MAYBE Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f49(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14, Ar_15, Ar_16, Ar_17, Ar_18, Ar_19) -> Com_1(f49(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14, Ar_15, Ar_16, Ar_17, Ar_18, Ar_19)) (Comp: ?, Cost: 1) f51(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14, Ar_15, Ar_16, Ar_17, Ar_18, Ar_19) -> Com_1(f54(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14, Ar_15, Ar_16, Ar_17, Ar_18, Ar_19)) (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14, Ar_15, Ar_16, Ar_17, Ar_18, Ar_19) -> Com_1(f49(Ar_0, Ar_1, 0, 0, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14, Ar_15, Ar_16, Ar_17, Ar_18, Ar_19)) [ Ar_0 >= Ar_1 ] (Comp: ?, Cost: 1) f35(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14, Ar_15, Ar_16, Ar_17, Ar_18, Ar_19) -> Com_1(f49(Ar_0, Ar_1, 0, 0, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14, Ar_15, Ar_16, Ar_17, Ar_18, Ar_19)) [ Ar_4 >= 3 ] (Comp: ?, Cost: 1) f35(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14, Ar_15, Ar_16, Ar_17, Ar_18, Ar_19) -> Com_1(f49(Ar_0, Ar_1, 0, 0, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14, Ar_15, Ar_16, Ar_17, Ar_18, Ar_19)) [ 1 >= Ar_4 ] (Comp: ?, Cost: 1) f35(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14, Ar_15, Ar_16, Ar_17, Ar_18, Ar_19) -> Com_1(f49(Ar_0, Ar_1, 0, 0, 2, Ar_6, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14, Ar_15, Ar_16, Ar_17, Ar_18, Ar_19)) [ Ar_4 = 2 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14, Ar_15, Ar_16, Ar_17, Ar_18, Ar_19) -> Com_1(f49(Ar_0, Ar_1, 0, 0, Ar_4, Ar_5, Fresh_14, Fresh_15, Fresh_16, Fresh_17, Ar_3, Fresh_14, Fresh_14, Ar_13, Ar_14, Ar_15, Ar_16, Ar_17, Ar_18, Ar_19)) [ Fresh_14 >= 1 /\ Ar_1 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14, Ar_15, Ar_16, Ar_17, Ar_18, Ar_19) -> Com_1(f35(Ar_0, Ar_1, Ar_2, Ar_3, Fresh_9, Ar_5, Fresh_10, Fresh_11, Fresh_12, Fresh_13, Ar_3, Fresh_10, Fresh_10, Fresh_10, Ar_15, 0, Fresh_9, Fresh_9, 0, Ar_19)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_10 /\ Fresh_9 >= 2 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14, Ar_15, Ar_16, Ar_17, Ar_18, Ar_19) -> Com_1(f35(Ar_0, Ar_1, Ar_2, Ar_3, Fresh_4, Ar_5, Fresh_5, Fresh_6, Fresh_7, Fresh_8, Ar_3, Fresh_5, Fresh_5, Fresh_5, Ar_15, 0, Fresh_4, Fresh_4, 0, Ar_19)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_5 /\ 0 >= Fresh_4 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14, Ar_15, Ar_16, Ar_17, Ar_18, Ar_19) -> Com_1(f11(Ar_0 + 1, Ar_1, Ar_2, Ar_3, 1, Ar_5, Fresh_0, Fresh_1, Fresh_2, Fresh_3, Ar_3, Fresh_0, Fresh_0, Fresh_0, Ar_15, Ar_15, 1, 1, 0, Ar_19)) [ 0 >= Fresh_0 /\ Ar_1 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14, Ar_15, Ar_16, Ar_17, Ar_18, Ar_19) -> Com_1(f11(Ar_0, Ar_1, 0, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14, Ar_15, Ar_16, Ar_17, Ar_18, 0)) (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, Ar_11, Ar_12, Ar_13, Ar_14, Ar_15, Ar_16, Ar_17, Ar_18, Ar_19) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5, Ar_6, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14, Ar_15, Ar_16, Ar_17, Ar_18, Ar_19)) [ 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_4]. We thus obtain the following problem: 2: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_4) -> Com_1(f0(Ar_0, Ar_1, Ar_4)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f0(Ar_0, Ar_1, Ar_4) -> Com_1(f11(Ar_0, Ar_1, Ar_4)) (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f11(Ar_0 + 1, Ar_1, 1)) [ 0 >= Fresh_0 /\ Ar_1 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f35(Ar_0, Ar_1, Fresh_4)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_5 /\ 0 >= Fresh_4 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f35(Ar_0, Ar_1, Fresh_9)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_10 /\ Fresh_9 >= 2 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ Fresh_14 >= 1 /\ Ar_1 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) f35(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, 2)) [ Ar_4 = 2 ] (Comp: ?, Cost: 1) f35(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ 1 >= Ar_4 ] (Comp: ?, Cost: 1) f35(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ Ar_4 >= 3 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ Ar_0 >= Ar_1 ] (Comp: ?, Cost: 1) f51(Ar_0, Ar_1, Ar_4) -> Com_1(f54(Ar_0, Ar_1, Ar_4)) (Comp: ?, Cost: 