MAYBE Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f12(Fresh_3, Fresh_3, Fresh_3, 0, Ar_4, Ar_5)) (Comp: ?, Cost: 1) f12(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f12(Ar_0, Ar_1, Ar_2, Ar_3 + 1, Fresh_2, Ar_5)) [ Ar_2 >= Ar_3 + 1 ] (Comp: ?, Cost: 1) f27(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f27(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Fresh_1)) [ 0 >= H + 1 ] (Comp: ?, Cost: 1) f27(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f27(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Fresh_0)) (Comp: ?, Cost: 1) f42(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f42(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ G >= H + 1 ] (Comp: ?, Cost: 1) f42(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f42(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) (Comp: ?, Cost: 1) f55(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f55(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ G >= H + 1 ] (Comp: ?, Cost: 1) f55(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f55(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) (Comp: ?, Cost: 1) f55(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f66(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) (Comp: ?, Cost: 1) f42(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f55(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) (Comp: ?, Cost: 1) f27(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f42(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) (Comp: ?, Cost: 1) f12(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f27(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ Ar_3 >= Ar_2 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Slicing away variables that do not contribute to conditions from problem 1 leaves variables [Ar_2, Ar_3]. We thus obtain the following problem: 2: T: (Comp: 1, Cost: 0) koat_start(Ar_2, Ar_3) -> Com_1(f0(Ar_2, Ar_3)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f12(Ar_2, Ar_3) -> Com_1(f27(Ar_2, Ar_3)) [ Ar_3 >= Ar_2 ] (Comp: ?, Cost: 1) f27(Ar_2, Ar_3) -> Com_1(f42(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f42(Ar_2, Ar_3) -> Com_1(f55(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f55(Ar_2, Ar_3) -> Com_1(f66(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f55(Ar_2, Ar_3) -> Com_1(f55(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f55(Ar_2, Ar_3) -> Com_1(f55(Ar_2, Ar_3)) [ G >= H + 1 ] (Comp: ?, Cost: 1) f42(Ar_2, Ar_3) -> Com_1(f42(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f42(Ar_2, Ar_3) -> Com_1(f42(Ar_2, Ar_3)) [ G >= H + 1 ] (Comp: ?, Cost: 1) f27(Ar_2, Ar_3) -> Com_1(f27(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f27(Ar_2, Ar_3) -> Com_1(f27(Ar_2, Ar_3)) [ 0 >= H + 1 ] (Comp: ?, Cost: 1) f12(Ar_2, Ar_3) -> Com_1(f12(Ar_2, Ar_3 + 1)) [ Ar_2 >= Ar_3 + 1 ] (Comp: ?, Cost: 1) f0(Ar_2, Ar_3) -> Com_1(f12(Fresh_3, 0)) 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_2, Ar_3) -> Com_1(f0(Ar_2, Ar_3)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f12(Ar_2, Ar_3) -> Com_1(f27(Ar_2, Ar_3)) [ Ar_3 >= Ar_2 ] (Comp: ?, Cost: 1) f27(Ar_2, Ar_3) -> Com_1(f42(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f42(Ar_2, Ar_3) -> Com_1(f55(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f55(Ar_2, Ar_3) -> Com_1(f66(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f55(Ar_2, Ar_3) -> Com_1(f55(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f55(Ar_2, Ar_3) -> Com_1(f55(Ar_2, Ar_3)) [ G >= H + 1 ] (Comp: ?, Cost: 1) f42(Ar_2, Ar_3) -> Com_1(f42(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f42(Ar_2, Ar_3) -> Com_1(f42(Ar_2, Ar_3)) [ G >= H + 1 ] (Comp: ?, Cost: 1) f27(Ar_2, Ar_3) -> Com_1(f27(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f27(Ar_2, Ar_3) -> Com_1(f27(Ar_2, Ar_3)) [ 0 >= H + 1 ] (Comp: ?, Cost: 1) f12(Ar_2, Ar_3) -> Com_1(f12(Ar_2, Ar_3 + 1)) [ Ar_2 >= Ar_3 + 1 ] (Comp: 1, Cost: 1) f0(Ar_2, Ar_3) -> Com_1(f12(Fresh_3, 0)) start location: koat_start leaf cost: 0 A polynomial rank function with Pol(koat_start) = 1 Pol(f0) = 1 Pol(f12) = 1 Pol(f27) = 0 Pol(f42) = -1 Pol(f55) = -2 Pol(f66) = -3 orients all transitions weakly and the transition f12(Ar_2, Ar_3) -> Com_1(f27(Ar_2, Ar_3)) [ Ar_3 >= Ar_2 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 0) koat_start(Ar_2, Ar_3) -> Com_1(f0(Ar_2, Ar_3)) [ 0 <= 0 ] (Comp: 1, Cost: 1) f12(Ar_2, Ar_3) -> Com_1(f27(Ar_2, Ar_3)) [ Ar_3 >= Ar_2 ] (Comp: ?, Cost: 1) f27(Ar_2, Ar_3) -> Com_1(f42(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f42(Ar_2, Ar_3) -> Com_1(f55(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f55(Ar_2, Ar_3) -> Com_1(f66(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f55(Ar_2, Ar_3) -> Com_1(f55(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f55(Ar_2, Ar_3) -> Com_1(f55(Ar_2, Ar_3)) [ G >= H + 1 ] (Comp: ?, Cost: 1) f42(Ar_2, Ar_3) -> Com_1(f42(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f42(Ar_2, Ar_3) -> Com_1(f42(Ar_2, Ar_3)) [ G >= H + 1 ] (Comp: ?, Cost: 1) f27(Ar_2, Ar_3) -> Com_1(f27(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f27(Ar_2, Ar_3) -> Com_1(f27(Ar_2, Ar_3)) [ 0 >= H + 1 ] (Comp: ?, Cost: 1) f12(Ar_2, Ar_3) -> Com_1(f12(Ar_2, Ar_3 + 1)) [ Ar_2 >= Ar_3 + 1 ] (Comp: 1, Cost: 1) f0(Ar_2, Ar_3) -> Com_1(f12(Fresh_3, 0)) start location: koat_start leaf cost: 0 A polynomial rank function with Pol(koat_start) = 3 Pol(f0) = 3 Pol(f12) = 3 Pol(f27) = 3 Pol(f42) = 2 Pol(f55) = 1 Pol(f66) = 0 orients all transitions weakly and the