MAYBE Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Fresh_0, 1, 0, Ar_3, Ar_4)) (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 - 10, Ar_1 - 1, Ar_2, Ar_3, Ar_4)) [ Ar_1 >= 1 /\ Ar_0 >= 101 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 + 11, Ar_1 + 1, Ar_2, Ar_3, Ar_4)) [ Ar_1 >= 1 /\ 100 >= Ar_0 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 - 10, Ar_1 - 1, 1, Ar_0, Ar_1)) [ Ar_0 >= 101 /\ 0 >= Ar_2 /\ Ar_1 >= 1 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 + 11, Ar_1 + 1, 1, Ar_0, Ar_1)) [ 100 >= Ar_0 /\ 0 >= Ar_2 /\ Ar_1 >= 1 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_3 >= Ar_0 /\ Ar_2 >= 1 /\ Ar_1 >= Ar_4 ] (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 Repeatedly propagating knowledge in problem 1 produces the following problem: 2: T: (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Fresh_0, 1, 0, Ar_3, Ar_4)) (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 - 10, Ar_1 - 1, Ar_2, Ar_3, Ar_4)) [ Ar_1 >= 1 /\ Ar_0 >= 101 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 + 11, Ar_1 + 1, Ar_2, Ar_3, Ar_4)) [ Ar_1 >= 1 /\ 100 >= Ar_0 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 - 10, Ar_1 - 1, 1, Ar_0, Ar_1)) [ Ar_0 >= 101 /\ 0 >= Ar_2 /\ Ar_1 >= 1 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 + 11, Ar_1 + 1, 1, Ar_0, Ar_1)) [ 100 >= Ar_0 /\ 0 >= Ar_2 /\ Ar_1 >= 1 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_3 >= Ar_0 /\ Ar_2 >= 1 /\ Ar_1 >= Ar_4 ] (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 A polynomial rank function with Pol(f0) = 1 Pol(f1) = 1 Pol(f2) = 0 Pol(koat_start) = 1 orients all transitions weakly and the transition f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_3 >= Ar_0 /\ Ar_2 >= 1 /\ Ar_1 >= Ar_4 ] strictly and produces the following problem: 3: T: (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Fresh_0, 1, 0, Ar_3, Ar_4)) (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 - 10, Ar_1 - 1, Ar_2, Ar_3, Ar_4)) [ Ar_1 >= 1 /\ Ar_0 >= 101 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 + 11, Ar_1 + 1, Ar_2, Ar_3, Ar_4)) [ Ar_1 >= 1 /\ 100 >= Ar_0 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 - 10, Ar_1 - 1, 1, Ar_0, Ar_1)) [ Ar_0 >= 101 /\ 0 >= Ar_2 /\ Ar_1 >= 1 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 + 11, Ar_1 + 1, 1, Ar_0, Ar_1)) [ 100 >= Ar_0 /\ 0 >= Ar_2 /\ Ar_1 >= 1 ] (Comp: 1, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_3 >= Ar_0 /\ Ar_2 >= 1 /\ Ar_1 >= Ar_4 ] (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 A polynomial rank function with Pol(f0) = 1 Pol(f1) = -V_3 + 1 Pol(f2) = -V_3 + 1 Pol(koat_start) = 1 orients all transitions weakly and the transition f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 - 10, Ar_1 - 1, 1, Ar_0, Ar_1)) [ Ar_0 >= 101 /\ 0 >= Ar_2 /\ Ar_1 >= 1 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Fresh_0, 1, 0, Ar_3, Ar_4)) (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 - 10, Ar_1 - 1, Ar_2, Ar_3, Ar_4)) [ Ar_1 >= 1 /\ Ar_0 >= 101 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 + 11, Ar_1 + 1, Ar_2, Ar_3, Ar_4)) [ Ar_1 >= 1 /\ 100 >= Ar_0 ] (Comp: 1, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 - 10, Ar_1 - 1, 1, Ar_0, Ar_1)) [ Ar_0 >= 101 /\ 0 >= Ar_2 /\ Ar_1 >= 1 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 + 11, Ar_1 + 1, 1, Ar_0, Ar_1)) [ 100 >= Ar_0 /\ 0 >= Ar_2 /\ Ar_1 >= 1 ] (Comp: 1, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_3 >= Ar_0 /\ Ar_2 >= 1 /\ Ar_1 >= Ar_4 ] (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 A polynomial rank function with Pol(f0) = 1 Pol(f1) = -V_3 + 1 Pol(f2) = -V_3 + 1 Pol(koat_start) = 1 orients all transitions weakly and the transition f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 + 11, Ar_1 + 1, 1, Ar_0, Ar_1)) [ 100 >= Ar_0 /\ 0 >= Ar_2 /\ Ar_1 >= 1 ] strictly and produces the following problem: 5: T: (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Fresh_0, 1, 0, Ar_3, Ar_4)) (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 - 10, Ar_1 - 1, Ar_2, Ar_3, Ar_4)) [ Ar_1 >= 1 /\ Ar_0 >= 101 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 + 11, Ar_1 + 1, Ar_2, Ar_3, Ar_4)) [ Ar_1 >= 1 /\ 100 >= Ar_0 ] (Comp: 1, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 - 10, Ar_1 - 1, 1, Ar_0, Ar_1)) [ Ar_0 >= 101 /\ 0 >= Ar_2 /\ Ar_1 >= 1 ] (Comp: 1, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 + 11, Ar_1 + 1, 1, Ar_0, Ar_1)) [ 100 >= Ar_0 /\ 0 >= Ar_2 /\ Ar_1 >= 1 ] (Comp: 1, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_3 >= Ar_0 /\ Ar_2 >= 1 /\ Ar_1 >= Ar_4 ] (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 Applied AI with 'oct' on problem 5 to obtain the following invariants: For symbol f1: X_3 >= 0 This yielded the following problem: 6: T: (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 ] (Comp: 1, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_2 >= 0 /\ Ar_3 >= Ar_0 /\ Ar_2 >= 1 /\ Ar_1 >= Ar_4 ] (Comp: 1, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 + 11, Ar_1 + 1, 1, Ar_0, Ar_1)) [ Ar_2 >= 0 /\ 100 >= Ar_0 /\ 0 >= Ar_2 /\ Ar_1 >= 1 ] (Comp: 1, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 - 10, Ar_1 - 1, 1, Ar_0, Ar_1)) [ Ar_2 >= 0 /\ Ar_0 >= 101 /\ 0 >= Ar_2 /\ Ar_1 >= 1 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 + 11, Ar_1 + 1, Ar_2, Ar_3, Ar_4)) [ Ar_2 >= 0 /\ Ar_1 >= 1 /\ 100 >= Ar_0 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Ar_0 - 10, Ar_1 - 1, Ar_2, Ar_3, Ar_4)) [ Ar_2 >= 0 /\ Ar_1 >= 1 /\ Ar_0 >= 101 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f1(Fresh_0, 1, 0, Ar_3, Ar_4)) start location: koat_start leaf cost: 0 Complexity upper bound ? Time: 4.130 sec (SMT: 4.018 sec)