MAYBE Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f9(Fresh_1, Fresh_1, Fresh_1, 0, Ar_4)) (Comp: ?, Cost: 1) f9(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f9(Ar_0, Ar_1, Ar_2, Ar_3 + 1, Fresh_0)) [ Ar_2 >= Ar_3 + 1 ] (Comp: ?, Cost: 1) f24(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f24(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 >= F + 1 ] (Comp: ?, Cost: 1) f24(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f24(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: ?, Cost: 1) f37(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f37(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ F >= G + 1 ] (Comp: ?, Cost: 1) f37(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f37(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: ?, Cost: 1) f37(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f48(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: ?, Cost: 1) f24(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f37(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: ?, Cost: 1) f9(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f24(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_3 >= Ar_2 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Slicing away variables that do not contribute to conditions from problem 1 leaves variables [Ar_2, Ar_3]. We thus obtain the following problem: 2: T: (Comp: 1, Cost: 0) koat_start(Ar_2, Ar_3) -> Com_1(f0(Ar_2, Ar_3)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f9(Ar_2, Ar_3) -> Com_1(f24(Ar_2, Ar_3)) [ Ar_3 >= Ar_2 ] (Comp: ?, Cost: 1) f24(Ar_2, Ar_3) -> Com_1(f37(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f37(Ar_2, Ar_3) -> Com_1(f48(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f37(Ar_2, Ar_3) -> Com_1(f37(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f37(Ar_2, Ar_3) -> Com_1(f37(Ar_2, Ar_3)) [ F >= G + 1 ] (Comp: ?, Cost: 1) f24(Ar_2, Ar_3) -> Com_1(f24(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f24(Ar_2, Ar_3) -> Com_1(f24(Ar_2, Ar_3)) [ 0 >= F + 1 ] (Comp: ?, Cost: 1) f9(Ar_2, Ar_3) -> Com_1(f9(Ar_2, Ar_3 + 1)) [ Ar_2 >= Ar_3 + 1 ] (Comp: ?, Cost: 1) f0(Ar_2, Ar_3) -> Com_1(f9(Fresh_1, 0)) 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_2, Ar_3) -> Com_1(f0(Ar_2, Ar_3)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f9(Ar_2, Ar_3) -> Com_1(f24(Ar_2, Ar_3)) [ Ar_3 >= Ar_2 ] (Comp: ?, Cost: 1) f24(Ar_2, Ar_3) -> Com_1(f37(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f37(Ar_2, Ar_3) -> Com_1(f48(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f37(Ar_2, Ar_3) -> Com_1(f37(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f37(Ar_2, Ar_3) -> Com_1(f37(Ar_2, Ar_3)) [ F >= G + 1 ] (Comp: ?, Cost: 