 ; : : : ; ` .
n
n
Proof
Given a proof of:
` 1 ! (2 ! : : : ! ( ! ))
n
Prove:
; 1 ` 2 ! : : : ! ( ! )
n
Proof of ` 1 ! (2 ! : : : ! ( ! ))
1 : premise
2 ! : : : ! ( ! ) : !-e
By induction, ; 1 ; 2 ; : : : ; ` .
I
I
I
n
n
n
Take = ; to prove the original theorem.
Summary
Soundness: Show that the deduction rules are justied by truth
tables.
Completeness:
1. Eliminate premises:
1 ; 2 ; : : : ; j= implies
j= 1 ! (2 ! : : : ! ( ! )).
n
n
2. Show provability:
j= 1 ! (2 ! : : : ! ( ! )) implies
` 1 ! (2 ! : : : ! ( ! )).
n
n
3. Reintroduce premises:
` 1 ! (2 ! : : : ! ( ! )) implies
1 ; 2 ; : : : ; ` .
Conclusion: can freely use either natural deduction or truth tables.
n
n