MAYBE Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f1(Fresh_3, Fresh_4)) [ 0 >= Fresh_3 + 1 /\ Ar_0 = 14 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Fresh_2, Ar_1)) [ Fresh_2 >= 0 /\ Ar_0 = 14 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f1(Ar_0 - 1, Fresh_1)) [ Ar_0 >= 15 /\ 0 >= Ar_0 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f1(Ar_0 - 1, Fresh_0)) [ 13 >= Ar_0 /\ 0 >= Ar_0 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - 1, Ar_1)) [ Ar_0 >= 15 /\ Ar_0 >= 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - 1, Ar_1)) [ 13 >= Ar_0 /\ Ar_0 >= 1 ] (Comp: ?, Cost: 1) f300(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f300(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Testing for reachability in the complexity graph removes the following transition from problem 1: f2(Ar_0, Ar_1) -> Com_1(f1(Ar_0 - 1, Fresh_1)) [ Ar_0 >= 15 /\ 0 >= Ar_0 ] We thus obtain the following problem: 2: T: (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - 1, Ar_1)) [ 13 >= Ar_0 /\ Ar_0 >= 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - 1, Ar_1)) [ Ar_0 >= 15 /\ Ar_0 >= 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f1(Ar_0 - 1, Fresh_0)) [ 13 >= Ar_0 /\ 0 >= Ar_0 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Fresh_2, Ar_1)) [ Fresh_2 >= 0 /\ Ar_0 = 14 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f1(Fresh_3, Fresh_4)) [ 0 >= Fresh_3 + 1 /\ Ar_0 = 14 ] (Comp: ?, Cost: 1) f300(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f300(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 2 produces the following problem: 3: T: (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - 1, Ar_1)) [ 13 >= Ar_0 /\ Ar_0 >= 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - 1, Ar_1)) [ Ar_0 >= 15 /\ Ar_0 >= 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f1(Ar_0 - 1, Fresh_0)) [ 13 >= Ar_0 /\ 0 >= Ar_0 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Fresh_2, Ar_1)) [ Fresh_2 >= 0 /\ Ar_0 = 14 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f1(Fresh_3, Fresh_4)) [ 0 >= Fresh_3 + 1 /\ Ar_0 = 14 ] (Comp: 1, Cost: 1) f300(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f300(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f2) = 1 Pol(f1) = 0 Pol(f300) = 1 Pol(koat_start) = 1 orients all transitions weakly and the transitions f2(Ar_0, Ar_1) -> Com_1(f1(Fresh_3, Fresh_4)) [ 0 >= Fresh_3 + 1 /\ Ar_0 = 14 ] f2(Ar_0, Ar_1) -> Com_1(f1(Ar_0 - 1, Fresh_0)) [ 13 >= Ar_0 /\ 0 >= Ar_0 ] strictly and produces the following problem: 4: T: (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - 1, Ar_1)) [ 13 >= Ar_0 /\ Ar_0 >= 1 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Ar_0 - 1, Ar_1)) [ Ar_0 >= 15 /\ Ar_0 >= 1 ] (Comp: 1, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f1(Ar_0 - 1, Fresh_0)) [ 13 >= Ar_0 /\ 0 >= Ar_0 ] (Comp: ?, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f2(Fresh_2, Ar_1)) [ Fresh_2 >= 0 /\ Ar_0 = 14 ] (Comp: 1, Cost: 1) f2(Ar_0, Ar_1) -> Com_1(f1(Fresh_3, Fresh_4)) [ 0 >= Fresh_3 + 1 /\ Ar_0 = 14 ] (Comp: 1, Cost: 1) f300(Ar_0, Ar_1) -> Com_1(f2(Ar_0, Ar_1)) (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1) -> Com_1(f300(Ar_0, Ar_1)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Complexity upper bound ? Time: 1.203 sec (SMT: 1.161 sec)