Complexity, Part 2 —

EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder

Description Logics: a Nice Family of Logics — Complexity, Part 2 —

Uli Sattler1 Thomas Schneider 2

1School of Computer Science, University of Manchester, UK

2Department of Computer Science, University of Bremen, Germany

ESSLLI, 18 August 2016

Uli Sattler, Thomas Schneider DL: Complexity (2) 1 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Goal for today

Automated reasoning plays an important role for DLs. It allows the development of intelligent applications. The expressivity of DLs is strongly tailored towards this goal.

Requirements for automated reasoning: Decidability of the relevant decision problems Low complexity if possible Algorithms that perform well in practice

Yesterday & today: 1 & 3

Now: 2

Uli Sattler, Thomas Schneider DL: Complexity (2) 2 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Plan for today

1 EXPTIME-membership

2 EXPTIME-hardness

3 PSPACE-hardness (not covered in the lecture)

4 Undecidability and a NEXPTIME lower bound ; Uli

Thanks to Carsten Lutz for much of the material on these slides.

Uli Sattler, Thomas Schneider DL: Complexity (2) 3 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder And now . . .

1 EXPTIME-membership

2 EXPTIME-hardness

3 PSPACE-hardness (not covered in the lecture)

4 Undecidability and a NEXPTIME lower bound ; Uli

Uli Sattler, Thomas Schneider DL: Complexity (2) 4 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder EXPTIME-membership

We start with an EXPTIME upper bound for concept satisﬁability in ALC relative to TBoxes.

Theorem The following problem is in EXPTIME.

Input: an ALC concept C0 and an ALC TBox T I Question: is there a model I |= T with C0 6= ∅ ?

We’ll use a technique known from modal logic: type elimination [Pratt 1978] The basis is a syntactic notion of a type.

Uli Sattler, Thomas Schneider DL: Complexity (2) 5 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Syntactic types

We assume that

the input concept C0 is in NNF the input TBox is T = {> v CT } with CT in NNF

Let sub(C0, T ) be the set of subconcepts of C0 and CT . A type for C0 and T is a subset t ⊆ sub(C0, T ) such that

1. A ∈ t iff ¬A ∈/ t for all ¬A ∈ sub(C0, T ) 2. C u D ∈ t iff C ∈ t and D ∈ t for all C u D ∈ sub(C0, T ) 3. C t D ∈ t iff C ∈ t or D ∈ t for all C t D ∈ sub(C0, T ) 4. CT ∈ t

Intuition: Types describe domain elements completely, up to sub(C0, T ). •

Uli Sattler, Thomas Schneider DL: Complexity (2) 6 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder General idea

General idea of type elimination for input C0, T :

Generate all types for C0 and T (exponentially many). Repeatedly eliminate types that cannot occur in any model of C0 and T .

Check whether some type containing C0 has survived. If yes, return “satisﬁable”; otherwise “unsatisﬁable”.

Uli Sattler, Thomas Schneider DL: Complexity (2) 7 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Bad types

The precise notion of “types that cannot occur in a model”:

Let Γ be a set of types and t ∈ Γ. Then t is bad in Γ if, for some ∃r.C ∈ t,

there is no t0 ∈ Γ with {C} ∪ {D | ∀r.D ∈ t} ⊆ t0.

Intuition: A type is bad if it contains an existential restriction for which there is no “witness” in Γ. •

Uli Sattler, Thomas Schneider DL: Complexity (2) 8 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder The type elimination algorithm

procedure ALC-Elim(C0, T )

compute Γ0: set of all types for C0 and T i := 0 repeat i := i + 1 Γi := {t ∈ Γi−1 | t not bad in Γi−1} until Γi = Γi−1

if there is t ∈ Γi with C0 ∈ t then return “satisﬁable” else return “unsatisﬁable”

Uli Sattler, Thomas Schneider DL: Complexity (2) 9 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Termination

. . . is easy (cf. the tableau algorithm): Lemma O(|C |+|T |) ALC-Elim(C0, T ) terminates after 2 0 steps.

