MAYBE Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f0(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_0)) [ Ar_0 = Ar_1 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0 + 1, Ar_1 + 1)) (Comp: ?, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) (Comp: ?, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f10000(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f0(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) f0(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_0)) [ Ar_0 = Ar_1 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0 + 1, Ar_1 + 1)) (Comp: ?, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) (Comp: ?, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f10000(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f0) = 1 Pol(f1) = 1 Pol(f2) = 0 Pol(f10000) = 0 Pol(koat_start) = 1 orients all transitions weakly and the transitions f1(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) f1(Ar_0, Ar_1) -> Com_1(f10000(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ] strictly and produces the following problem: 3: T: (Comp: 1, Cost: 1) f0(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_0)) [ Ar_0 = Ar_1 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0 + 1, Ar_1 + 1)) (Comp: 1, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) (Comp: 1, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f10000(Ar_0, Ar_1)) [ Ar_0 >= Ar_1 + 1 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Applied AI with 'oct' on problem 3 to obtain the following invariants: For symbol f1: X_1 - X_2 >= 0 /\ -X_1 + X_2 >= 0 This yielded the following problem: 4: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ] (Comp: 1, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f10000(Ar_0, Ar_1)) [ Ar_0 - Ar_1 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 >= Ar_1 + 1 ] (Comp: 1, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) [ Ar_0 - Ar_1 >= 0 /\ -Ar_0 + Ar_1 >= 0 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0 + 1, Ar_1 + 1)) [ Ar_0 - Ar_1 >= 0 /\ -Ar_0 + Ar_1 >= 0 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_0)) [ Ar_0 = Ar_1 ] start location: koat_start leaf cost: 0 Testing for unsatisfiable constraints removes the following transition from problem 4: f1(Ar_0, Ar_1) -> Com_1(f10000(Ar_0, Ar_1)) [ Ar_0 - Ar_1 >= 0 /\ -Ar_0 + Ar_1 >= 0 /\ Ar_0 >= Ar_1 + 1 ] We thus obtain the following problem: 5: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ] (Comp: 1, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) [ Ar_0 - Ar_1 >= 0 /\ -Ar_0 + Ar_1 >= 0 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0 + 1, Ar_1 + 1)) [ Ar_0 - Ar_1 >= 0 /\ -Ar_0 + Ar_1 >= 0 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_0)) [ Ar_0 = Ar_1 ] start location: koat_start leaf cost: 0 Complexity upper bound ? Time: 0.828 sec (SMT: 0.783 sec)