1) f49(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) start location: koat_start leaf cost: 0 Testing for reachability in the complexity graph removes the following transition from problem 2: f51(Ar_0, Ar_1, Ar_4) -> Com_1(f54(Ar_0, Ar_1, Ar_4)) We thus obtain the following problem: 3: T: (Comp: ?, Cost: 1) f49(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) (Comp: ?, Cost: 1) f35(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ Ar_4 >= 3 ] (Comp: ?, Cost: 1) f35(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, 2)) [ Ar_4 = 2 ] (Comp: ?, Cost: 1) f35(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ 1 >= Ar_4 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ Ar_0 >= Ar_1 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ Fresh_14 >= 1 /\ Ar_1 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f35(Ar_0, Ar_1, Fresh_9)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_10 /\ Fresh_9 >= 2 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f35(Ar_0, Ar_1, Fresh_4)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_5 /\ 0 >= Fresh_4 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f11(Ar_0 + 1, Ar_1, 1)) [ 0 >= Fresh_0 /\ Ar_1 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) f0(Ar_0, Ar_1, Ar_4) -> Com_1(f11(Ar_0, Ar_1, Ar_4)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_4) -> Com_1(f0(Ar_0, Ar_1, Ar_4)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 3 produces the following problem: 4: T: (Comp: ?, Cost: 1) f49(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) (Comp: ?, Cost: 1) f35(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ Ar_4 >= 3 ] (Comp: ?, Cost: 1) f35(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, 2)) [ Ar_4 = 2 ] (Comp: ?, Cost: 1) f35(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ 1 >= Ar_4 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ Ar_0 >= Ar_1 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ Fresh_14 >= 1 /\ Ar_1 >= Ar_0 + 1 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f35(Ar_0, Ar_1, Fresh_9)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_10 /\ Fresh_9 >= 2 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f35(Ar_0, Ar_1, Fresh_4)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_5 /\ 0 >= Fresh_4 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f11(Ar_0 + 1, Ar_1, 1)) [ 0 >= Fresh_0 /\ Ar_1 >= Ar_0 + 1 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_4) -> Com_1(f11(Ar_0, Ar_1, Ar_4)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_4) -> Com_1(f0(Ar_0, Ar_1, Ar_4)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f49) = 0 Pol(f35) = 1 Pol(f11) = 2 Pol(f0) = 2 Pol(koat_start) = 2 orients all transitions weakly and the transitions f35(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ 1 >= Ar_4 ] f35(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ Ar_4 >= 3 ] f11(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ Fresh_14 >= 1 /\ Ar_1 >= Ar_0 + 1 ] f11(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ Ar_0 >= Ar_1 ] f11(Ar_0, Ar_1, Ar_4) -> Com_1(f35(Ar_0, Ar_1, Fresh_9)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_10 /\ Fresh_9 >= 2 ] f11(Ar_0, Ar_1, Ar_4) -> Com_1(f35(Ar_0, Ar_1, Fresh_4)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_5 /\ 0 >= Fresh_4 ] strictly and produces the following problem: 5: T: (Comp: ?, Cost: 1) f49(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) (Comp: 2, Cost: 1) f35(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ Ar_4 >= 3 ] (Comp: ?, Cost: 1) f35(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, 2)) [ Ar_4 = 2 ] (Comp: 2, Cost: 1) f35(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ 1 >= Ar_4 ] (Comp: 2, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ Ar_0 >= Ar_1 ] (Comp: 2, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ Fresh_14 >= 1 /\ Ar_1 >= Ar_0 + 1 ] (Comp: 2, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f35(Ar_0, Ar_1, Fresh_9)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_10 /\ Fresh_9 >= 2 ] (Comp: 2, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f35(Ar_0, Ar_1, Fresh_4)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_5 /\ 0 >= Fresh_4 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f11(Ar_0 + 1, Ar_1, 1)) [ 0 >= Fresh_0 /\ Ar_1 >= Ar_0 + 1 