Complexity of intuitionistic logic. Martin Mundhenk
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1 Complexity of intuitionistic logic Mrtin Mundhenk
2 2 The formul evlution problem k the model checking problem Given: formul α for the logic L nd model M ccording to the semntics of L Decide: does M stisfy α? Exmple for propositionl logic: formul α = ( c) ( b) model M = {, c} (ssignment to the toms) c b
3 2 The formul evlution problem k the model checking problem Given: formul α for the logic L nd model M ccording to the semntics of L Decide: does M stisfy α? Exmple for propositionl logic: formul α = ( c) ( b) model M = {, c} (ssignment to the toms) c b
4 2 The formul evlution problem k the model checking problem Given: formul α for the logic L nd model M ccording to the semntics of L Decide: does M stisfy α? Exmple for propositionl logic: formul α = ( c) ( b) model M = {, c} (ssignment to the toms) c b
5 2 The formul evlution problem k the model checking problem Given: formul α for the logic L nd model M ccording to the semntics of L Decide: does M stisfy α? Exmple for propositionl logic: formul α = ( c) ( b) model M = {, c} (ssignment to the toms) 0 c b
6 2 The formul evlution problem k the model checking problem Given: formul α for the logic L nd model M ccording to the semntics of L Decide: does M stisfy α? Exmple for propositionl logic: formul α = ( c) ( b) model M = {, c} (ssignment to the toms) 0 c b
7 2 The formul evlution problem k the model checking problem Given: formul α for the logic L nd model M ccording to the semntics of L Decide: does M stisfy α? Exmple for propositionl logic: formul α = ( c) ( b) model M = {, c} (ssignment to the toms) 0 c b
8 2 The formul evlution problem k the model checking problem Given: formul α for the logic L nd model M ccording to the semntics of L Decide: does M stisfy α? Exmple for propositionl logic: formul α = ( c) ( b) model M = {, c} (ssignment to the toms) 0 c b 0
9 2 The formul evlution problem k the model checking problem Given: formul α for the logic L nd model M ccording to the semntics of L Decide: does M stisfy α? Exmple for propositionl logic: formul α = ( c) ( b) model M = {, c} (ssignment to the toms) 0 c 0 b 0
10 2 The formul evlution problem k the model checking problem Given: formul α for the logic L nd model M ccording to the semntics of L Decide: does M stisfy α? Exmple for propositionl logic: formul α = ( c) ( b) model M = {, c} (ssignment to the toms) 0 c 0 b 0
11 2 The formul evlution problem k the model checking problem Given: formul α for the logic L nd model M ccording to the semntics of L Decide: does M stisfy α? Exmple for propositionl logic: formul α = ( c) ( b) model M = {, c} (ssignment to the toms) 0 c 0 b 0
12 3 Complexity of formul evlution Theorem (Buss 987) The formul evlution problem for propositionl logic is NC -complete. Theorem The formul evlution problem for propositionl modl logic without toms is P-complete. Wht hppens in between? How is the trnsition from NC to P?
13 Overview: formul evlution problem The complexity of the formul evlution problem for intuitionistic logics with bounded number k of toms. k=0 k= PC LC KC IPC FPL BPL NC -complete P AC LOGCFL k=2 [Buss87] P-complete DLOGCFL L NC NL
14 Overview: formul evlution problem The complexity of the formul evlution problem for intuitionistic logics with bounded number k of toms. PC LC KC IPC FPL BPL k=0 NC -complete k= AC k=2 NC DLOGCFL [Buss87] NL LOGCFL P-complete P AC LOGCFL DLOGCFL NL L NC
15 Overview: stisfibility problem The complexity of the stisfibility problem for intuitionistic logics with bounded number k of toms. PC LC KC IPC FPL BPL k=0 NC -complete k= k=2 NP-complete
16 Overview: tutology problem The complexity of the tutology problem for intuitionistic logics with bounded number k of toms. PC LC KC IPC FPL BPL k=0 NC -complete k= k=2 conp-compl. PSPACE-complete [Rybkov 2006]
17 Intuitionistic propositionl logic IPC Kleene (952) used Modus Ponens nd the following xioms for (clssicl) propositionl logic PC. BPL: 7, wek MP. α (β α) 2. (α (β γ)) ((α β) (α γ)) 3. (α β) α nd (α β) β 4. α (β (α β)) 5. α (α β) nd β (α β) 6. (α γ) ((β γ) ((α β) γ)) 7. (α β) ((α β) α) 8. α α IPC: 7, MP KC: 7, α α, MP LC: 7, (α β) (β α), MP PC: 8, MP IPC is obtined by leving out xiom 8.
