MAYBE Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f11(Ar_0 - 1, Ar_1 - 1, Ar_2 + 1, Fresh_2, Ar_4, Ar_5)) [ Ar_0 >= 1 /\ Fresh_2 >= 1 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f11(Ar_0 - 1, Ar_1, Ar_2, Fresh_1, Ar_4, Ar_5)) [ 0 >= Fresh_1 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ] (Comp: ?, Cost: 1) f21(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f21(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) (Comp: ?, Cost: 1) f23(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f26(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) (Comp: ?, Cost: 1) f11(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f21(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3, Ar_4, Ar_5) -> Com_1(f11(4, Fresh_0, 0, Ar_3, Fresh_0, 4)) [ Fresh_0 >= 1 ] (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_0, Ar_1]. We thus obtain the following problem: 2: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f0(Ar_0, Ar_1) -> Com_1(f11(4, Fresh_0)) [ Fresh_0 >= 1 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) f23(Ar_0, Ar_1) -> Com_1(f26(Ar_0, Ar_1)) (Comp: ?, Cost: 1) f21(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) (Comp: ?, Cost: 1) f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1)) [ 0 >= Fresh_1 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1 - 1)) [ Ar_0 >= 1 /\ Fresh_2 >= 1 ] start location: koat_start leaf cost: 0 Testing for reachability in the complexity graph removes the following transition from problem 2: f23(Ar_0, Ar_1) -> Com_1(f26(Ar_0, Ar_1)) We thus obtain the following problem: 3: T: (Comp: ?, Cost: 1) f21(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) (Comp: ?, Cost: 1) f11(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1 - 1)) [ Ar_0 >= 1 /\ Fresh_2 >= 1 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1)) [ 0 >= Fresh_1 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ] (Comp: ?, Cost: 1) f0(Ar_0, Ar_1) -> Com_1(f11(4, Fresh_0)) [ Fresh_0 >= 1 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 3 produces the following problem: 4: T: (Comp: ?, Cost: 1) f21(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) (Comp: ?, Cost: 1) f11(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1 - 1)) [ Ar_0 >= 1 /\ Fresh_2 >= 1 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1)) [ 0 >= Fresh_1 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1) -> Com_1(f11(4, Fresh_0)) [ Fresh_0 >= 1 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f21) = 0 Pol(f11) = 1 Pol(f0) = 1 Pol(koat_start) = 1 orients all transitions weakly and the transition f11(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ 0 >= Ar_0 ] strictly and produces the following problem: 5: T: (Comp: ?, Cost: 1) f21(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) (Comp: 1, Cost: 1) f11(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ 0 >= Ar_0 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1 - 1)) [ Ar_0 >= 1 /\ Fresh_2 >= 1 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1)) [ 0 >= Fresh_1 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1) -> Com_1(f11(4, Fresh_0)) [ Fresh_0 >= 1 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f21) = V_1 Pol(f11) = V_1 Pol(f0) = 4 Pol(koat_start) = 4 orients all transitions weakly and the transition f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1 - 1)) [ Ar_0 >= 1 /\ Fresh_2 >= 1 ] strictly and produces the following problem: 6: T: (Comp: ?, Cost: 1) f21(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) (Comp: 1, Cost: 1) f11(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ 0 >= Ar_0 ] (Comp: 4, Cost: 1) f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1 - 1)) [ Ar_0 >= 1 /\ Fresh_2 >= 1 ] (Comp: ?, Cost: 1) f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1)) [ 0 >= Fresh_1 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1) -> Com_1(f11(4, Fresh_0)) [ Fresh_0 >= 1 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f21) = V_1 Pol(f11) = V_1 Pol(f0) = 4 Pol(koat_start) = 4 orients all transitions weakly and the transition f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1)) [ 0 >= Fresh_1 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ] strictly and produces the following problem: 7: T: (Comp: ?, Cost: 1) f21(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) (Comp: 1, Cost: 1) f11(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ 0 >= Ar_0 ] (Comp: 4, Cost: 1) f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1 - 1)) [ Ar_0 >= 1 /\ Fresh_2 >= 1 ] (Comp: 4, Cost: 1) f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1)) [ 0 >= Fresh_1 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1) -> Com_1(f11(4, Fresh_0)) [ Fresh_0 >= 1 ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Applied AI with 'oct' on problem 7 to obtain the following invariants: For symbol f11: -X_1 + X_2 + 3 >= 0 /\ -X_1 + 4 >= 0 For symbol f21: -X_1 + X_2 + 3 >= 0 /\ -X_1 >= 0 This yielded the following problem: 8: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f0(Ar_0, Ar_1)) [ 0 <= 0 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1) -> Com_1(f11(4, Fresh_0)) [ Fresh_0 >= 1 ] (Comp: 4, Cost: 1) f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1)) [ -Ar_0 + Ar_1 + 3 >= 0 /\ -Ar_0 + 4 >= 0 /\ 0 >= Fresh_1 /\ Ar_0 >= 1 /\ Ar_0 >= Ar_1 + 1 ] (Comp: 4, Cost: 1) f11(Ar_0, Ar_1) -> Com_1(f11(Ar_0 - 1, Ar_1 - 1)) [ -Ar_0 + Ar_1 + 3 >= 0 /\ -Ar_0 + 4 >= 0 /\ Ar_0 >= 1 /\ Fresh_2 >= 1 ] (Comp: 1, Cost: 1) f11(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ -Ar_0 + Ar_1 + 3 >= 0 /\ -Ar_0 + 4 >= 0 /\ 0 >= Ar_0 ] (Comp: ?, Cost: 1) f21(Ar_0, Ar_1) -> Com_1(f21(Ar_0, Ar_1)) [ -Ar_0 + Ar_1 + 3 >= 0 /\ -Ar_0 >= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound ? Time: 1.945 sec (SMT: 1.880 sec)