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: 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 Semantics A truth-assignment h is a function that assigns either True or 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 the Antecedent Weakening the Consequent 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 QRS 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 11 2k-1 2k 00 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 11 2k-1 2k 00
Soundness Relative Recursive Conditional Reading III 11 2k-1 2k 00 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 11 2k-1 2k 00
Soundness Soundness of Insertion 11 2k-1 2k 00 11 2k-1 2k 00 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 11 2k-1 2k 00 11 2k-1 2k 00 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 11 2k-1 2k 11 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 11 2k-1 2k 11 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 Completeness A graph is Alpha inconsistent iff EG or EG A graph is Alpha complete iff for any graph : EG
Completeness Either-Or Lemma For 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 Theorem For any graphs and : EG iff EG Proof: ‘if’: EG ‘only if’: EG
Completeness Transposition Lemma For 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 Theorem For 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 EG A 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 ii Step: Case 1: = 1 2 By inductive assumption: ’ EG i or ’ EG If ’ EG 1 and ’ EG 2 then ’ ’ ’ EG 1 2 If ’ EG 11 or ’ EG 22 then ’ EG 1 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 Lemma For 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 Theorem For 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 Theorem For any : if TF then EG Proof: Suppose TF Thenis inconsistent. Hence (Central Theorem):is Alpha inconsistent. Therefore EG So (Transposition Theorem): EG
Completeness Completeness Theorem For any and : if TF then EG Proof: Suppose TF Then TF Hence (Basic Completeness Theorem): EG Therefore (Deduction Theorem): EG