MAYBE Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2) -> Com_1(f300(Ar_0, Ar_1, Fresh_0)) [ Ar_0 >= 3 /\ Ar_1 >= 2 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(1, 2, Ar_2)) [ 1 >= D /\ Ar_1 = 1 /\ Ar_0 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 + 1, 2, Ar_2)) [ Ar_0 >= 2 /\ 1 >= D /\ Ar_1 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 + 1, 2, Ar_2)) [ 0 >= Ar_0 /\ 1 >= D /\ Ar_1 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(1, Ar_1 + 1, Ar_2)) [ Ar_1 >= 2 /\ 1 >= Ar_1 /\ Ar_0 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(1, Ar_1 + 1, Ar_2)) [ 0 >= Ar_1 /\ 1 >= Ar_1 /\ Ar_0 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1, Ar_2)) [ 1 >= Ar_1 /\ Ar_0 >= 2 /\ Ar_1 >= 2 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1, Ar_2)) [ 1 >= Ar_1 /\ Ar_0 >= 2 /\ 0 >= Ar_1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1, Ar_2)) [ 1 >= Ar_1 /\ 0 >= Ar_0 /\ Ar_1 >= 2 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1, Ar_2)) [ 1 >= Ar_1 /\ 0 >= Ar_0 /\ 0 >= Ar_1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(1, Ar_1 + 1, Ar_2)) [ Ar_1 >= 2 /\ 2 >= D /\ Ar_0 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(1, Ar_1 + 1, Ar_2)) [ Ar_1 >= 2 /\ 0 >= Ar_1 /\ 2 >= D /\ Ar_0 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1, Ar_2)) [ Ar_1 >= 2 /\ Ar_0 = 2 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1, Ar_2)) [ Ar_1 >= 2 /\ 0 >= Ar_1 /\ Ar_0 = 2 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1, Ar_2)) [ Ar_1 >= 2 /\ 2 >= Ar_0 /\ 0 >= Ar_0 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1, Ar_2)) [ Ar_1 >= 2 /\ 2 >= Ar_0 /\ 0 >= Ar_0 /\ 0 >= Ar_1 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1, Ar_2) -> Com_1(f2(Ar_0, Ar_1, Ar_2)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2) -> Com_1(f1(Ar_0, Ar_1, Ar_2)) [ 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(f1(Ar_0, Ar_1)) [ 0 <= 0 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ Ar_1 >= 2 /\ 2 >= Ar_0 /\ 0 >= Ar_0 /\ 0 >= Ar_1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ Ar_1 >= 2 /\ 2 >= Ar_0 /\ 0 >= Ar_0 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ Ar_1 >= 2 /\ 0 >= Ar_1 /\ Ar_0 = 2 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ Ar_1 >= 2 /\ Ar_0 = 2 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(1, Ar_1 + 1)) [ Ar_1 >= 2 /\ 0 >= Ar_1 /\ 2 >= D /\ Ar_0 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(1, Ar_1 + 1)) [ Ar_1 >= 2 /\ 2 >= D /\ Ar_0 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ 1 >= Ar_1 /\ 0 >= Ar_0 /\ 0 >= Ar_1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ 1 >= Ar_1 /\ 0 >= Ar_0 /\ Ar_1 >= 2 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ 1 >= Ar_1 /\ Ar_0 >= 2 /\ 0 >= Ar_1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ 1 >= Ar_1 /\ Ar_0 >= 2 /\ Ar_1 >= 2 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(1, Ar_1 + 1)) [ 0 >= Ar_1 /\ 1 >= Ar_1 /\ Ar_0 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(1, Ar_1 + 1)) [ Ar_1 >= 2 /\ 1 >= Ar_1 /\ Ar_0 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, 2)) [ 0 >= Ar_0 /\ 1 >= D /\ Ar_1 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, 2)) [ Ar_0 >= 2 /\ 1 >= D /\ Ar_1 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(1, 2)) [ 1 >= D /\ Ar_1 = 1 /\ Ar_0 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f300(Ar_0, Ar_1)) [ Ar_0 >= 3 /\ Ar_1 >= 2 ] start location: koat_start leaf cost: 0 Testing for reachability in the complexity graph removes the following transitions from problem 2: f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ Ar_1 >= 2 /\ 2 >= Ar_0 /\ 0 >= Ar_0 /\ 0 >= Ar_1 ] f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ Ar_1 >= 2 /\ 0 >= Ar_1 /\ Ar_0 = 2 ] f2(Ar_0, Ar_1) -> Com_1(f2(1, Ar_1 + 1)) [ Ar_1 >= 2 /\ 0 >= Ar_1 /\ 2 >= D /\ Ar_0 = 1 ] f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ 1 >= Ar_1 /\ 0 >= Ar_0 /\ Ar_1 >= 