YES(?, 4*a + 6) Initial complexity problem: 1: T: (1, 1) evaleasy2start(a) -> evaleasy2entryin(a) (?, 1) evaleasy2entryin(a) -> evaleasy2bb1in(a) (?, 1) evaleasy2bb1in(a) -> evaleasy2bbin(a) [ a >= 1 ] (?, 1) evaleasy2bb1in(a) -> evaleasy2returnin(a) [ 0 >= a ] (?, 1) evaleasy2bbin(a) -> evaleasy2bb1in(a - 1) (?, 1) evaleasy2returnin(a) -> evaleasy2stop(a) start location: evaleasy2start leaf cost: 0 Repeatedly removing leaves of the complexity graph in problem 1 produces the following problem: 2: T: (1, 1) evaleasy2start(a) -> evaleasy2entryin(a) (?, 1) evaleasy2entryin(a) -> evaleasy2bb1in(a) (?, 1) evaleasy2bb1in(a) -> evaleasy2bbin(a) [ a >= 1 ] (?, 1) evaleasy2bbin(a) -> evaleasy2bb1in(a - 1) start location: evaleasy2start leaf cost: 2 Repeatedly propagating knowledge in problem 2 produces the following problem: 3: T: (1, 1) evaleasy2start(a) -> evaleasy2entryin(a) (1, 1) evaleasy2entryin(a) -> evaleasy2bb1in(a) (?, 1) evaleasy2bb1in(a) -> evaleasy2bbin(a) [ a >= 1 ] (?, 1) evaleasy2bbin(a) -> evaleasy2bb1in(a - 1) start location: evaleasy2start leaf cost: 2 A polynomial rank function with Pol(evaleasy2start) = 2*V_1 + 1 Pol(evaleasy2entryin) = 2*V_1 + 1 Pol(evaleasy2bb1in) = 2*V_1 + 1 Pol(evaleasy2bbin) = 2*V_1 orients all transitions weakly and the transition evaleasy2bb1in(a) -> evaleasy2bbin(a) [ a >= 1 ] strictly and produces the following problem: 4: T: (1, 1) evaleasy2start(a) -> evaleasy2entryin(a) (1, 1) evaleasy2entryin(a) -> evaleasy2bb1in(a) (2*a + 1, 1) evaleasy2bb1in(a) -> evaleasy2bbin(a) [ a >= 1 ] (?, 1) evaleasy2bbin(a) -> evaleasy2bb1in(a - 1) start location: evaleasy2start leaf cost: 2 Repeatedly propagating knowledge in problem 4 produces the following problem: 5: T: (1, 1) evaleasy2start(a) -> evaleasy2entryin(a) (1, 1) evaleasy2entryin(a) -> evaleasy2bb1in(a) (2*a + 1, 1) evaleasy2bb1in(a) -> evaleasy2bbin(a) [ a >= 1 ] (2*a + 1, 1) evaleasy2bbin(a) -> evaleasy2bb1in(a - 1) start location: evaleasy2start leaf cost: 2 Complexity upper bound 4*a + 6 Time: 0.079 sec (SMT: 0.076 sec)