Proof. |C |+|T | There are 2 0 subsets of sub(C0, T ). |C |+|T | Thus Γ0 has at most 2 0 elements, and all of them can be generated in time 2O(|C0|+|T |). Determining whether a type t is bad in Γ takes time polynomial in |t| · |Γ|.

Since Γ0 ⊇ · · · ⊇ Γi ⊇ Γi+1, there are at most 2|C0|+|T | iterations.

Uli Sattler, Thomas Schneider DL: Complexity (2) 10 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Soundness and completeness

Lemma ALC-Elim(C0, T ) returns “satisfiable” iff C0 is satisfiable w.r.t. T .

Proof of “⇒” Construct model I from the set of remaining types Γ: ∆I = Γ AI = {t | A ∈ t} rI = {(t, t0) | ∀r.D ∈ t implies D ∈ t0} The following claim can be proven inductively, exploiting the fact that no type in Γ is bad: I For all t ∈ Γ and C ∈ sub(C0, T ): C ∈ t iﬀ t ∈ C

I |= T because all types contain CT and using the claim I C0 6= ∅ because some type containing C0 has survived

Uli Sattler, Thomas Schneider DL: Complexity (2) 11 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Soundness and completeness

Lemma ALC-Elim(C0, T ) returns “satisfiable” iff C0 is satisfiable w.r.t. T .

Proof of “⇐” I Let I |= T with d0 ∈ C0 . I I For every d ∈ ∆ , let tp(d) = {C ∈ sub(C0, T ) | d ∈ C }, i.e., the type of d in I relative to C0 and T . Let Γ = {tp(d) | d ∈ ∆I}, i.e., the set of all types occurring in I. Since I is a model, no type in Γ can be bad.

Now by easy induction: Γ ⊆ Γi for every i; hence all types in Γ survive.

Since C0 ∈ tp(d0) ∈ Γ, the algo. returns “satisﬁable”.

Uli Sattler, Thomas Schneider DL: Complexity (2) 12 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Harvest

Direct consequence from the previous lemmas: Theorem Satisﬁability of and subsumption between ALC concepts w.r.t. TBoxes is decidable in EXPTIME.

Uli Sattler, Thomas Schneider DL: Complexity (2) 13 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Type elimination for more expressive DLs

Type elimination can be extended to give the same EXPTIME upper bound for ALCI Exercise: Extend the deﬁnition of “bad” to inverse roles. Adapt the soundness and completeness proof. Verify the new algorithm on the following inputs. − C0 = >, T = {> v ∃r.A u ∀r .¬A} − C0 = >, T = {> v ∃r .A u ∀r.¬A} ALCQI

Uli Sattler, Thomas Schneider DL: Complexity (2) 14 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Comparison with tableau algorithms

Similarities u-, t-, and GCI-rule are captured by the definition of a type. ∃- and ∀-rule are captured by the definition of “bad”. Clash-freeness is captured by the definition of a type.

Diﬀerence In the worst case, type elimination runs in exponential time; the tableau algorithm in exponential space (and 2-exp. time). Should we thus prefer type elimination over tableaux for reasoning over real-world ontologies? No! Type elimination requires exp. time in the best case (for every input: initial generation of all types). Tableau algorithms lend themselves to numerous optimisations avoiding the worst-case behaviour for many real-world inputs.

Uli Sattler, Thomas Schneider DL: Complexity (2) 15 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Tight complexity bounds for ALC

Can we do better than EXPTIME?

No, concept satisﬁability and subsumption w.r.t. ALC TBoxes is EXPTIME-complete. We’ll next show the lower bound (hardness).

Uli Sattler, Thomas Schneider DL: Complexity (2) 16 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder And now . . .

