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, Ar_10) -> Com_1(f2(Ar_0, Ar_2, Ar_2, Fresh_22, Fresh_23, Fresh_24, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10)) [ Ar_0 >= 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, Ar_10) -> Com_1(f1(Ar_0, Fresh_15, Fresh_16, Fresh_17, Fresh_18, Ar_5, Ar_6, Fresh_15, Fresh_19, Fresh_20, Fresh_21)) [ 0 >= Ar_6 ] (Comp: ?, Cost: 1) f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10) -> Com_1(f2(Ar_0, Fresh_8, Fresh_8, Fresh_9, Fresh_10, Fresh_11, Ar_6, Fresh_12, Fresh_13, Fresh_14, Ar_10)) [ Ar_0 >= 1 /\ Ar_6 >= 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, Ar_10) -> Com_1(f1(Ar_0, Fresh_0, Fresh_0, Fresh_1, Fresh_2, Fresh_3, Ar_6, Fresh_4, Fresh_5, Fresh_6, Fresh_7)) [ 0 >= Ar_0 /\ Ar_6 >= 1 ] (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, Ar_10) -> Com_1(f300(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10)) [ 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_6]. We thus obtain the following problem: 2: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_6) -> Com_1(f300(Ar_0, Ar_6)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f300(Ar_0, Ar_6) -> Com_1(f1(Ar_0, Ar_6)) [ 0 >= Ar_0 /\ Ar_6 >= 1 ] (Comp: ?, Cost: 1) f300(Ar_0, Ar_6) -> Com_1(f2(Ar_0, Ar_6)) [ Ar_0 >= 1 /\ Ar_6 >= 1 ] (Comp: ?, Cost: 1) f300(Ar_0, Ar_6) -> Com_1(f1(Ar_0, Ar_6)) [ 0 >= Ar_6 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_6) -> Com_1(f2(Ar_0, Ar_6)) [ Ar_0 >= 1 ] 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_0, Ar_6) -> Com_1(f300(Ar_0, Ar_6)) [ 0 <= 0 ] (Comp: 1, Cost: 1) f300(Ar_0, Ar_6) -> Com_1(f1(Ar_0, Ar_6)) [ 0 >= Ar_0 /\ Ar_6 >= 1 ] (Comp: 1, Cost: 1) f300(Ar_0, Ar_6) -> Com_1(f2(Ar_0, Ar_6)) [ Ar_0 >= 1 /\ Ar_6 >= 1 ] (Comp: 1, Cost: 1) f300(Ar_0, Ar_6) -> Com_1(f1(Ar_0, Ar_6)) [ 0 >= Ar_6 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_6) -> Com_1(f2(Ar_0, Ar_6)) [ Ar_0 >= 1 ] start location: koat_start leaf cost: 0 Applied AI with 'oct' on problem 3 to obtain the following invariants: For symbol f2: X_2 - 1 >= 0 /\ X_1 + X_2 - 2 >= 0 /\ X_1 - 1 >= 0 This yielded the following problem: 4: T: (Comp: ?, Cost: 1) f2(Ar_0, Ar_6) -> Com_1(f2(Ar_0, Ar_6)) [ Ar_6 - 1 >= 0 /\ Ar_0 + Ar_6 - 2 >= 0 /\ Ar_0 - 1 >= 0 /\ Ar_0 >= 1 ] (Comp: 1, Cost: 1) f300(Ar_0, Ar_6) -> Com_1(f1(Ar_0, Ar_6)) [ 0 >= Ar_6 ] (Comp: 1, Cost: 1) f300(Ar_0, Ar_6) -> Com_1(f2(Ar_0, Ar_6)) [ Ar_0 >= 1 /\ Ar_6 >= 1 ] (Comp: 1, Cost: 1) f300(Ar_0, Ar_6) -> Com_1(f1(Ar_0, Ar_6)) [ 0 >= Ar_0 /\ Ar_6 >= 1 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_6) -> Com_1(f300(Ar_0, Ar_6)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound ? Time: 0.595 sec (SMT: 0.564 sec)