18 8 Semntics for IPC reflexive nd trnsitive Kripke model with monotone vlution Semntics of conjunction nd disjunction is s usul. Semntics of impliction nd negtion is s follows: s = α β iff t s : if t = α, then t = β s = α }{{} = α iff t s : t = α nd re not vlid in IPC =,,, = = =,,,
19 8 Semntics for IPC reflexive nd trnsitive Kripke model with monotone vlution Semntics of conjunction nd disjunction is s usul. Semntics of impliction nd negtion is s follows: s = α β iff t s : if t = α, then t = β s = α }{{} = α iff t s : t = α b nd b re not equivlent, b =, b, b, b =, b = =,, b, b, b b
20 8 Semntics for IPC reflexive nd trnsitive Kripke model with monotone vlution Semntics of conjunction nd disjunction is s usul. Semntics of impliction nd negtion is s follows: s = α β iff t s : if t = α, then t = β s = α} {{ } iff t s : t = α = α ( b) (b ) does not hold =, b = b, b b = b, b =, b =, b, b, b, ( b) (b )
21 IPC formuls with only one tom Theorem (Rieger, Nishimur 960) There re infinitely mny equivlence clsses of IPC formuls. [ ].. ϕ 0 ϕ 9 [......] [......] ϕ 6 [( ) ( )] ϕ 2 [ ] ϕ 7 ϕ 8 [......] [......] ϕ 3 ϕ 4 [ ] [ ] ϕ 0 [ ] ϕ 5 [( ) ( )] ϕ [] Rules: [α β] = [α] [β] [α β] = [α] [β] [α β] = mx{[δ] [α] [δ] [β]} Ex. : 9
22 IPC formuls with only one tom Theorem (Rieger, Nishimur 960) There re infinitely mny equivlence clsses of IPC formuls. [ ].. ϕ 0 ϕ 9 [......] [......] ϕ 6 [( ) ( )] ϕ 2 [ ] ϕ 7 ϕ 8 [......] [......] ϕ 3 ϕ 4 [ ] [ ] ϕ 0 [ ] ϕ 5 [( ) ( )] ϕ [] Rules: [α β] = [α] [β] [α β] = [α] [β] [α β] = mx{[δ] [α] [δ] [β]} Ex. : ϕ ϕ 0 9
23 IPC formuls with only one tom Theorem (Rieger, Nishimur 960) There re infinitely mny equivlence clsses of IPC formuls. [ ].. ϕ 0 ϕ 9 [......] [......] ϕ 6 [( ) ( )] ϕ 2 [ ] ϕ 7 ϕ 8 [......] [......] ϕ 3 ϕ 4 [ ] [ ] ϕ 0 [ ] ϕ 5 [( ) ( )] ϕ [] Rules: [α β] = [α] [β] [α β] = [α] [β] [α β] = mx{[δ] [α] [δ] [β]} Ex. : ϕ ϕ 2 ϕ 0 ϕ 0 9
24 IPC formuls with only one tom Theorem (Rieger, Nishimur 960) There re infinitely mny equivlence clsses of IPC formuls. [ ].. ϕ 0 ϕ 9 [......] [......] ϕ 6 [( ) ( )] ϕ 2 [ ] ϕ 7 ϕ 8 [......] [......] ϕ 3 ϕ 4 [ ] [ ] ϕ 0 [ ] ϕ 5 [( ) ( )] ϕ [] Rules: [α β] = [α] [β] [α β] = [α] [β] [α β] = mx{[δ] [α] [δ] [β]} Ex. : ϕ ϕ 2 ϕ 4 ϕ 0 ϕ 0 9
25 IPC formuls with only one tom (IPC ) Lemm Given n IPC formul α, the i with α [ϕ i ] cn be computed with the ressources of DLOGCFL. Open question The tutology problem for IPC is in DLOGCFL. Is this optiml? (The word problem for free Heyting lgebrs on one genertor.)
26 Models for IPC formuls Theorem (Gbby 98) Every IPC model is homomorphic to some H i. H 0 H H 2 H 3 H 4 H 5 H 6
27 Complexity of model homomorphism Theorem The following problem is P-complete. Given: IPC model K, nd integer i Asked: is K homomorphic to H i? Theorem The following problem is AC -complete. Given: IPC model K, nd string 2i/2 Asked: is K homomorphic to H i?
28 3 Models for IPC formuls Theorem (Gbby 98) Model K stisfies ϕ 2i+ iff K is homomorphic to some H i. Model K stisfies ϕ 2i iff K is homomorphic to... H 0 H H 2 H 3 H 4 H 5 H 6 : : ( ) ( ) :
29 4 Complexity of model checking for IPC Lemm For α [ϕ i ] holds α 2 i/2. Theorem The following problem is AC -complete. Given: IPC model K, nd IPC formul α (insted of 2i/2 ) Asked: does K stisfy α?
30 5 The other logics BPL IPC, but wek modus ponens IPC FPL BPL + modus ponens BPL + Löb s rule KC IPC + α α LC IPC + (α β) (β α) PC IPC + α α
31 5 The other logics BPL trnsitive IPC FPL trnsitive & reflexive trnsitive & irreflexive KC directed semi-order LC liner order PC equivlence reltion
32 5 The other logics BPL K4 IPC FPL S4 PrL KC S4.2 LC S4.3 PC S5
33 6 FPL without toms: FPL 0 FPL hs trnsitive nd irreflexive Kripke models [ ]. [ ( )] [ ] [ ] The equivlence clsses of formuls The bsic models
34 Open problems Clculte the index of the equivlence clss of given formul α. PC: index equls 0 or, solvble in NC [Buss 987] IPC: index log α FPL: index α
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