A Proof from 1988 that PDL has Interpolation? (1) EDV-Beratung Manfred Borzechowski, Berlin, Germany (2) University of Groningen, Groningen, The Netherlands Advances in Modal Logic 2020 2020-08-26 15:00 1 Manfred Borzechowski (1) & Malvin Gattinger (2)
Propositional Dynamic Logic (PDL) x: π β³, π₯ β¨ β¨π΅β©(β¨π΅β©Β¬π β§ β¨πΆβ©[πΆ β ]π) β³, π₯ β¨ [πΆ β ]π β³, π₯ β¨ [πΆ]π β³, π₯ β¨ β¨π΅; πΆβ©π B B A A w: π Syntax v: π Example Also expressible: βifβ¦thenβ¦elseβ¦β and βwhileβ¦doβ¦β βΆβΆ= π π β£ Β¬π β£ π β§ π β£ [π]π βΆβΆ= π 2 π΅ β£ π; π β£ π βͺ π β£ π β β£ π?
Propositional Dynamic Logic (PDL) x: π β³, π₯ β¨ β¨π΅β©(β¨π΅β©Β¬π β§ β¨πΆβ©[πΆ β ]π) β³, π₯ β¨ [πΆ β ]π β³, π₯ β¨ [πΆ]π β³, π₯ β¨ β¨π΅; πΆβ©π B B A A w: π Syntax v: π Example Also expressible: βifβ¦thenβ¦elseβ¦β and βwhileβ¦doβ¦β βΆβΆ= π π β£ Β¬π β£ π β§ π β£ [π]π βΆβΆ= π 2 π΅ β£ π; π β£ π βͺ π β£ π β β£ π?
Craig Interpolation π βΆ (π β¨ π ) π (π β§ π) Example (1918 β 2016) William Craig βΆ π . We then write π π(π) βΆ= {π β£ π occurs in π} - π(π) β π(π) β© π(π) π β π is valid - is valid - π β π A logic has Craig Interpolation iff for any valid Definition 3 π β π there is an interpolant π such that:
Logics that (we know) have Craig Interpolation β’ Propositional Logic β’ First-Order Logic β’ Intuitionistic Logic β’ Basic and Multi-modal logic (MadarΓ‘sz 1995) β’ π -Calculus (DβAgostino and Hollenberg 2000) What about PDL? Language of a formula := all atomic propositions and programs . Example [(π΅ βͺ πΆ) β ](π β§ π) [πΆ β ]π βΆ [(πΆ; πΆ) β ](π β¨ [π·]π ) Problem: How to find interpolants for β systematically? 4
Logics that (we know) have Craig Interpolation β’ Propositional Logic β’ First-Order Logic β’ Intuitionistic Logic β’ Basic and Multi-modal logic (MadarΓ‘sz 1995) β’ π -Calculus (DβAgostino and Hollenberg 2000) What about PDL? Language of a formula := all atomic propositions and programs . Example [(π΅ βͺ πΆ) β ](π β§ π) [πΆ β ]π βΆ [(πΆ; πΆ) β ](π β¨ [π·]π ) Problem: How to find interpolants for β systematically? 4
History β’ Daniel Leivant: Proof theoretic methodology for propositional dynamic logic . Conference paper in LNCS, 1981. β’ Manfred Borzechowski: TableauβKalkΓΌl fΓΌr PDL und Interpolation . Diploma thesis, FU Berlin, 1988. β’ Tomasz Kowalski: PDL has interpolation . Journal of Symbolic Logic, 2002. Revoked in 2004. β’ Marcus Kracht: Chapter The open question in Tools and Techniques in Modal Logic , 1999. 5
History β’ Daniel Leivant: Proof theoretic methodology for propositional dynamic logic . Conference paper in LNCS, 1981. β’ Manfred Borzechowski: TableauβKalkΓΌl fΓΌr PDL und Interpolation . Diploma thesis, FU Berlin, 1988. β’ Tomasz Kowalski: PDL has interpolation . Journal of Symbolic Logic, 2002. Revoked in 2004. β’ Marcus Kracht: Chapter The open question in Tools and Techniques in Modal Logic , 1999. 5
History β’ Daniel Leivant: Proof theoretic methodology for propositional dynamic logic . Conference paper in LNCS, 1981. β’ Manfred Borzechowski: TableauβKalkΓΌl fΓΌr PDL und Interpolation . Diploma thesis, FU Berlin, 1988. β’ Tomasz Kowalski: PDL has interpolation . Journal of Symbolic Logic, 2002. Revoked in 2004. β’ Marcus Kracht: Chapter The open question in Tools and Techniques in Modal Logic , 1999. 5
Borzechowski 1988: Why look at it now? β’ It seems it was never really published. β’ We now make an English translation available. β’ Kracht (1999): not βpossible to verify the argumentβ 6
Outline of the proof attempt 1. Define a tableaux system. 2. Show soundness and completeness. 3. Define interpolants for each node βbottom-upβ. Problems caused by β : β’ How to ensure finite tableaux? β’ How to define interpolants for β steps? 7
Tableaux Rules: Part 1/2 (Β¬?) π; π; [π][π (π) ]π (π) π; [π β ]π β£ π; Β¬[π][π (π) ]π π; Β¬π (Β¬π) π; Β¬[π β ]π π; [π; π]π π; [π ?]π π; [π βͺ π]π π; Β¬[π; π]π classical rules: π; π ; Β¬π π; Β¬[π ?]π π; Β¬(π β§ π ) π; ¬¬π (Β¬) π; π π; π β§ π ( β§ ) π; π; π 8 β£ π; Β¬[π]π β£ π; Β¬π local rules: π; Β¬[π βͺ π]π (Β¬β§) π; Β¬π (Β¬βͺ) π; Β¬[π]π (Β¬; ) π; Β¬[π][π]π (βͺ) π; [π]π; [π]π (?) π¦; Β¬π β£ π; π (; ) π; [π][π]π
Tableaux Rules: Part 2/2 (π΅π’) where (β¦) (Β¬?) (Β¬π) (Β¬; ) (Β¬βͺ) PDL rules: marked rules: the critical rule. 9 π free π; Β¬[π]π (πβ) the loading rule , π; Β¬[π 0 ] β¦ [π π ]π (π+) the liberation rule , π; Β¬[π 0 ] β¦ [π π ]π π π; Β¬[π]π π π; Β¬[π΅]π π π π΅ ; Β¬π π\π π; Β¬[π βͺ π]π π π; Β¬[π; π]π π π; Β¬[π]π π β£ π; Β¬[π]π π π; Β¬[π][π]π π π; Β¬[π β ]π π π; Β¬[π ?]π π π; Β¬π π\π β£ π; Β¬[π][π (π) ]π π π; π ; Β¬π π\π π\π indicates that π is removed iff π = π .
