MAYBE Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Fresh_2, 0, Ar_2, Ar_3)) (Comp: ?, Cost: 1) f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f10(Ar_0 - 1, Ar_1 + 1, Ar_2, Ar_3)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) f10(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Ar_0, Ar_1 - 1, Ar_0 - 1, Ar_3)) [ Ar_1 >= 1 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Ar_0, Ar_1, Ar_2 - 1, 0)) [ Ar_2 >= 1 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Ar_0 - 1, Ar_1 + 1, Ar_2 - 1, Fresh_1)) [ Ar_2 >= 1 /\ 0 >= Fresh_1 + 1 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Ar_0 - 1, Ar_1 + 1, Ar_2 - 1, Fresh_0)) [ Ar_2 >= 1 /\ Fresh_0 >= 1 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f10(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 >= Ar_2 ] (Comp: ?, Cost: 1) f10(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f6(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 >= Ar_1 ] (Comp: ?, Cost: 1) f6(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f25(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 >= Ar_0 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 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_1, Ar_2]. We thus obtain the following problem: 2: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f6(Ar_0, Ar_1, Ar_2) -> Com_1(f25(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) f10(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_1 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 - 1, Ar_1 + 1, Ar_2 - 1)) [ Ar_2 >= 1 /\ Fresh_0 >= 1 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 - 1, Ar_1 + 1, Ar_2 - 1)) [ Ar_2 >= 1 /\ 0 >= Fresh_1 + 1 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2 - 1)) [ Ar_2 >= 1 ] (Comp: ?, Cost: 1) f10(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1 - 1, Ar_0 - 1)) [ Ar_1 >= 1 ] (Comp: ?, Cost: 1) f6(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0 - 1, Ar_1 + 1, Ar_2)) [ Ar_0 >= 1 ] (Comp: ?, Cost: 1) f0(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Fresh_2, 0, Ar_2)) 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_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f6(Ar_0, Ar_1, Ar_2) -> Com_1(f25(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) f10(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_1 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 - 1, Ar_1 + 1, Ar_2 - 1)) [ Ar_2 >= 1 /\ Fresh_0 >= 1 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 - 1, Ar_1 + 1, Ar_2 - 1)) [ Ar_2 >= 1 /\ 0 >= Fresh_1 + 1 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2 - 1)) [ Ar_2 >= 1 ] (Comp: ?, Cost: 1) f10(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1 - 1, Ar_0 - 1)) [ Ar_1 >= 1 ] (Comp: ?, Cost: 1) f6(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0 - 1, Ar_1 + 1, Ar_2)) [ Ar_0 >= 1 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Fresh_2, 0, Ar_2)) start location: koat_start leaf cost: 0 A polynomial rank function with Pol(koat_start) = 1 Pol(f0) = 1 Pol(f6) = 1 Pol(f25) = 0 Pol(f10) = 1 Pol(f14) = 1 orients all transitions weakly and the transition f6(Ar_0, Ar_1, Ar_2) -> Com_1(f25(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_0 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] (Comp: 1, Cost: 1) f6(Ar_0, Ar_1, Ar_2) -> Com_1(f25(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) f10(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_1 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ 0 >= Ar_2 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 - 1, Ar_1 + 1, Ar_2 - 1)) [ Ar_2 >= 1 /\ Fresh_0 >= 1 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 - 1, Ar_1 + 1, Ar_2 - 1)) [ Ar_2 >= 1 /\ 0 >= Fresh_1 + 1 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2 - 1)) [ Ar_2 >= 1 ] (Comp: ?, Cost: 1) f10(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1 - 1, Ar_0 - 1)) [ Ar_1 >= 1 ] (Comp: ?, Cost: 1) f6(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0 - 1, Ar_1 + 1, Ar_2)) [ Ar_0 >= 1 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Fresh_2, 0, Ar_2)) start location: koat_start leaf cost: 0 Applied AI with 'oct' on problem 4 to obtain the following invariants: For symbol f10: X_1 >= 0 For symbol f14: X_1 - X_3 - 1 >= 0 /\ X_3 + 1 >= 0 /\ X_2 + X_3 + 1 >= 0 /\ X_1 + X_3 + 1 >= 0 /\ X_2 >= 0 /\ X_1 + X_2 >= 0 /\ X_1 >= 0 For symbol f6: -X_2 >= 0 This yielded the following problem: 5: T: (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Fresh_2, 0, Ar_2)) (Comp: ?, Cost: 1) f6(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0 - 1, Ar_1 + 1, Ar_2)) [ -Ar_1 >= 0 /\ Ar_0 >= 1 ] (Comp: ?, Cost: 1) f10(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1 - 1, Ar_0 - 1)) [ Ar_0 >= 0 /\ Ar_1 >= 1 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0, Ar_1, Ar_2 - 1)) [ Ar_0 - Ar_2 - 1 >= 0 /\ Ar_2 + 1 >= 0 /\ Ar_1 + Ar_2 + 1 >= 0 /\ Ar_0 + Ar_2 + 1 >= 0 /\ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 /\ Ar_2 >= 1 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 - 1, Ar_1 + 1, Ar_2 - 1)) [ Ar_0 - Ar_2 - 1 >= 0 /\ Ar_2 + 1 >= 0 /\ Ar_1 + Ar_2 + 1 >= 0 /\ Ar_0 + Ar_2 + 1 >= 0 /\ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 /\ Ar_2 >= 1 /\ 0 >= Fresh_1 + 1 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2) -> Com_1(f14(Ar_0 - 1, Ar_1 + 1, Ar_2 - 1)) [ Ar_0 - Ar_2 - 1 >= 0 /\ Ar_2 + 1 >= 0 /\ Ar_1 + Ar_2 + 1 >= 0 /\ Ar_0 + Ar_2 + 1 >= 0 /\ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 /\ Ar_2 >= 1 /\ Fresh_0 >= 1 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2) -> Com_1(f10(Ar_0, Ar_1, Ar_2)) [ Ar_0 - Ar_2 - 1 >= 0 /\ Ar_2 + 1 >= 0 /\ Ar_1 + Ar_2 + 1 >= 0 /\ Ar_0 + Ar_2 + 1 >= 0 /\ Ar_1 >= 0 /\ Ar_0 + Ar_1 >= 0 /\ Ar_0 >= 0 /\ 0 >= Ar_2 ] (Comp: ?, Cost: 1) f10(Ar_0, Ar_1, Ar_2) -> Com_1(f6(Ar_0, Ar_1, Ar_2)) [ Ar_0 >= 0 /\ 0 >= Ar_1 ] (Comp: 1, Cost: 1) f6(Ar_0, Ar_1, Ar_2) -> Com_1(f25(Ar_0, Ar_1, Ar_2)) [ -Ar_1 >= 0 /\ 0 >= Ar_0 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f0(Ar_0, Ar_1, Ar_2)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound ? Time: 4.023 sec (SMT: 3.923 sec)