1 EXPTIME-membership

2 EXPTIME-hardness

3 PSPACE-hardness (not covered in the lecture)

4 Undecidability and a NEXPTIME lower bound ; Uli

Uli Sattler, Thomas Schneider DL: Complexity (2) 17 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Lower bound

Standard approach to showing EXPTIME-hardness: Reduction from the word problem for polynomially space-bounded alternating Turing machines

More convenient approach: Reduction from a related, but much more intuitive problem: a game-theoretic problem whose EXPTIME-completeness is known

Uli Sattler, Thomas Schneider DL: Complexity (2) 18 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder EXPTIME games

Two players 1, 2 play on a given propositional formula ϕ.

Every propositional variable in ϕ belongs to exactly one player.

The game begins with a given initial assignment π0 of the vars.

Player 1 begins; the players take turns.

In every move the player changes the truth value of one of her variables or passes.

Player 1 wins as soon as ϕ becomes true (no matter which player’s turn it is).

Player 2 wins if the game continues inﬁnitely without ϕ becoming true.

Uli Sattler, Thomas Schneider DL: Complexity (2) 19 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Example

Consider ϕ = (p ↔ ¬q) ∧ (p0 ↔ ¬q0) p q p0q0 0 0 where P1 owns p, p and P2 owns q, q , and π0 = 0 0 0 0 . A run of the game where Player 1 wins: P1 sets p to 1 ; π = 1000 P2 passes (obviously he hasn’t understood the game yet! :-)) P1 sets p0 to 1 ; π = 1010 WIN!

A run of the game where Player 2 wins: Whenever P1 changes the value of p (or p0), in the subsequent move P2 changes the value of q (or q0). Whenever P1 passes, so does P2 in the subsequent move. This ensures that, always, either p and q or p0 and q0 have the same truth value ; P1 cannot win! Uli Sattler, Thomas Schneider DL: Complexity (2) 20 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder EXPTIME games: basics

Basic notions

We represent a game as a tuple (ϕ, Γ1, Γ2, π0) with

Γ1, Γ2 a partitioning of the variables in ϕ

π0 the initial assignment

A conﬁguration is a pair (i, π) with i ∈ {1, 2} the active player and π an assignment. (a “snapshot” of the running game)

π is a j-variation of π0 for j ∈ {1, 2} if 0 0 π = π or π and π diﬀer in exactly one variable p ∈ Γj . (i.e., Player j can transform π into π0 or vice versa)

The decision problem from which we will reduce refers to winning strategies.

Uli Sattler, Thomas Schneider DL: Complexity (2) 21 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Winning strategies

Intuitively: A winning strategy speciﬁes a next move for Player 2 in every possible situation of the game. (situation = a ﬁnite sequence of the players’ previous moves)

If Player 2 has a winning strategy, then she can win the game regardless of Player 1’s moves.

To take into account the “history” of game situations, winning strategies are represented using trees.

Uli Sattler, Thomas Schneider DL: Complexity (2) 22 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Winning strategies

A winning strategy for Player 2 in the game (ϕ, Γ1, Γ2, π0) is an inﬁnite node-labelled tree (V , E, `), where ` assigns to each node v ∈ V a conﬁguration `(v) such that:

The root is labelled (1, π0). If `(v) = (2, π), then v has a successor v0 with `(v0) = (1, π0), where π0 is some 2-variation of π.

If `(v) = (1, π), then v has successors v ,..., v with 0 |Γ1| `(vi ) = (2, πi ), where the πi are all 1-variations of π. If `(v) = (i, π), then π 6|= ϕ.

Uli Sattler, Thomas Schneider DL: Complexity (2) 23 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Example

Consider again ϕ = (p ↔ ¬q) ∧ (p0 ↔ ¬q0) p q p0q0 0 0 with Γ1 = {p, p } and Γ2 = {q, q } and π0 = 0 0 0 0 .

...

Uli Sattler, Thomas Schneider DL: Complexity (2) 24 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder EXPTIME games as a decision problem

Game1 is the following decision problem:

Input: a game (ϕ, Γ1, Γ2, π0) Question: does Player 2 have a winning strategy?

