Formalizing Alpha:Soundness and Completeness

Bram van Heuveln

Dept. of Cognitive Science

RPI

Overview

• This presentation discusses:– Formal Syntax of Alpha– Formal Semantics of Alpha– Soundness of Alpha– Completeness of Alpha

Formalizing Alpha

Formal Syntax

1.

is an Alpha Graph, with an Alpha Graph.

2.

Alpha Graphs are recursively defined as follows:

3.

4.

is an Alpha Graph.

is an Alpha Graph, with P an atomic statement.

is an Alpha Graph, with and Alpha Graphs.

P

Formalizing Alpha

Formal SemanticsA truth-assignment h is a function that assigns either Trueor False to any Alpha Graph in accordance with:

1. h(

2. h(

3. h(

4. h(

P

) = True iff is h( ) = False.

) = True.

) = True or False.

) = True iff h( ) = True and h( ) = True.

Formalizing Alpha

Consequence• Let us use the following notation:

– h |= iff h() = True

– |= iff for any truth-assignment h: h |=

• Let us define: is a truth-functional consequence of iff for any

truth-assignment h: if h |= then h |= .

Formalizing Alpha

Soundness and Completeness• Let us use the symbol TF to indicate truth-functional

consequence: TF iff is a truth-functional consequence of .

• Let us use the symbol EG to indicate truth-functional provability in Alpha: EG iff there exists a formal proof from to .

• We will show two things:– Alpha is truth-functionally sound:

• For any and : if EG then TF – Alpha is truth-functionally complete:

• For any and : if TF then EG

Soundness

Proving Soundness• A straightforward way to demonstrate the soundness of some

system S is to demonstrate that each of the inference rules of S is sound.

• If the inference rules deal with subproofs, such a proof can actually become rather technical, even if the soundness of each inference rules is intuitive clear (see e.g. proof of soundness of F in “Language, Proof, and Logic”).

• The good news is that Alpha does not deal with subproofs.• However, the bad news is that, except for double cut, the

soundness of Alpha’s inference rules is not intuitive.• To make the soundness of the inference rules more intuitive, we

use the notion of a recursive conditional reading as well as 3 less commonly used inference rules.

Soundness

2 Rules of Inference and 1 Rule of Equivalence

p r (p q) r

Strengthening theAntecedent

Weakening theConsequent

Absorption Case 1

Absorption Case 2

p (q r)p r

p and q r p and (q p) r

p and q r p and q (r p)

Soundness

Recursive Conditional Reading

P Q R S

This graph can be read as: ‘if P is true then Q is true, but if P is true, then it is also true that if R is true then S true’. Hence the conditional ‘if R then S’ is conditionally true. Thus, as more and more conditions get added, the truth of more and more statements can be asserted.

One can see graphs with multiple cuts inside each other as expressing recursively conditioned conditionals. For example:

Soundness

Relative Recursive Conditional Reading I

1 2k-1 2k0

Relative to any subgraph existing at an odd level 2k-1, one can see the graph as a whole in the above manner (note: if there is no cut next to , then one has to add an empty double cut next to to make it work).

Soundness

Relative Recursive Conditional Reading II

The Recursive Conditional Reading (RCR) of a graph relative to a subgraph existing at odd level 2k – 1 is defined as follows:0 &1 2 &(1 3) 4

(1 2k-1 ) 2k

1 2k-1 2k0

Soundness

Relative Recursive Conditional Reading III

1 2k-1 2k0

Relative to any subgraph existing at an even level 2k, one can see the graph as a whole in the above manner.

Soundness

Relative Recursive Conditional Reading IV

The Recursive Conditional Reading (RCR) of a graph relative to a subgraph existing at even level 2k is defined as follows:0 &1 2 &(1 3) 4

(1 2k-1) 2k

1 2k-1 2k0

Soundness

Soundness of Insertion

1 2k-1 2k0

1 2k-1 2k0

IN

0 &1 2 &

(1 2k-1) 2k

0 &1 2 &

(1 2k-1 ) 2k

p r (p q) r

?

Insertion corresponds to Strengthening the Antecedent

Soundness

Soundness of Erasure

1 2k-1 2k0

1 2k-1 2k0

E

0 &1 2 &

(1 2k-1) 2k

p (q r)p r

0 &1 2 &

(1 2k-1) (2k )

?

Erasure corresponds to Weakening the Consequent

Soundness

Soundness of Iteration/Deiteration

• To demonstrate the soundness of iteration and deiteration, it suffices to demonstrate that the subgraph that has the original at level 0 is equivalent to the corresponding subgraph in the result of the iteration or deiteration.

IT/DE

?

Soundness

Soundness of Iteration/Deiteration Case 1

1 2k-1 2k

1 2k-1 2k

IT/DE

&1 2 &

(1 2k-1) 2k

&1 2 &

(1 2k-1 ) 2k

p and q r p and (q p) r

?

