MAYBE Initial complexity problem: 1: T: (Comp: ?, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Fresh_3, Fresh_3, 0, Ar_3)) [ 0 >= Fresh_3 ] (Comp: ?, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Fresh_2, Fresh_2, 0, Ar_3)) [ Fresh_2 >= 1024 ] (Comp: ?, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Fresh_0, Fresh_0, 0, Fresh_1)) [ 1023 >= Fresh_0 /\ Fresh_0 >= 1 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Ar_0, Ar_1, Ar_2 + 1, Ar_3)) [ E >= Ar_2 + 1 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f22(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_2 >= E ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Repeatedly propagating knowledge in problem 1 produces the following problem: 2: T: (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Fresh_3, Fresh_3, 0, Ar_3)) [ 0 >= Fresh_3 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Fresh_2, Fresh_2, 0, Ar_3)) [ Fresh_2 >= 1024 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Fresh_0, Fresh_0, 0, Fresh_1)) [ 1023 >= Fresh_0 /\ Fresh_0 >= 1 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Ar_0, Ar_1, Ar_2 + 1, Ar_3)) [ E >= Ar_2 + 1 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f22(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_2 >= E ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 A polynomial rank function with Pol(f0) = 1 Pol(f14) = 1 Pol(f22) = 0 Pol(koat_start) = 1 orients all transitions weakly and the transition f14(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f22(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_2 >= E ] strictly and produces the following problem: 3: T: (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Fresh_3, Fresh_3, 0, Ar_3)) [ 0 >= Fresh_3 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Fresh_2, Fresh_2, 0, Ar_3)) [ Fresh_2 >= 1024 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Fresh_0, Fresh_0, 0, Fresh_1)) [ 1023 >= Fresh_0 /\ Fresh_0 >= 1 ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Ar_0, Ar_1, Ar_2 + 1, Ar_3)) [ E >= Ar_2 + 1 ] (Comp: 1, Cost: 1) f14(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f22(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_2 >= E ] (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ] start location: koat_start leaf cost: 0 Applied AI with 'oct' on problem 3 to obtain the following invariants: For symbol f14: X_3 >= 0 This yielded the following problem: 4: T: (Comp: 1, Cost: 0) koat_start(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f0(Ar_0, Ar_1, Ar_2, Ar_3)) [ 0 <= 0 ] (Comp: 1, Cost: 1) f14(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f22(Ar_0, Ar_1, Ar_2, Ar_3)) [ Ar_2 >= 0 /\ Ar_2 >= E ] (Comp: ?, Cost: 1) f14(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Ar_0, Ar_1, Ar_2 + 1, Ar_3)) [ Ar_2 >= 0 /\ E >= Ar_2 + 1 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Fresh_0, Fresh_0, 0, Fresh_1)) [ 1023 >= Fresh_0 /\ Fresh_0 >= 1 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Fresh_2, Fresh_2, 0, Ar_3)) [ Fresh_2 >= 1024 ] (Comp: 1, Cost: 1) f0(Ar_0, Ar_1, Ar_2, Ar_3) -> Com_1(f14(Fresh_3, Fresh_3, 0, Ar_3)) [ 0 >= Fresh_3 ] start location: koat_start leaf cost: 0 Complexity upper bound ? Time: 1.634 sec (SMT: 1.583 sec)