Theorem (Stockmeyer & Chandra 1979) Game1 is EXPTIME-complete.

Uli Sattler, Thomas Schneider DL: Complexity (2) 25 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder EXPTIME-hardness

We want to show via a reduction from Game1 : Theorem Satisﬁability of concepts w.r.t. ALC-TBoxes is EXPTIME-hard.

That is, given a game G = (ϕ, Γ1, Γ2, π0), we want to construct (in polytime) a concept CG and TBox TG such that:

Player 2 has a winning strategy in G iﬀ CG satisﬁable w.r.t. TG

Main idea: (tree) models of CG and TG encode winning strategies!

TG restricts (tree) models such that they represent only winning strategies.

CG simply says that there is a root.

Uli Sattler, Thomas Schneider DL: Complexity (2) 26 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Details of the reduction

Let the variable partitioning be Γ1 = {p1,..., pm} and Γ2 = {pm+1,..., pn}.

We build CG and TG from the following concept and role names. role name r for the edges in the strategy tree concept name R for the root

concept names P1,..., Pn for the variables concept names S1, S2 for the active player concept names V1,..., Vn for the variable whose value was changed to reach the current conﬁguration

Uli Sattler, Thomas Schneider DL: Complexity (2) 27 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder

TG consists of the following GCIs

1 The initial conﬁguration is correctly represented.

R v S1 u l ¬Pi u l Pi i≤n, π0(pi )=0 i≤n, π0(pi )=1 2 Whenever it is Player 1’s turn, there are m + 1 successors representing all her possible moves:

S1 v ∃r.(¬V1 u · · · u ¬Vn) u l ∃r.Vi i≤m

3 Whenever it is Player 2’s turn, there is one successor: G S2 v ∃r.(¬V0 u · · · u ¬Vn) t ∃r.Vi m

4 In every move, at most one variable is changed:

> v l ¬(Vi u Vj ) i

Uli Sattler, Thomas Schneider DL: Complexity (2) 28 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder

TG consists of the following GCIs

5 The selected variable changes its truth value: > v l (Pi → ∀r.(Vi → ¬Pi )) u (¬Pi → ∀r.(Vi → Pi )) i≤n

6 All other variables keep their truth value: > v l (Pi → ∀r.(¬Vi → Pi )) u (¬Pi → ∀r.(¬Vi → ¬Pi )) i≤n

7 The players take turns:

S1 v ∀r.S2, S2 v ∀r.S1, S1 v ¬S2

8 The formula ϕ never becomes true: > v ¬ϕ

Additionally, set CG = R. Uli Sattler, Thomas Schneider DL: Complexity (2) 29 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Correctness of the reduction

Lemma Player 2 has a winning strategy in G iﬀ CG satisﬁable w.r.t. TG

To prove the lemma, we have to transform

“⇒” . . . a winning strategy (V , E, `) into a model I of CG and TG : use V as the domain and interpret r as E; interpret r as the root;

read oﬀ interpretation of the Pi , Si , Vi from the labels `(v). It is then easy to check that I satisﬁes all GCIs 1–8.

“⇐” . . . a tree model I of CG & TG into a winning strat. (V , E, `):

analogously read oﬀ the `(v) from the interpretations of the Pi . GCIs 1–8 ensure that the resulting tree is a winning strategy.

Uli Sattler, Thomas Schneider DL: Complexity (2) 30 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Correctness, proof of “⇒”

Player 2 has a winning strategy in G ⇒ CG satisﬁable w.r.t. TG

Construct model I from winning strat. (V , E, `) with root d0: I I I ∆ = V r = ER = {d0} I Pi = {v ∈ V | `(v) = (j, π) and π(pi ) = 1} I Sj = {v ∈ V | `(v) = (j, π)} I 0 0 Vi = {v ∈ V \{d0} | `(v) = (j, π) and `(v ) = (j, π ) 0 0 for the parent node v of v and π(pi ) 6= π (pi )}

I |= TG : check each GCI 1–8 separately. (Exercise!) I I CG 6= ∅ because R = {d0}.

