MAYBE Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f10(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f16(Ar_0, 0, Fresh_2, Fresh_2, Ar_4)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) f16(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f16(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_3 >= 1 ] (Comp: ?, Cost: 1) f25(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f25(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: ?, Cost: 1) f27(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f30(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) (Comp: ?, Cost: 1) f16(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f10(Fresh_1, Ar_1, Fresh_1, Ar_3, 0)) [ 0 >= Ar_3 ] (Comp: ?, Cost: 1) f10(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f25(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4) -> Com_1(f10(Fresh_0, 0, Fresh_0, Ar_3, 0)) (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_0, Ar_3]. We thus obtain the following problem: 2: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_3) -> Com_1(f0(Ar_0, Ar_3)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f0(Ar_0, Ar_3) -> Com_1(f10(Fresh_0, Ar_3)) (Comp: ?, Cost: 1) f10(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) f16(Ar_0, Ar_3) -> Com_1(f10(Fresh_1, Ar_3)) [ 0 >= Ar_3 ] (Comp: ?, Cost: 1) f27(Ar_0, Ar_3) -> Com_1(f30(Ar_0, Ar_3)) (Comp: ?, Cost: 1) f25(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) (Comp: ?, Cost: 1) f16(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Ar_3)) [ Ar_3 >= 1 ] (Comp: ?, Cost: 1) f10(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Fresh_2)) [ 0 >= Ar_0 ] start location: koat_start leaf cost: 0 Testing for reachability in the complexity graph removes the following transition from problem 2: f27(Ar_0, Ar_3) -> Com_1(f30(Ar_0, Ar_3)) We thus obtain the following problem: 3: T: (Comp: ?, Cost: 1) f16(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Ar_3)) [ Ar_3 >= 1 ] (Comp: ?, Cost: 1) f16(Ar_0, Ar_3) -> Com_1(f10(Fresh_1, Ar_3)) [ 0 >= Ar_3 ] (Comp: ?, Cost: 1) f25(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) (Comp: ?, Cost: 1) f10(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Fresh_2)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) f10(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) f0(Ar_0, Ar_3) -> Com_1(f10(Fresh_0, Ar_3)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_3) -> Com_1(f0(Ar_0, Ar_3)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 3 produces the following problem: 4: T: (Comp: ?, Cost: 1) f16(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Ar_3)) [ Ar_3 >= 1 ] (Comp: ?, Cost: 1) f16(Ar_0, Ar_3) -> Com_1(f10(Fresh_1, Ar_3)) [ 0 >= Ar_3 ] (Comp: ?, Cost: 1) f25(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) (Comp: ?, Cost: 1) f10(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Fresh_2)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) f10(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) [ Ar_0 >= 1 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_3) -> Com_1(f10(Fresh_0, Ar_3)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_3) -> Com_1(f0(Ar_0, Ar_3)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f16) = 1 Pol(f10) = 1 Pol(f25) = 0 Pol(f0) = 1 Pol(koat_start) = 1 orients all transitions weakly and the transition f10(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) [ Ar_0 >= 1 ] strictly and produces the following problem: 5: T: (Comp: ?, Cost: 1) f16(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Ar_3)) [ Ar_3 >= 1 ] (Comp: ?, Cost: 1) f16(Ar_0, Ar_3) -> Com_1(f10(Fresh_1, Ar_3)) [ 0 >= Ar_3 ] (Comp: ?, Cost: 1) f25(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) (Comp: ?, Cost: 1) f10(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Fresh_2)) [ 0 >= Ar_0 ] (Comp: 1, Cost: 1) f10(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) [ Ar_0 >= 1 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_3) -> Com_1(f10(Fresh_0, Ar_3)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_3) -> Com_1(f0(Ar_0, Ar_3)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Applied AI with 'oct' on problem 5 to obtain the following invariants: For symbol f16: -X_1 >= 0 For symbol f25: X_1 - 1 >= 0 This yielded the following problem: 6: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_3) -> Com_1(f0(Ar_0, Ar_3)) [ 0 <= 0 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_3) -> Com_1(f10(Fresh_0, Ar_3)) (Comp: 1, Cost: 1) f10(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) f10(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Fresh_2)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) f25(Ar_0, Ar_3) -> Com_1(f25(Ar_0, Ar_3)) [ Ar_0 - 1 >= 0 ] (Comp: ?, Cost: 1) f16(Ar_0, Ar_3) -> Com_1(f10(Fresh_1, Ar_3)) [ -Ar_0 >= 0 /\ 0 >= Ar_3 ] (Comp: ?, Cost: 1) f16(Ar_0, Ar_3) -> Com_1(f16(Ar_0, Ar_3)) [ -Ar_0 >= 0 /\ Ar_3 >= 1 ] start location: koat_start leaf cost: 0 Complexity upper bound ? Time: 0.984 sec (SMT: 0.942 sec)