MAYBE Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f1(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) -> Com_1(f1(Ar_0, Ar_1 + 1, Ar_3, Fresh_38, Ar_3, Fresh_39, Ar_1, Ar_7, Ar_8, Ar_9, Ar_10, Ar_11, Ar_12, Ar_13, Ar_14, Ar_15)) [ Ar_0 >= Ar_1 + 1 /\ Ar_1 >= 0 ] (Comp: ?, Cost: 1) f1(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) -> Com_1(f4(Fresh_28, Fresh_29, Fresh_30, Fresh_31, Fresh_32, Ar_5, Ar_6, Fresh_33, Fresh_34, Fresh_35, Fresh_36, Fresh_37, Ar_2, Ar_13, Ar_14, Ar_15)) [ Ar_1 >= Ar_0 /\ Ar_1 >= 0 /\ Fresh_29 >= Fresh_33 /\ Fresh_33 >= 2 ] (Comp: ?, Cost: 1) f3(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) -> Com_1(f4(Fresh_17, Fresh_18, Fresh_19, Fresh_20, Fresh_21, Ar_5, Ar_6, Fresh_22, Fresh_23, Fresh_24, Fresh_25, Fresh_26, Ar_14, Fresh_27, Ar_14, Ar_15)) [ 0 >= Fresh_22 /\ 0 >= A1 ] (Comp: ?, Cost: 1) f3(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) -> Com_1(f1(Fresh_11, 2, Fresh_12, Fresh_13, Fresh_12, Ar_5, Ar_6, Fresh_11, Fresh_14, Fresh_12, Ar_10, Ar_11, Ar_12, Fresh_15, Fresh_14, Fresh_16)) [ Fresh_11 >= 2 ] (Comp: ?, Cost: 1) f3(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) -> Com_1(f4(Fresh_0, Fresh_1, Fresh_2, Fresh_3, Fresh_4, Ar_5, Ar_6, 1, Fresh_5, Fresh_6, Fresh_7, Fresh_8, Ar_3, Fresh_9, Fresh_10, Ar_15)) (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) -> Com_1(f3(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)) [ 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]. We thus obtain the following problem: 2: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_1)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f3(Ar_0, Ar_1) -> Com_1(f4(Fresh_0, Fresh_1)) (Comp: ?, Cost: 1) f3(Ar_0, Ar_1) -> Com_1(f1(Fresh_11, 2)) [ Fresh_11 >= 2 ] (Comp: ?, Cost: 1) f3(Ar_0, Ar_1) -> Com_1(f4(Fresh_17, Fresh_18)) [ 0 >= Fresh_22 /\ 0 >= A1 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f4(Fresh_28, Fresh_29)) [ Ar_1 >= Ar_0 /\ Ar_1 >= 0 /\ Fresh_29 >= Fresh_33 /\ Fresh_33 >= 2 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1 + 1)) [ Ar_0 >= Ar_1 + 1 /\ Ar_1 >= 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_0, Ar_1) -> Com_1(f3(Ar_0, Ar_1)) [ 0 <= 0 ] (Comp: 1, Cost: 1) f3(Ar_0, Ar_1) -> Com_1(f4(Fresh_0, Fresh_1)) (Comp: 1, Cost: 1) f3(Ar_0, Ar_1) -> Com_1(f1(Fresh_11, 2)) [ Fresh_11 >= 2 ] (Comp: 1, Cost: 1) f3(Ar_0, Ar_1) -> Com_1(f4(Fresh_17, Fresh_18)) [ 0 >= Fresh_22 /\ 0 >= A1 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f4(Fresh_28, Fresh_29)) [ Ar_1 >= Ar_0 /\ Ar_1 >= 0 /\ Fresh_29 >= Fresh_33 /\ Fresh_33 >= 2 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1 + 1)) [ Ar_0 >= Ar_1 + 1 /\ Ar_1 >= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(koat_start) = 1 Pol(f3) = 1 Pol(f4) = 0 Pol(f1) = 1 orients all transitions weakly and the transition f1(Ar_0, Ar_1) -> Com_1(f4(Fresh_28, Fresh_29)) [ Ar_1 >= Ar_0 /\ Ar_1 >= 0 /\ Fresh_29 >= Fresh_33 /\ Fresh_33 >= 2 ] strictly and produces the following problem: 4: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_1)) [ 0 <= 0 ] (Comp: 1, Cost: 1) f3(Ar_0, Ar_1) -> Com_1(f4(Fresh_0, Fresh_1)) (Comp: 1, Cost: 1) f3(Ar_0, Ar_1) -> Com_1(f1(Fresh_11, 2)) [ Fresh_11 >= 2 ] (Comp: 1, Cost: 1) f3(Ar_0, Ar_1) -> Com_1(f4(Fresh_17, Fresh_18)) [ 0 >= Fresh_22 /\ 0 >= A1 ] (Comp: 1, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f4(Fresh_28, Fresh_29)) [ Ar_1 >= Ar_0 /\ Ar_1 >= 0 /\ Fresh_29 >= Fresh_33 /\ Fresh_33 >= 2 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1 + 1)) [ Ar_0 >= Ar_1 + 1 /\ Ar_1 >= 0 ] start location: koat_start leaf cost: 0 Applied AI with 'oct' on problem 4 to obtain the following invariants: For symbol f1: X_1 - X_2 >= 0 /\ X_2 - 2 >= 0 /\ X_1 + X_2 - 4 >= 0 /\ X_1 - 2 >= 0 This yielded the following problem: 5: T: (Comp: ?, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1 + 1)) [ Ar_0 - Ar_1 >= 0 /\ Ar_1 - 2 >= 0 /\ Ar_0 + Ar_1 - 4 >= 0 /\ Ar_0 - 2 >= 0 /\ Ar_0 >= Ar_1 + 1 /\ Ar_1 >= 0 ] (Comp: 1, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f4(Fresh_28, Fresh_29)) [ Ar_0 - Ar_1 >= 0 /\ Ar_1 - 2 >= 0 /\ Ar_0 + Ar_1 - 4 >= 0 /\ Ar_0 - 2 >= 0 /\ Ar_1 >= Ar_0 /\ Ar_1 >= 0 /\ Fresh_29 >= Fresh_33 /\ Fresh_33 >= 2 ] (Comp: 1, Cost: 1) f3(Ar_0, Ar_1) -> Com_1(f4(Fresh_17, Fresh_18)) [ 0 >= Fresh_22 /\ 0 >= A1 ] (Comp: 1, Cost: 1) f3(Ar_0, Ar_1) -> Com_1(f1(Fresh_11, 2)) [ Fresh_11 >= 2 ] (Comp: 1, Cost: 1) f3(Ar_0, Ar_1) -> Com_1(f4(Fresh_0, Fresh_1)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f3(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound ? Time: 1.366 sec (SMT: 1.321 sec)