1) f24(Ar_2, Ar_3) -> Com_1(f24(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f24(Ar_2, Ar_3) -> Com_1(f24(Ar_2, Ar_3)) [ 0 >= F + 1 ] (Comp: ?, Cost: 1) f9(Ar_2, Ar_3) -> Com_1(f9(Ar_2, Ar_3 + 1)) [ Ar_2 >= Ar_3 + 1 ] (Comp: 1, Cost: 1) f0(Ar_2, Ar_3) -> Com_1(f9(Fresh_1, 0)) start location: koat_start leaf cost: 0 A polynomial rank function with Pol(koat_start) = 3 Pol(f0) = 3 Pol(f9) = 3 Pol(f24) = 2 Pol(f37) = 1 Pol(f48) = 0 orients all transitions weakly and the transitions f9(Ar_2, Ar_3) -> Com_1(f24(Ar_2, Ar_3)) [ Ar_3 >= Ar_2 ] f37(Ar_2, Ar_3) -> Com_1(f48(Ar_2, Ar_3)) f24(Ar_2, Ar_3) -> Com_1(f37(Ar_2, Ar_3)) strictly and produces the following problem: 4: T: (Comp: 1, Cost: 0) koat_start(Ar_2, Ar_3) -> Com_1(f0(Ar_2, Ar_3)) [ 0 <= 0 ] (Comp: 3, Cost: 1) f9(Ar_2, Ar_3) -> Com_1(f24(Ar_2, Ar_3)) [ Ar_3 >= Ar_2 ] (Comp: 3, Cost: 1) f24(Ar_2, Ar_3) -> Com_1(f37(Ar_2, Ar_3)) (Comp: 3, Cost: 1) f37(Ar_2, Ar_3) -> Com_1(f48(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f37(Ar_2, Ar_3) -> Com_1(f37(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f37(Ar_2, Ar_3) -> Com_1(f37(Ar_2, Ar_3)) [ F >= G + 1 ] (Comp: ?, Cost: 1) f24(Ar_2, Ar_3) -> Com_1(f24(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f24(Ar_2, Ar_3) -> Com_1(f24(Ar_2, Ar_3)) [ 0 >= F + 1 ] (Comp: ?, Cost: 1) f9(Ar_2, Ar_3) -> Com_1(f9(Ar_2, Ar_3 + 1)) [ Ar_2 >= Ar_3 + 1 ] (Comp: 1, Cost: 1) f0(Ar_2, Ar_3) -> Com_1(f9(Fresh_1, 0)) start location: koat_start leaf cost: 0 Applied AI with 'oct' on problem 4 to obtain the following invariants: For symbol f24: X_2 >= 0 /\ -X_1 + X_2 >= 0 For symbol f37: X_2 >= 0 /\ -X_1 + X_2 >= 0 For symbol f9: X_2 >= 0 This yielded the following problem: 5: T: (Comp: 1, Cost: 1) f0(Ar_2, Ar_3) -> Com_1(f9(Fresh_1, 0)) (Comp: ?, Cost: 1) f9(Ar_2, Ar_3) -> Com_1(f9(Ar_2, Ar_3 + 1)) [ Ar_3 >= 0 /\ Ar_2 >= Ar_3 + 1 ] (Comp: ?, Cost: 1) f24(Ar_2, Ar_3) -> Com_1(f24(Ar_2, Ar_3)) [ Ar_3 >= 0 /\ -Ar_2 + Ar_3 >= 0 /\ 0 >= F + 1 ] (Comp: ?, Cost: 1) f24(Ar_2, Ar_3) -> Com_1(f24(Ar_2, Ar_3)) [ Ar_3 >= 0 /\ -Ar_2 + Ar_3 >= 0 ] (Comp: ?, Cost: 1) f37(Ar_2, Ar_3) -> Com_1(f37(Ar_2, Ar_3)) [ Ar_3 >= 0 /\ -Ar_2 + Ar_3 >= 0 /\ F >= G + 1 ] (Comp: ?, Cost: 1) f37(Ar_2, Ar_3) -> Com_1(f37(Ar_2, Ar_3)) [ Ar_3 >= 0 /\ -Ar_2 + Ar_3 >= 0 ] (Comp: 3, Cost: 1) f37(Ar_2, Ar_3) -> Com_1(f48(Ar_2, Ar_3)) [ Ar_3 >= 0 /\ -Ar_2 + Ar_3 >= 0 ] (Comp: 3, Cost: 1) f24(Ar_2, Ar_3) -> Com_1(f37(Ar_2, Ar_3)) [ Ar_3 >= 0 /\ -Ar_2 + Ar_3 >= 0 ] (Comp: 3, Cost: 1) f9(Ar_2, Ar_3) -> Com_1(f24(Ar_2, Ar_3)) [ Ar_3 >= 0 /\ Ar_3 >= Ar_2 ] (Comp: 1, Cost: 0) koat_start(Ar_2, Ar_3) -> Com_1(f0(Ar_2, Ar_3)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound ? Time: 1.707 sec (SMT: 1.646 sec)