Iteration/Deiteration (Case 1) corresponds to Absorption (Case 1)

Soundness

Soundness of Iteration/Deiteration Case 2

1 2k-1 2k

1 2k-1 2k

IT/DE

&1 2 &

(1 2k-1) 2k

p and q r p and q (r p)

&1 2 &

(1 2k-1) (2k )

?

Iteration/Deiteration (Case 2) corresponds to Absorption (Case 2)

Completeness

Proving Completeness• On the next slides, we will provide a direct proof

of the completeness of Alpha that very much follows the strategy used in “Language, Proof, and Logic” to prove that system F is complete.

Completeness

Alpha Consistency and Alpha CompletenessA graph is Alpha inconsistent iff

EG or EG

A graph is Alpha complete iff for any graph :

EG

Completeness

Either-Or LemmaFor any graph : if is Alpha consistent and

Alpha complete, then for any graph :

EG or EG but not both EG and EG

Proof: Suppose is Alpha consistent and Alpha complete. Then:

By Alpha completeness: EG or EG

Now suppose EG and EG

EG

Contradiction, so not both

. Then:

i.e. is Alpha inconsistent.

EG and EG

Completeness

Deduction TheoremFor any graphs and :

EG iff EG

Proof:

‘if’:

EG

‘only if’:

EG

Completeness

Transposition LemmaFor any graphs and :

EG iff EG

Proof:

‘only if’: Suppose EG . Then (Ded. Thm.): EG

So: EG

‘if’: Suppose

EG . Then (‘only if’): EG

So: EG

Completeness

Expansion TheoremFor any graph : if is Alpha consistent, then there exists a

’ = for some such that ’ is Alpha consistent and Alpha complete.

= ’

As = Set of all atomic statements

While As :

Select and remove some A from As

’ = { if EG A or EGA

A otherwise

Proof: Obtain ’ from the following routine:

Next two slides: ’ is Alpha consistent and Alpha complete.

Completeness

Expansion Theorem (Alpha Consistency of ’)To prove: ’ as obtained by the routine is Alpha consistent.

Step: Suppose is Alpha consistent, and suppose that adding A As

Base: is Alpha consistent

to makes ’ = A Alpha inconsistent, i.e: A EG

Then: A A EG

So (Transposition Theorem): EG A A

Hence (Ded. Thm): EG A So, A would not be added to .

Contradiction, so adding A keeps Alpha consistent.

Proof: By induction we’ll show that at any point ’ is Alpha consistent

Completeness

Expansion Theorem (Alpha Completeness of ’)

Base: For every atomic statement A, ’ EG A or ’ EG

i

Step: Case 1: = 1 2

By inductive assumption: ’ EG i or ’ EG

If ’ EG 1 and ’ EG 2 then ’ ’ ’ EG 1 2

If ’ EG 1 or ’ EG 2 then ’ EG1 2 by IN

Case 2: = ’

By inductive assumption: ’ EG ’ or ’ EG ’

Hence, ’ EG ’

A

’

To prove: ’ as obtained by the routine is Alpha complete.

Proof: By induction on the composition of any statement .

or ’ EG

Completeness

Consistency LemmaFor any graph : if is Alpha consistent and Alpha complete,

then is consistent.Proof: Suppose is Alpha consistent and Alpha complete.

Define truth-value assignment h as follows:

For any atomic statement A: h(A) = True iff EG A (the Either-Or Lemma guarantees that h is well-defined).

On the next slide we’ll show that for any graph :

h() = True iff EG Finally, because EG , h() = True. Hence, is consistent.

Completeness

Consistency Lemma (Continued)

Base: For every atomic statement A, h(A) = True iff EG A (def. h)

Step: Case 1: = 1 2

By inductive assumption: h(i) = True iff EG i

So, h(1 2) = True iff h(1) = True and h(1) = True iff

EG 1 and EG 2 iff (trivial) EG 1 2 iff EG

Case 2: = ’ By ind. assumption: h(’) = true iff EG ’

Hence, h() = True iff h(’) = false iff not EG ’ iff

To prove: for any graph : h() = True iff EG Proof: By induction on the composition of any statement .

(Either-Or Lemma) EG ’ iff EG

Completeness

The Central TheoremFor any graph :

if is Alpha consistent, then is consistent.

Proof:Suppose is Alpha consistent.

By the Expansion Theorem there exists a ’ = for some such that ’ is Alpha consistent and Alpha complete.

By the Consistency Lemma, this ’ is consistent.

So, is consistent.

Therefore, is consistent.

Completeness

Basic Completeness TheoremFor any : if TF then EG

Proof:Suppose TF

Then is inconsistent.

Hence (Central Theorem): is Alpha inconsistent.

Therefore EG

So (Transposition Theorem): EG

Completeness

Completeness TheoremFor any and : if TF then EG

Proof:Suppose TF

Then TF

Hence (Basic Completeness Theorem): EG

Therefore (Deduction Theorem): EG