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_4) -> Com_1(f11(Ar_0, Ar_1, Ar_4)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_4) -> Com_1(f0(Ar_0, Ar_1, Ar_4)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 5 produces the following problem: 6: T: (Comp: ?, Cost: 1) f49(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) (Comp: 2, Cost: 1) f35(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ Ar_4 >= 3 ] (Comp: 2, Cost: 1) f35(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, 2)) [ Ar_4 = 2 ] (Comp: 2, Cost: 1) f35(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ 1 >= Ar_4 ] (Comp: 2, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ Ar_0 >= Ar_1 ] (Comp: 2, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ Fresh_14 >= 1 /\ Ar_1 >= Ar_0 + 1 ] (Comp: 2, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f35(Ar_0, Ar_1, Fresh_9)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_10 /\ Fresh_9 >= 2 ] (Comp: 2, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f35(Ar_0, Ar_1, Fresh_4)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_5 /\ 0 >= Fresh_4 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f11(Ar_0 + 1, Ar_1, 1)) [ 0 >= Fresh_0 /\ Ar_1 >= Ar_0 + 1 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_4) -> Com_1(f11(Ar_0, Ar_1, Ar_4)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_4) -> Com_1(f0(Ar_0, Ar_1, Ar_4)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f49) = -V_1 + V_2 Pol(f35) = -V_1 + V_2 Pol(f11) = -V_1 + V_2 Pol(f0) = -V_1 + V_2 Pol(koat_start) = -V_1 + V_2 orients all transitions weakly and the transition f11(Ar_0, Ar_1, Ar_4) -> Com_1(f11(Ar_0 + 1, Ar_1, 1)) [ 0 >= Fresh_0 /\ Ar_1 >= Ar_0 + 1 ] strictly and produces the following problem: 7: T: (Comp: ?, Cost: 1) f49(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) (Comp: 2, Cost: 1) f35(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ Ar_4 >= 3 ] (Comp: 2, Cost: 1) f35(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, 2)) [ Ar_4 = 2 ] (Comp: 2, Cost: 1) f35(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ 1 >= Ar_4 ] (Comp: 2, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ Ar_0 >= Ar_1 ] (Comp: 2, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ Fresh_14 >= 1 /\ Ar_1 >= Ar_0 + 1 ] (Comp: 2, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f35(Ar_0, Ar_1, Fresh_9)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_10 /\ Fresh_9 >= 2 ] (Comp: 2, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f35(Ar_0, Ar_1, Fresh_4)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_5 /\ 0 >= Fresh_4 ] (Comp: Ar_0 + Ar_1, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f11(Ar_0 + 1, Ar_1, 1)) [ 0 >= Fresh_0 /\ Ar_1 >= Ar_0 + 1 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_4) -> Com_1(f11(Ar_0, Ar_1, Ar_4)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_4) -> Com_1(f0(Ar_0, Ar_1, Ar_4)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Applied AI with 'oct' on problem 7 to obtain the following invariants: For symbol f35: -X_1 + X_2 - 1 >= 0 This yielded the following problem: 8: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_4) -> Com_1(f0(Ar_0, Ar_1, Ar_4)) [ 0 <= 0 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_4) -> Com_1(f11(Ar_0, Ar_1, Ar_4)) (Comp: Ar_0 + Ar_1, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f11(Ar_0 + 1, Ar_1, 1)) [ 0 >= Fresh_0 /\ Ar_1 >= Ar_0 + 1 ] (Comp: 2, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f35(Ar_0, Ar_1, Fresh_4)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_5 /\ 0 >= Fresh_4 ] (Comp: 2, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f35(Ar_0, Ar_1, Fresh_9)) [ Ar_1 >= Ar_0 + 1 /\ 0 >= Fresh_10 /\ Fresh_9 >= 2 ] (Comp: 2, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ Fresh_14 >= 1 /\ Ar_1 >= Ar_0 + 1 ] (Comp: 2, Cost: 1) f11(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ Ar_0 >= Ar_1 ] (Comp: 2, Cost: 1) f35(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ -Ar_0 + Ar_1 - 1 >= 0 /\ 1 >= Ar_4 ] (Comp: 2, Cost: 1) f35(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, 2)) [ -Ar_0 + Ar_1 - 1 >= 0 /\ Ar_4 = 2 ] (Comp: 2, Cost: 1) f35(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) [ -Ar_0 + Ar_1 - 1 >= 0 /\ Ar_4 >= 3 ] (Comp: ?, Cost: 1) f49(Ar_0, Ar_1, Ar_4) -> Com_1(f49(Ar_0, Ar_1, Ar_4)) start location: koat_start leaf cost: 0 Complexity upper bound ? Time: 2.272 sec (SMT: 2.191 sec)