transitions f55(Ar_2, Ar_3) -> Com_1(f66(Ar_2, Ar_3)) f42(Ar_2, Ar_3) -> Com_1(f55(Ar_2, Ar_3)) f27(Ar_2, Ar_3) -> Com_1(f42(Ar_2, Ar_3)) strictly and produces the following problem: 5: T: (Comp: 1, Cost: 0) koat_start(Ar_2, Ar_3) -> Com_1(f0(Ar_2, Ar_3)) [ 0 <= 0 ] (Comp: 1, Cost: 1) f12(Ar_2, Ar_3) -> Com_1(f27(Ar_2, Ar_3)) [ Ar_3 >= Ar_2 ] (Comp: 3, Cost: 1) f27(Ar_2, Ar_3) -> Com_1(f42(Ar_2, Ar_3)) (Comp: 3, Cost: 1) f42(Ar_2, Ar_3) -> Com_1(f55(Ar_2, Ar_3)) (Comp: 3, Cost: 1) f55(Ar_2, Ar_3) -> Com_1(f66(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f55(Ar_2, Ar_3) -> Com_1(f55(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f55(Ar_2, Ar_3) -> Com_1(f55(Ar_2, Ar_3)) [ G >= H + 1 ] (Comp: ?, Cost: 1) f42(Ar_2, Ar_3) -> Com_1(f42(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f42(Ar_2, Ar_3) -> Com_1(f42(Ar_2, Ar_3)) [ G >= H + 1 ] (Comp: ?, Cost: 1) f27(Ar_2, Ar_3) -> Com_1(f27(Ar_2, Ar_3)) (Comp: ?, Cost: 1) f27(Ar_2, Ar_3) -> Com_1(f27(Ar_2, Ar_3)) [ 0 >= H + 1 ] (Comp: ?, Cost: 1) f12(Ar_2, Ar_3) -> Com_1(f12(Ar_2, Ar_3 + 1)) [ Ar_2 >= Ar_3 + 1 ] (Comp: 1, Cost: 1) f0(Ar_2, Ar_3) -> Com_1(f12(Fresh_3, 0)) start location: koat_start leaf cost: 0 Applied AI with 'oct' on problem 5 to obtain the following invariants: For symbol f12: X_2 >= 0 For symbol f27: X_2 >= 0 /\ -X_1 + X_2 >= 0 For symbol f42: X_2 >= 0 /\ -X_1 + X_2 >= 0 For symbol f55: X_2 >= 0 /\ -X_1 + X_2 >= 0 This yielded the following problem: 6: T: (Comp: 1, Cost: 1) f0(Ar_2, Ar_3) -> Com_1(f12(Fresh_3, 0)) (Comp: ?, Cost: 1) f12(Ar_2, Ar_3) -> Com_1(f12(Ar_2, Ar_3 + 1)) [ Ar_3 >= 0 /\ Ar_2 >= Ar_3 + 1 ] (Comp: ?, Cost: 1) f27(Ar_2, Ar_3) -> Com_1(f27(Ar_2, Ar_3)) [ Ar_3 >= 0 /\ -Ar_2 + Ar_3 >= 0 /\ 0 >= H + 1 ] (Comp: ?, Cost: 1) f27(Ar_2, Ar_3) -> Com_1(f27(Ar_2, Ar_3)) [ Ar_3 >= 0 /\ -Ar_2 + Ar_3 >= 0 ] (Comp: ?, Cost: 1) f42(Ar_2, Ar_3) -> Com_1(f42(Ar_2, Ar_3)) [ Ar_3 >= 0 /\ -Ar_2 + Ar_3 >= 0 /\ G >= H + 1 ] (Comp: ?, Cost: 1) f42(Ar_2, Ar_3) -> Com_1(f42(Ar_2, Ar_3)) [ Ar_3 >= 0 /\ -Ar_2 + Ar_3 >= 0 ] (Comp: ?, Cost: 1) f55(Ar_2, Ar_3) -> Com_1(f55(Ar_2, Ar_3)) [ Ar_3 >= 0 /\ -Ar_2 + Ar_3 >= 0 /\ G >= H + 1 ] (Comp: ?, Cost: 1) f55(Ar_2, Ar_3) -> Com_1(f55(Ar_2, Ar_3)) [ Ar_3 >= 0 /\ -Ar_2 + Ar_3 >= 0 ] (Comp: 3, Cost: 1) f55(Ar_2, Ar_3) -> Com_1(f66(Ar_2, Ar_3)) [ Ar_3 >= 0 /\ -Ar_2 + Ar_3 >= 0 ] (Comp: 3, Cost: 1) f42(Ar_2, Ar_3) -> Com_1(f55(Ar_2, Ar_3)) [ Ar_3 >= 0 /\ -Ar_2 + Ar_3 >= 0 ] (Comp: 3, Cost: 1) f27(Ar_2, Ar_3) -> Com_1(f42(Ar_2, Ar_3)) [ Ar_3 >= 0 /\ -Ar_2 + Ar_3 >= 0 ] (Comp: 1, Cost: 1) f12(Ar_2, Ar_3) -> Com_1(f27(Ar_2, Ar_3)) [ Ar_3 >= 0 /\ Ar_3 >= Ar_2 ] (Comp: 1, Cost: 0) koat_start(Ar_2, Ar_3) -> Com_1(f0(Ar_2, Ar_3)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound ? Time: 2.121 sec (SMT: 2.025 sec)