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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 Alpha Graphs are recursively defined as follows: 1. 2. is an Alpha Graph. P is an Alpha Graph, with P an atomic statement. 3. is an Alpha Graph, with and Alpha Graphs. 4. is an Alpha Graph, with an Alpha Graph. 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( ) = True. P ) = True or False. 3. h( ) = True iff h( ) = True and h( ) = True. 4. h( ) = True iff is h( ) = False. 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 truthfunctional consequence: – TF iff is a truth-functional consequence of . • Let us use the symbol EG to indicate truthfunctional 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 Strengthening the Antecedent pr (p q) r Absorption Case 1 Weakening the Consequent p (q r) pr Absorption Case 2 p and q r p and (q p) r p and q r p and q (r p) Soundness Recursive Conditional Reading One can see graphs with multiple cuts inside each other as expressing recursively conditioned conditionals. For example: 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. Soundness Relative Recursive Conditional Reading I 0 1 2k-1 2k 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 0 1 2k-1 2k 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 Soundness Relative Recursive Conditional Reading III 0 1 2k-1 2k 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 0 1 2k-1 2k 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 Soundness Soundness of Insertion 0 1 2k-1 2k 0 & 1 2 & (1 2k-1) 2k ? IN 0 1 2k-1 2k pr (p q) r 0 & 1 2 & (1 2k-1 ) 2k Insertion corresponds to Strengthening the Antecedent Soundness Soundness of Erasure 0 1 2k-1 2k 0 & 1 2 & (1 2k-1) (2k ) ? E 0 1 2k-1 2k p (q r) pr 0 & 1 2 & (1 2k-1) 2k Erasure corresponds to Weakening the Consequent Soundness Soundness of Iteration/Deiteration ? IT/DE • 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. Soundness Soundness of Iteration/Deiteration Case 1 1 2k-1 2k & 1 2 & (1 2k-1) 2k ? IT/DE 1 2k-1 2k p and q r p and (q p) r & 1 2 & (1 2k-1 ) 2k Iteration/Deiteration (Case 1) corresponds to Absorption (Case 1) Soundness Soundness of Iteration/Deiteration Case 2 1 2k-1 2k & 1 2 & (1 2k-1) 2k ? IT/DE 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 A graph is Alpha complete iff for any graph : EG or 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 . Then: EG i.e. is Alpha inconsistent. Contradiction, so not both 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’: So: ‘if’: Suppose EG . Then (Ded. Thm.): EG Suppose So: EG EG . Then (‘only if’): EG 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. Proof: Obtain ’ from the following routine: As = Set of all atomic statements While As : Select and remove some A from As if EG A or EG A ’ = A otherwise { = ’ 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. Proof: By induction we’ll show that at any point ’ is Alpha consistent Base: is Alpha consistent Step: Suppose is Alpha consistent, and suppose that adding A As to makes ’ = A Alpha inconsistent, i.e: A EG Then: A A EG So (Transposition Theorem): EG A Hence (Ded. Thm): EG A A So, A would not be added to . Contradiction, so adding A keeps Alpha consistent. Completeness Expansion Theorem (Alpha Completeness of ’) To prove: ’ as obtained by the routine is Alpha complete. Proof: By induction on the composition of any statement . Base: For every atomic statement A, ’ EG A or ’ EG A Step: Case 1: = 1 2 By inductive assumption: ’ EG i or ’ EG i If ’ EG 1 and ’ EG 2 then ’ ’ ’ EG 1 2 If ’ EG 1 or ’ EG 2 then ’ EG 1 2 by IN Case 2: = ’ By inductive assumption: ’ EG ’ or ’ EG ’ Hence, ’ EG ’ ’ 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) To prove: for any graph : h() = True iff EG Proof: By induction on the composition of any statement . 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 (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 Then is inconsistent. Hence (Central Theorem): Therefore is Alpha inconsistent. 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