WORST_CASE(?, O(1)) Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f12(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f5(4, 0, 0, Ar_3, Ar_4, Ar_5)) (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f11(Ar_0, Ar_1, 0, 0, 0, Ar_5)) [ Ar_1 >= Ar_0 /\ Ar_2 = 0 ] (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f10(Ar_0, Ar_1, Ar_2, Ar_2, Ar_2, Ar_5)) [ Ar_2 >= 1 ] (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f10(Ar_0, Ar_1, Ar_2, Ar_2, Ar_2, Ar_5)) [ 0 >= Ar_2 + 1 ] (Comp: ?, Cost: 1) f7(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_4, Fresh_5, Ar_4, Fresh_4)) (Comp: ?, Cost: 1) f8(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_2, Fresh_3, Ar_4, Fresh_2)) (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_0, Fresh_1, Ar_4, Fresh_0)) [ Ar_0 >= Ar_1 + 1 /\ Ar_2 = 0 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f12(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Slicing away variables that do not contribute to conditions from problem 1 leaves variables [Ar_0, Ar_1, Ar_2]. We thus obtain the following problem: 2: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f12(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_0)) [ Ar_0 >= Ar_1 + 1 /\ Ar_2 = 0 ] (Comp: ?, Cost: 1) f8(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_2)) (Comp: ?, Cost: 1) f7(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_4)) (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 + 1 ] (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Ar_1, 0)) [ Ar_1 >= Ar_0 /\ Ar_2 = 0 ] (Comp: ?, Cost: 1) f12(Ar_0, Ar_1, Ar_2) -> Com_1(f5(4, 0, 0)) start location: koat_start leaf cost: 0 Testing for reachability in the complexity graph removes the following transitions from problem 2: f8(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_2)) f7(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_4)) We thus obtain the following problem: 3: T: (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Ar_1, 0)) [ Ar_1 >= Ar_0 /\ Ar_2 = 0 ] (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 + 1 ] (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_0)) [ Ar_0 >= Ar_1 + 1 /\ Ar_2 = 0 ] (Comp: ?, Cost: 1) f12(Ar_0, Ar_1, Ar_2) -> Com_1(f5(4, 0, 0)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f12(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 3 produces the following problem: 4: T: (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Ar_1, 0)) [ Ar_1 >= Ar_0 /\ Ar_2 = 0 ] (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 + 1 ] (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_0)) [ Ar_0 >= Ar_1 + 1 /\ Ar_2 = 0 ] (Comp: 1, Cost: 1) f12(Ar_0, Ar_1, Ar_2) -> Com_1(f5(4, 0, 0)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f12(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f5) = 1 Pol(f11) = 0 Pol(f10) = 0 Pol(f12) = 1 Pol(koat_start) = 1 orients all transitions weakly and the transitions f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 + 1 ] strictly and produces the following problem: 5: T: (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Ar_1, 0)) [ Ar_1 >= Ar_0 /\ Ar_2 = 0 ] (Comp: 1, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] (Comp: 1, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 + 1 ] (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_0)) [ Ar_0 >= Ar_1 + 1 /\ Ar_2 = 0 ] (Comp: 1, Cost: 1) f12(Ar_0, Ar_1, Ar_2) -> Com_1(f5(4, 0, 0)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f12(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f5) = 1 Pol(f11) = 0 Pol(f10) = 1 Pol(f12) = 1 Pol(koat_start) = 1 orients all transitions weakly and the transition f5(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Ar_1, 0)) [ Ar_1 >= Ar_0 /\ Ar_2 = 0 ] strictly and produces the following problem: 6: T: (Comp: 1, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Ar_1, 0)) [ Ar_1 >= Ar_0 /\ Ar_2 = 0 ] (Comp: 1, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] (Comp: 1, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 + 1 ] (Comp: ?, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_0)) [ Ar_0 >= Ar_1 + 1 /\ Ar_2 = 0 ] (Comp: 1, Cost: 1) f12(Ar_0, Ar_1, Ar_2) -> Com_1(f5(4, 0, 0)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f12(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f5) = V_1 - V_2 Pol(f11) = V_1 - V_2 Pol(f10) = V_1 - V_2 Pol(f12) = 4 Pol(koat_start) = 4 orients all transitions weakly and the transition f5(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_0)) [ Ar_0 >= Ar_1 + 1 /\ Ar_2 = 0 ] strictly and produces the following problem: 7: T: (Comp: 1, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f11(Ar_0, Ar_1, 0)) [ Ar_1 >= Ar_0 /\ Ar_2 = 0 ] (Comp: 1, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ Ar_2 >= 1 ] (Comp: 1, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 + 1 ] (Comp: 4, Cost: 1) f5(Ar_0, Ar_1, Ar_2) -> Com_1(f5(Ar_0, Ar_1 + 1, Fresh_0)) [ Ar_0 >= Ar_1 + 1 /\ Ar_2 = 0 ] (Comp: 1, Cost: 1) f12(Ar_0, Ar_1, Ar_2) -> Com_1(f5(4, 0, 0)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f12(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound 8 Time: 0.554 sec (SMT: 0.535 sec)