MAYBE Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f3(Ar_0, Ar_1) -> Com_1(f0(0, Ar_1)) (Comp: ?, Cost: 1) f0(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1 - 1)) [ Ar_1 >= 1 ] (Comp: ?, Cost: 1) f4(Ar_0, Ar_1) -> Com_1(f4(Ar_0, Ar_1)) (Comp: ?, Cost: 1) f0(Ar_0, Ar_1) -> Com_1(f4(-1, Ar_1)) [ 0 >= Ar_1 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 1 produces the following problem: 2: T: (Comp: 1, Cost: 1) f3(Ar_0, Ar_1) -> Com_1(f0(0, Ar_1)) (Comp: ?, Cost: 1) f0(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1 - 1)) [ Ar_1 >= 1 ] (Comp: ?, Cost: 1) f4(Ar_0, Ar_1) -> Com_1(f4(Ar_0, Ar_1)) (Comp: ?, Cost: 1) f0(Ar_0, Ar_1) -> Com_1(f4(-1, Ar_1)) [ 0 >= Ar_1 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f3) = 1 Pol(f0) = 1 Pol(f4) = 0 Pol(koat_start) = 1 orients all transitions weakly and the transition f0(Ar_0, Ar_1) -> Com_1(f4(-1, Ar_1)) [ 0 >= Ar_1 ] strictly and produces the following problem: 3: T: (Comp: 1, Cost: 1) f3(Ar_0, Ar_1) -> Com_1(f0(0, Ar_1)) (Comp: ?, Cost: 1) f0(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1 - 1)) [ Ar_1 >= 1 ] (Comp: ?, Cost: 1) f4(Ar_0, Ar_1) -> Com_1(f4(Ar_0, Ar_1)) (Comp: 1, Cost: 1) f0(Ar_0, Ar_1) -> Com_1(f4(-1, Ar_1)) [ 0 >= Ar_1 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f3) = V_2 Pol(f0) = V_2 Pol(f4) = V_2 Pol(koat_start) = V_2 orients all transitions weakly and the transition f0(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1 - 1)) [ Ar_1 >= 1 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 1) f3(Ar_0, Ar_1) -> Com_1(f0(0, Ar_1)) (Comp: Ar_1, Cost: 1) f0(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1 - 1)) [ Ar_1 >= 1 ] (Comp: ?, Cost: 1) f4(Ar_0, Ar_1) -> Com_1(f4(Ar_0, Ar_1)) (Comp: 1, Cost: 1) f0(Ar_0, Ar_1) -> Com_1(f4(-1, Ar_1)) [ 0 >= Ar_1 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Applied AI with 'oct' on problem 4 to obtain the following invariants: For symbol f0: -X_1 >= 0 /\ X_1 >= 0 For symbol f4: -X_2 >= 0 /\ X_1 - X_2 + 1 >= 0 /\ -X_1 - X_2 - 1 >= 0 /\ -X_1 - 1 >= 0 /\ X_1 + 1 >= 0 This yielded the following problem: 5: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_1)) [ 0 <= 0 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1) -> Com_1(f4(-1, Ar_1)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ 0 >= Ar_1 ] (Comp: ?, Cost: 1) f4(Ar_0, Ar_1) -> Com_1(f4(Ar_0, Ar_1)) [ -Ar_1 >= 0 /\ Ar_0 - Ar_1 + 1 >= 0 /\ -Ar_0 - Ar_1 - 1 >= 0 /\ -Ar_0 - 1 >= 0 /\ Ar_0 + 1 >= 0 ] (Comp: Ar_1, Cost: 1) f0(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1 - 1)) [ -Ar_0 >= 0 /\ Ar_0 >= 0 /\ Ar_1 >= 1 ] (Comp: 1, Cost: 1) f3(Ar_0, Ar_1) -> Com_1(f0(0, Ar_1)) start location: koat_start leaf cost: 0 Complexity upper bound ? Time: 1.274 sec (SMT: 1.229 sec)