MAYBE Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9) -> Com_1(f300(Fresh_6, Fresh_7, Fresh_8, Fresh_9, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9)) (Comp: ?, Cost: 1) f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9) -> Com_1(f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Fresh_3, Fresh_4, Fresh_5, Ar_9)) [ Ar_5 >= Ar_4 + 1 ] (Comp: ?, Cost: 1) f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9) -> Com_1(f1(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Fresh_0, Fresh_1, Ar_8, Fresh_2)) [ Ar_4 >= Ar_5 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9) -> Com_1(f2(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Slicing away variables that do not contribute to conditions from problem 1 leaves variables [Ar_4, Ar_5]. We thus obtain the following problem: 2: T: (Comp: 1, Cost: 0) koat_start(Ar_4, Ar_5) -> Com_1(f2(Ar_4, Ar_5)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f300(Ar_4, Ar_5) -> Com_1(f1(Ar_4, Ar_5)) [ Ar_4 >= Ar_5 ] (Comp: ?, Cost: 1) f300(Ar_4, Ar_5) -> Com_1(f300(Ar_4, Ar_5)) [ Ar_5 >= Ar_4 + 1 ] (Comp: ?, Cost: 1) f2(Ar_4, Ar_5) -> Com_1(f300(Ar_4, Ar_5)) 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_4, Ar_5) -> Com_1(f2(Ar_4, Ar_5)) [ 0 <= 0 ] (Comp: 1, Cost: 1) f300(Ar_4, Ar_5) -> Com_1(f1(Ar_4, Ar_5)) [ Ar_4 >= Ar_5 ] (Comp: ?, Cost: 1) f300(Ar_4, Ar_5) -> Com_1(f300(Ar_4, Ar_5)) [ Ar_5 >= Ar_4 + 1 ] (Comp: 1, Cost: 1) f2(Ar_4, Ar_5) -> Com_1(f300(Ar_4, Ar_5)) start location: koat_start leaf cost: 0 Complexity upper bound ? Time: 0.385 sec (SMT: 0.361 sec)