Tableaux Rules: Extra Conditions 1. On reaching π; Β¬[π΅]π or π; [π΅]π , change (π) back to β in π . 2. Instead of π; [π (π) ]π we always obtain π . 3. A rule must be applied to an π -formula whenever it is possible. 4. No rule may be applied to a Β¬[π (π) ] -node. 5. To a node obtained using (π+) we may not apply (πβ) . 6. If a normal node π’ has a predecessor π‘ with the same formulas and the path π‘ β¦ π’ uses (π΅π’) and is loaded if π‘ is loaded, then π’ is an end node. 7. Every loaded node that is not an end note by 6 has a successor. 10
Tableaux Rules: Extra Conditions 1. On reaching π; Β¬[π΅]π or π; [π΅]π , change (π) back to β in π . 2. Instead of π; [π (π) ]π we always obtain π . 3. A rule must be applied to an π -formula whenever it is possible. 4. No rule may be applied to a Β¬[π (π) ] -node. 5. To a node obtained using (π+) we may not apply (πβ) . path π‘ β¦ π’ uses (π΅π’) and is loaded if π‘ is loaded, then π’ is an end node. 7. Every loaded node that is not an end note by 6 has a successor. 10 6. If a normal node π’ has a predecessor π‘ with the same formulas and the
A Full Proof Example (Β¬π) x (Β¬?) π ; Β¬[(π΅ βͺ π?) x (π΅π’) Β¬[(π΅ βͺ π?) (Β¬βͺ) Β¬[π?][(π΅ βͺ π?) (Β¬βͺ) Β¬[(π΅ βͺ π?) Β¬[π΅][(π΅ βͺ π?) 11 (π+) x β ]π ; [π΅ β ]π (Β¬π)1. Β¬π ; π ; [π΅][π΅ β ]π Β¬[(π΅ βͺ π?) (π) Β¬[(π΅ βͺ π?)][(π΅ βͺ π?) Β¬[(π΅ βͺ π?) β ]π π ; [π΅ β ]π β ]π π ; π ; [π΅][π΅ β ]π (π) ]π π ; π ; [π΅][π΅ β ]π β ]π π ; π ; [π΅][π΅ β ]π (π) ]π π ; π ; [π΅][π΅ β ]π π β ]π π ; [π΅ β ] (π) ]π π ; π ; [π΅][π΅ β ]π
A Full Proof Example (Β¬π) closed x (Β¬?) π ; Β¬[(π΅ βͺ π?) x (π΅π’) Β¬[(π΅ βͺ π?) (Β¬βͺ) Β¬[π?][(π΅ βͺ π?) (Β¬βͺ) Β¬[(π΅ βͺ π?) Β¬[π΅][(π΅ βͺ π?) 11 β ]π ; [π΅ β ]π x (Β¬π)1. Β¬[(π΅ βͺ π?) Β¬π ; π ; [π΅][π΅ β ]π (π) Β¬[(π΅ βͺ π?)][(π΅ βͺ π?) Β¬[(π΅ βͺ π?) (π+) β ]π π ; [π΅ β ]π β ]π π ; π ; [π΅][π΅ β ]π (π) ]π π ; π ; [π΅][π΅ β ]π β ]π π ; π ; [π΅][π΅ β ]π (π) ]π π ; π ; [π΅][π΅ β ]π π β ]π π ; [π΅ β ] (π) ]π π ; π ; [π΅][π΅ β ]π
A Full Proof Example (Β¬π) closed by condition 6 x (Β¬?) π ; Β¬[(π΅ βͺ π?) x (π΅π’) Β¬[(π΅ βͺ π?) (Β¬βͺ) Β¬[π?][(π΅ βͺ π?) (Β¬βͺ) Β¬[(π΅ βͺ π?) Β¬[π΅][(π΅ βͺ π?) 11 β ]π ; [π΅ β ]π x (Β¬π)1. Β¬[(π΅ βͺ π?) Β¬π ; π ; [π΅][π΅ β ]π (π) Β¬[(π΅ βͺ π?)][(π΅ βͺ π?) Β¬[(π΅ βͺ π?) (π+) β ]π π ; [π΅ β ]π β ]π π ; π ; [π΅][π΅ β ]π (π) ]π π ; π ; [π΅][π΅ β ]π β ]π π ; π ; [π΅][π΅ β ]π (π) ]π π ; π ; [π΅][π΅ β ]π π β ]π π ; [π΅ β ] (π) ]π π ; π ; [π΅][π΅ β ]π
Recommend
More recommend