MAYBE Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f12(Fresh_1, Fresh_2, Fresh_3, 0, Ar_4, Ar_5, Ar_6)) (Comp: ?, Cost: 1) f12(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f12(Ar_0, Ar_1, Ar_2, Ar_3 + 1, Ar_4, Ar_5, Ar_6)) [ Ar_2 >= Ar_3 + 1 ] (Comp: ?, Cost: 1) f25(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f25(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5 + 1, Ar_6)) [ Ar_4 >= Ar_5 + 1 ] (Comp: ?, Cost: 1) f25(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f34(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ Ar_5 >= Ar_4 ] (Comp: ?, Cost: 1) f12(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f25(Ar_0, Ar_1, Ar_2, Ar_3, Ar_0, 0, Fresh_0)) [ Ar_3 >= Ar_2 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6)) [ 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_2, Ar_3, Ar_4, Ar_5]. We thus obtain the following problem: 2: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f12(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f25(Ar_0, Ar_2, Ar_3, Ar_0, 0)) [ Ar_3 >= Ar_2 ] (Comp: ?, Cost: 1) f25(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f34(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5)) [ Ar_5 >= Ar_4 ] (Comp: ?, Cost: 1) f25(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f25(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5 + 1)) [ Ar_4 >= Ar_5 + 1 ] (Comp: ?, Cost: 1) f12(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f12(Ar_0, Ar_2, Ar_3 + 1, Ar_4, Ar_5)) [ Ar_2 >= Ar_3 + 1 ] (Comp: ?, Cost: 1) f0(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f12(Fresh_1, Fresh_3, 0, Ar_4, Ar_5)) start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 2 produces the following problem: 3: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f12(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f25(Ar_0, Ar_2, Ar_3, Ar_0, 0)) [ Ar_3 >= Ar_2 ] (Comp: ?, Cost: 1) f25(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f34(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5)) [ Ar_5 >= Ar_4 ] (Comp: ?, Cost: 1) f25(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f25(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5 + 1)) [ Ar_4 >= Ar_5 + 1 ] (Comp: ?, Cost: 1) f12(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f12(Ar_0, Ar_2, Ar_3 + 1, Ar_4, Ar_5)) [ Ar_2 >= Ar_3 + 1 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f12(Fresh_1, Fresh_3, 0, Ar_4, Ar_5)) start location: koat_start leaf cost: 0 A polynomial rank function with Pol(koat_start) = 2 Pol(f0) = 2 Pol(f12) = 2 Pol(f25) = 1 Pol(f34) = 0 orients all transitions weakly and the transitions f25(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f34(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5)) [ Ar_5 >= Ar_4 ] f12(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f25(Ar_0, Ar_2, Ar_3, Ar_0, 0)) [ Ar_3 >= Ar_2 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ] (Comp: 2, Cost: 1) f12(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f25(Ar_0, Ar_2, Ar_3, Ar_0, 0)) [ Ar_3 >= Ar_2 ] (Comp: 2, Cost: 1) f25(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f34(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5)) [ Ar_5 >= Ar_4 ] (Comp: ?, Cost: 1) f25(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f25(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5 + 1)) [ Ar_4 >= Ar_5 + 1 ] (Comp: ?, Cost: 1) f12(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f12(Ar_0, Ar_2, Ar_3 + 1, Ar_4, Ar_5)) [ Ar_2 >= Ar_3 + 1 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f12(Fresh_1, Fresh_3, 0, Ar_4, Ar_5)) start location: koat_start leaf cost: 0 Applied AI with 'oct' on problem 4 to obtain the following invariants: For symbol f12: X_3 >= 0 For symbol f25: X_5 >= 0 /\ X_3 + X_5 >= 0 /\ X_1 - X_4 >= 0 /\ -X_1 + X_4 >= 0 /\ X_3 >= 0 /\ -X_2 + X_3 >= 0 This yielded the following problem: 5: T: (Comp: 1, Cost: 1) f0(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f12(Fresh_1, Fresh_3, 0, Ar_4, Ar_5)) (Comp: ?, Cost: 1) f12(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f12(Ar_0, Ar_2, Ar_3 + 1, Ar_4, Ar_5)) [ Ar_3 >= 0 /\ Ar_2 >= Ar_3 + 1 ] (Comp: ?, Cost: 1) f25(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f25(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5 + 1)) [ Ar_5 >= 0 /\ Ar_3 + Ar_5 >= 0 /\ Ar_0 - Ar_4 >= 0 /\ -Ar_0 + Ar_4 >= 0 /\ Ar_3 >= 0 /\ -Ar_2 + Ar_3 >= 0 /\ Ar_4 >= Ar_5 + 1 ] (Comp: 2, Cost: 1) f25(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f34(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5)) [ Ar_5 >= 0 /\ Ar_3 + Ar_5 >= 0 /\ Ar_0 - Ar_4 >= 0 /\ -Ar_0 + Ar_4 >= 0 /\ Ar_3 >= 0 /\ -Ar_2 + Ar_3 >= 0 /\ Ar_5 >= Ar_4 ] (Comp: 2, Cost: 1) f12(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f25(Ar_0, Ar_2, Ar_3, Ar_0, 0)) [ Ar_3 >= 0 /\ Ar_3 >= Ar_2 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound ? Time: 1.656 sec (SMT: 1.590 sec)