Uli Sattler, Thomas Schneider DL: Complexity (2) 31 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Correctness, proof of “⇐”

Player 2 has a winning strategy in G ⇐ CG satisﬁable w.r.t. TG

Exercise. (use the hint on Slide 30)

Uli Sattler, Thomas Schneider DL: Complexity (2) 32 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Harvest

Theorem Satisﬁability of concepts w.r.t. ALC-TBoxes is EXPTIME-complete.

The lower bound follows from the previous lemma. For the upper bound see Slide 13 (type elimination).

Uli Sattler, Thomas Schneider DL: Complexity (2) 33 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Let’s do this again!

We’ve used EXPTIME games successfully to prove a matching lower bound for satisﬁability of ALC concepts w.r.t. TBoxes.

For satisﬁability/subsumption without TBoxes, we already have a PSPACE upper bound (tableau procedure).

We’ll now show a matching lower bound via a reduction from a very similar type of game.

Uli Sattler, Thomas Schneider DL: Complexity (2) 34 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder And now . . .

1 EXPTIME-membership

2 EXPTIME-hardness

3 PSPACE-hardness (not covered in the lecture)

4 Undecidability and a NEXPTIME lower bound ; Uli

Uli Sattler, Thomas Schneider DL: Complexity (2) 35 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Lower bound

We’ve seen on Wednesday: Concept satisﬁability and subsumption in ALC is PSPACE-complete.

Standard approach to showing PSPACE-hardness: Reduction from the validity problem of quantiﬁed boolean formulas (QBF)

More convenient approach: Reduction from a related, but much more intuitive problem: a game-theoretic problem whose PSPACE-completeness is known

Uli Sattler, Thomas Schneider DL: Complexity (2) 36 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder PSPACE games

Two players 1, 2 play on a given propositional formula ϕ.

Every propositional variable in ϕ belongs to exactly one player.

Both players own the same number of variables.

Each player’s variables are linearly ordered.

Player 1 begins; the players take turns.

In every move the player chooses the truth value of her next variable.

Player 1 wins if ϕ is true at the end; otherwise Player 2 wins.

Uli Sattler, Thomas Schneider DL: Complexity (2) 37 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Example

Consider the same formula ϕ = (p1 ↔ ¬p2) ∧ (p3 ↔ ¬p4) where P1 owns p1, p3 and P2 owns p2, p4.

A run of the game where Player 2 wins:

P1 sets p1 to 1 ; π = 1??? P2 sets p2 to 1 ; π = 11?? ; No matter how p3, p4 are set, P2 will win (since (p1 ↔ ¬p2) is false).

Uli Sattler, Thomas Schneider DL: Complexity (2) 38 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Diﬀerences in comparison with EXPTIME games

The game always ends; the number of moves is predetermined. The players have no freedom in choosing which variables to assign next. The players cannot pass. Each variable is assigned a truth value exactly once. No initial assignment is needed.

Uli Sattler, Thomas Schneider DL: Complexity (2) 39 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder PSPACE games: basics

A game is now just a propositional formula ϕ with an even number of variables p1,..., pn.

A conﬁguration is just a word π ∈ {0, 1}∗.

Intuition: Variables with odd index belong to P1, with even index to P2. Conﬁguration π is a partial assignment: i-th symbol of π is the truth value of pi .

Uli Sattler, Thomas Schneider DL: Complexity (2) 40 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Winning strategies

We’re interested in winning strategies of Player 1:

A winning strategy for Player 1 in the game ϕ is a ﬁnite node-labelled tree (V , E, `), where ` assigns to each node v ∈ V a conﬁguration `(v) such that:

The root is labelled ε (empty conﬁguration).