2 ] f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ 1 >= Ar_1 /\ Ar_0 >= 2 /\ Ar_1 >= 2 ] f2(Ar_0, Ar_1) -> Com_1(f2(1, Ar_1 + 1)) [ Ar_1 >= 2 /\ 1 >= Ar_1 /\ Ar_0 = 1 ] We thus obtain the following problem: 3: T: (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f300(Ar_0, Ar_1)) [ Ar_0 >= 3 /\ Ar_1 >= 2 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(1, 2)) [ 1 >= D /\ Ar_1 = 1 /\ Ar_0 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, 2)) [ Ar_0 >= 2 /\ 1 >= D /\ Ar_1 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, 2)) [ 0 >= Ar_0 /\ 1 >= D /\ Ar_1 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(1, Ar_1 + 1)) [ 0 >= Ar_1 /\ 1 >= Ar_1 /\ Ar_0 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ 1 >= Ar_1 /\ Ar_0 >= 2 /\ 0 >= Ar_1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ 1 >= Ar_1 /\ 0 >= Ar_0 /\ 0 >= Ar_1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(1, Ar_1 + 1)) [ Ar_1 >= 2 /\ 2 >= D /\ Ar_0 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ Ar_1 >= 2 /\ Ar_0 = 2 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ Ar_1 >= 2 /\ 2 >= Ar_0 /\ 0 >= Ar_0 ] (Comp: ?, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f1(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) f2(Ar_0, Ar_1) -> Com_1(f300(Ar_0, Ar_1)) [ Ar_0 >= 3 /\ Ar_1 >= 2 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(1, 2)) [ 1 >= D /\ Ar_1 = 1 /\ Ar_0 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, 2)) [ Ar_0 >= 2 /\ 1 >= D /\ Ar_1 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, 2)) [ 0 >= Ar_0 /\ 1 >= D /\ Ar_1 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(1, Ar_1 + 1)) [ 0 >= Ar_1 /\ 1 >= Ar_1 /\ Ar_0 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ 1 >= Ar_1 /\ Ar_0 >= 2 /\ 0 >= Ar_1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ 1 >= Ar_1 /\ 0 >= Ar_0 /\ 0 >= Ar_1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(1, Ar_1 + 1)) [ Ar_1 >= 2 /\ 2 >= D /\ Ar_0 = 1 ] (Comp: 1, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ Ar_1 >= 2 /\ Ar_0 = 2 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ Ar_1 >= 2 /\ 2 >= Ar_0 /\ 0 >= Ar_0 ] (Comp: 1, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f2) = 1 Pol(f300) = 0 Pol(f1) = 1 Pol(koat_start) = 1 orients all transitions weakly and the transition f2(Ar_0, Ar_1) -> Com_1(f300(Ar_0, Ar_1)) [ Ar_0 >= 3 /\ Ar_1 >= 2 ] strictly and produces the following problem: 5: T: (Comp: 1, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f300(Ar_0, Ar_1)) [ Ar_0 >= 3 /\ Ar_1 >= 2 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(1, 2)) [ 1 >= D /\ Ar_1 = 1 /\ Ar_0 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, 2)) [ Ar_0 >= 2 /\ 1 >= D /\ Ar_1 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, 2)) [ 0 >= Ar_0 /\ 1 >= D /\ Ar_1 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(1, Ar_1 + 1)) [ 0 >= Ar_1 /\ 1 >= Ar_1 /\ Ar_0 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ 1 >= Ar_1 /\ Ar_0 >= 2 /\ 0 >= Ar_1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ 1 >= Ar_1 /\ 0 >= Ar_0 /\ 0 >= Ar_1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(1, Ar_1 + 1)) [ Ar_1 >= 2 /\ 2 >= D /\ Ar_0 = 1 ] (Comp: 1, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ Ar_1 >= 2 /\ Ar_0 = 2 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ Ar_1 >= 2 /\ 2 >= Ar_0 /\ 0 >= Ar_0 ] (Comp: 1, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f2) = -V_2 + 3 Pol(f300) = -V_2 + 3 Pol(f1) = -V_2 + 3 Pol(koat_start) = -V_2 + 3 orients all transitions weakly and the transitions f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, 2)) [ 0 >= Ar_0 /\ 1 >= D /\ Ar_1 = 1 ] f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ 1 >= Ar_1 /\ Ar_0 >= 2 /\ 0 >= Ar_1 ] f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ 1 >= Ar_1 /\ 0 >= Ar_0 /\ 0 >= Ar_1 ] f2(Ar_0, Ar_1) -> Com_1(f2(1, 2)) [ 1 >= D /\ Ar_1 = 1 /\ Ar_0 = 1 ] f2(Ar_0, Ar_1) -> Com_1(f2(1, Ar_1 + 1)) [ 0 >= Ar_1 /\ 1 >= Ar_1 /\ Ar_0 = 1 ] strictly and produces the following problem: 6: T: (Comp: 1, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f300(Ar_0, Ar_1)) [ Ar_0 >= 3 /\ Ar_1 >= 2 ] (Comp: Ar_1 + 3, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(1, 2)) [ 1 >= D /\ Ar_1 = 1 /\ Ar_0 = 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, 2)) [ Ar_0 >= 2 /\ 1 >= D /\ Ar_1 = 1 ] (Comp: Ar_1 + 3, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, 2)) [ 0 >= Ar_0 /\ 1 >= D /\ Ar_1 = 1 ] (Comp: Ar_1 + 3, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(1, Ar_1 + 1)) [ 0 >= Ar_1 /\ 1 >= Ar_1 /\ Ar_0 = 1 ] (Comp: Ar_1 + 3, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ 1 >= Ar_1 /\ Ar_0 >= 2 /\ 0 >= Ar_1 ] (Comp: Ar_1 + 3, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ 1 >= Ar_1 /\ 0 >= Ar_0 /\ 0 >= Ar_1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(1, Ar_1 + 1)) [ Ar_1 >= 2 /\ 2 >= D /\ Ar_0 = 1 ] (Comp: 1, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ Ar_1 >= 2 /\ Ar_0 = 2 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ Ar_1 >= 2 /\ 2 >= Ar_0 /\ 0 >= Ar_0 ] (Comp: 1, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 6 produces the following problem: 7: T: (Comp: 1, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f300(Ar_0, Ar_1)) [ Ar_0 >= 3 /\ Ar_1 >= 2 ] (Comp: Ar_1 + 3, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(1, 2)) [ 1 >= D /\ Ar_1 = 1 /\ Ar_0 = 1 ] (Comp: Ar_1 + 4, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, 2)) [ Ar_0 >= 2 /\ 1 >= D /\ Ar_1 = 1 ] (Comp: Ar_1 + 3, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, 2)) [ 0 >= Ar_0 /\ 1 >= D /\ Ar_1 = 1 ] (Comp: Ar_1 + 3, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(1, Ar_1 + 1)) [ 0 >= Ar_1 /\ 1 >= Ar_1 /\ Ar_0 = 1 ] (Comp: Ar_1 + 3, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ 1 >= Ar_1 /\ Ar_0 >= 2 /\ 0 >= Ar_1 ] (Comp: Ar_1 + 3, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ 1 >= Ar_1 /\ 0 >= Ar_0 /\ 0 >= Ar_1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(1, Ar_1 + 1)) [ Ar_1 >= 2 /\ 2 >= D /\ Ar_0 = 1 ] (Comp: 1, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ Ar_1 >= 2 /\ Ar_0 = 2 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ Ar_1 >= 2 /\ 2 >= Ar_0 /\ 0 >= Ar_0 ] (Comp: 1, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f2) = -V_1 + 2 Pol(f300) = -V_1 + 2 Pol(f1) = -V_1 + 2 Pol(koat_start) = -V_1 + 2 orients all transitions weakly and the transition f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ Ar_1 >= 2 /\ 2 >= Ar_0 /\ 0 >= Ar_0 ] strictly and produces the following problem: 8: T: (Comp: 1, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f300(Ar_0, Ar_1)) [ Ar_0 >= 3 /\ Ar_1 >= 2 ] (Comp: Ar_1 + 3, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(1, 2)) [ 1 >= D /\ Ar_1 = 1 /\ Ar_0 = 1 ] (Comp: Ar_1 + 4, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, 2)) [ Ar_0 >= 2 /\ 1 >= D /\ Ar_1 = 1 ] (Comp: Ar_1 + 3, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, 2)) [ 0 >= Ar_0 /\ 1 >= D /\ Ar_1 = 1 ] (Comp: Ar_1 + 3, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(1, Ar_1 + 1)) [ 0 >= Ar_1 /\ 1 >= Ar_1 /\ Ar_0 = 1 ] (Comp: Ar_1 + 3, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ 1 >= Ar_1 /\ Ar_0 >= 2 /\ 0 >= Ar_1 ] (Comp: Ar_1 + 3, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ 1 >= Ar_1 /\ 0 >= Ar_0 /\ 0 >= Ar_1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(1, Ar_1 + 1)) [ Ar_1 >= 2 /\ 2 >= D /\ Ar_0 = 1 ] (Comp: 1, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ Ar_1 >= 2 /\ Ar_0 = 2 ] (Comp: Ar_0 + 2, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 + 1, Ar_1 + 1)) [ Ar_1 >= 2 /\ 2 >= Ar_0 /\ 0 >= Ar_0 ] (Comp: 1, Cost: 1) f1(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f1(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound ? Time: 5.126 sec (SMT: 4.996 sec)