If `(v) = π with |π| even and |π| < n (P1’s move), then v has a successor v0 with `(v0) = π0 or `(v0) = π1.

If `(v) = π with |π| odd (P2’s move), then v has successors v0, v00 with `(v0) = π0 & `(v00) = π1.

If `(v) = π with |π| = n, then π |= ϕ.

Uli Sattler, Thomas Schneider DL: Complexity (2) 41 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder PSPACE games as a decision problem

Game2 is the following decision problem: Input: a game ϕ Question: does Player 1 have a winning strategy?

Theorem (Stockmeyer & Chandra 1979) Game2 is PSPACE-complete.

Uli Sattler, Thomas Schneider DL: Complexity (2) 42 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder PSPACE-hardness

We want to show via a reduction from Game2 : Theorem Satisﬁability of ALC concepts is PSPACE-hard.

That is, given a game ϕ, we want to construct (in polytime) a concept Cϕ such that:

Player 1 has a winning strategy in game ϕ iﬀ Cϕ satisﬁable

Main idea, again: (tree) models of Cϕ encode winning strategies!

Now Cϕ restricts (tree) models such that they represent only winning strategies.

Uli Sattler, Thomas Schneider DL: Complexity (2) 43 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Details of the reduction

Let the variables in ϕ be p1,..., pn, n even.

We build Cϕ from the following concept and role names. role name r for the edges in the strategy tree

concept names P1,..., Pn for truth values of the variables in the partial assignments

We write ∀ri .C for ∀r. · · · ∀r.C | {z } i times

Uli Sattler, Thomas Schneider DL: Complexity (2) 44 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Details of the reduction

Cϕ consists of the following conjuncts.

1 If |π| is even (i.e., it is P1’s move), then there is a successor that selects the truth value of Pi+1 :

i C1 := l ∀r . ∃r.¬Pi+1 t ∃r.Pi+1 i∈{0,2,...,n−2}

2 If |π| is odd (i.e., it is P2’s move), then there are two successors for both truth values of Pi+1 :

i C2 := l ∀r . ∃r.¬Pi+1 u ∃r.Pi+1 i∈{1,3,...,n−1}

Uli Sattler, Thomas Schneider DL: Complexity (2) 45 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Details of the reduction

Cϕ consists of the following conjuncts.

1 The chosen truth values do not change:

j C3 := l ∀r . (Pi → ∀r.Pi ) u (¬Pi → ∀r.¬Pi ) 1≤i≤j

2 At the leaves ϕ is true:

n C4 := ∀r .ϕ

Set Cϕ = C1 u C2 u C3 u C4.

Uli Sattler, Thomas Schneider DL: Complexity (2) 46 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Correctness of the reduction

Lemma Player 1 has a winning strategy in ϕ iﬀ Cϕ satisﬁable.

Proof: very similar to the corresponding EXPTIME lemma.

As a consequence we get: Theorem Satisﬁability of ALC concepts is PSPACE-complete.

Uli Sattler, Thomas Schneider DL: Complexity (2) 47 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder Summary

Satisﬁability of and subsumption for ALC concepts w.r.t. TBoxes is EXPTIME-complete.

Satisﬁability of and subsumption for ALC concepts without TBoxes is PSPACE-complete.

The same holds for ALC, ALCI, ALCQI.

Tree models play an important rôle in all cases.

PSPACE versus EXPTIME: polynomially deep versus inﬁnite trees

Types are an important notion for developing worst-case optimal algorithms.

Uli Sattler, Thomas Schneider DL: Complexity (2) 48 EXPTIME-membership EXPTIME-hardness PSPACE-hardness harder And now . . .

1 EXPTIME-membership

2 EXPTIME-hardness

3 PSPACE-hardness (not covered in the lecture)

4 Undecidability and a NEXPTIME lower bound ; Uli

Uli Sattler, Thomas Schneider DL: Complexity (2) 49