3.40/1.67 WORST_CASE(?, O(1)) 3.40/1.68 proof of /export/starexec/sandbox2/benchmark/theBenchmark.koat 3.40/1.68 # AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty 3.40/1.68 3.40/1.68 3.40/1.68 The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(1, 1). 3.40/1.68 3.40/1.68 (0) CpxIntTrs 3.40/1.68 (1) Koat Proof [FINISHED, 36 ms] 3.40/1.68 (2) BOUNDS(1, 1) 3.40/1.68 3.40/1.68 3.40/1.68 ---------------------------------------- 3.40/1.68 3.40/1.68 (0) 3.40/1.68 Obligation: 3.40/1.68 Complexity Int TRS consisting of the following rules: 3.40/1.68 f4(A) -> Com_1(f5(A)) :|: 0 >= B + 1 3.40/1.68 f4(A) -> Com_1(f5(A)) :|: TRUE 3.40/1.68 f0(A) -> Com_1(f4(0)) :|: TRUE 3.40/1.68 f5(A) -> Com_1(f11(A)) :|: A >= 3 3.40/1.68 f4(A) -> Com_1(f11(A)) :|: TRUE 3.40/1.68 f5(A) -> Com_1(f4(A + 1)) :|: 2 >= A 3.40/1.68 f11(A) -> Com_1(f14(A)) :|: 1 >= A 3.40/1.68 f11(A) -> Com_1(f14(A)) :|: A >= 2 3.40/1.68 3.40/1.68 The start-symbols are:[f0_1] 3.40/1.68 3.40/1.68 3.40/1.68 ---------------------------------------- 3.40/1.68 3.40/1.68 (1) Koat Proof (FINISHED) 3.40/1.68 YES(?, 20) 3.40/1.68 3.40/1.68 3.40/1.68 3.40/1.68 Initial complexity problem: 3.40/1.68 3.40/1.68 1: T: 3.40/1.68 3.40/1.68 (Comp: ?, Cost: 1) f4(ar_0) -> Com_1(f5(ar_0)) [ 0 >= b + 1 ] 3.40/1.68 3.40/1.68 (Comp: ?, Cost: 1) f4(ar_0) -> Com_1(f5(ar_0)) 3.40/1.68 3.40/1.68 (Comp: ?, Cost: 1) f0(ar_0) -> Com_1(f4(0)) 3.40/1.68 3.40/1.68 (Comp: ?, Cost: 1) f5(ar_0) -> Com_1(f11(ar_0)) [ ar_0 >= 3 ] 3.40/1.68 3.40/1.68 (Comp: ?, Cost: 1) f4(ar_0) -> Com_1(f11(ar_0)) 3.40/1.68 3.40/1.68 (Comp: ?, Cost: 1) f5(ar_0) -> Com_1(f4(ar_0 + 1)) [ 2 >= ar_0 ] 3.40/1.68 3.40/1.68 (Comp: ?, Cost: 1) f11(ar_0) -> Com_1(f14(ar_0)) [ 1 >= ar_0 ] 3.40/1.68 3.40/1.68 (Comp: ?, Cost: 1) f11(ar_0) -> Com_1(f14(ar_0)) [ ar_0 >= 2 ] 3.40/1.68 3.40/1.68 (Comp: 1, Cost: 0) koat_start(ar_0) -> Com_1(f0(ar_0)) [ 0 <= 0 ] 3.40/1.68 3.40/1.68 start location: koat_start 3.40/1.68 3.40/1.68 leaf cost: 0 3.40/1.68 3.40/1.68 3.40/1.68 3.40/1.68 Repeatedly propagating knowledge in problem 1 produces the following problem: 3.40/1.68 3.40/1.68 2: T: 3.40/1.68 3.40/1.68 (Comp: ?, Cost: 1) f4(ar_0) -> Com_1(f5(ar_0)) [ 0 >= b + 1 ] 3.40/1.68 3.40/1.68 (Comp: ?, Cost: 1) f4(ar_0) -> Com_1(f5(ar_0)) 3.40/1.68 3.40/1.68 (Comp: 1, Cost: 1) f0(ar_0) -> Com_1(f4(0)) 3.40/1.68 3.40/1.68 (Comp: ?, Cost: 1) f5(ar_0) -> Com_1(f11(ar_0)) [ ar_0 >= 3 ] 3.40/1.68 3.40/1.68 (Comp: ?, Cost: 1) f4(ar_0) -> Com_1(f11(ar_0)) 3.40/1.68 3.40/1.68 (Comp: ?, Cost: 1) f5(ar_0) -> Com_1(f4(ar_0 + 1)) [ 2 >= ar_0 ] 3.40/1.68 3.40/1.68 (Comp: ?, Cost: 1) f11(ar_0) -> Com_1(f14(ar_0)) [ 1 >= ar_0 ] 3.40/1.68 3.40/1.68 (Comp: ?, Cost: 1) f11(ar_0) -> Com_1(f14(ar_0)) [ ar_0 >= 2 ] 3.40/1.68 3.40/1.68 (Comp: 1, Cost: 0) koat_start(ar_0) -> Com_1(f0(ar_0)) [ 0 <= 0 ] 3.40/1.68 3.40/1.68 start location: koat_start 3.40/1.68 3.40/1.68 leaf cost: 0 3.40/1.68 3.40/1.68 3.40/1.68 3.40/1.68 A polynomial rank function with 3.40/1.68 3.40/1.68 Pol(f4) = 2 3.40/1.68 3.40/1.68 Pol(f5) = 2 3.40/1.68 3.40/1.68 Pol(f0) = 2 3.40/1.68 3.40/1.68 Pol(f11) = 1 3.40/1.68 3.40/1.68 Pol(f14) = 0 3.40/1.68 3.40/1.68 Pol(koat_start) = 2 3.40/1.68 3.40/1.68 orients all transitions weakly and the transitions 3.40/1.68 3.40/1.68 f5(ar_0) -> Com_1(f11(ar_0)) [ ar_0 >= 3 ] 3.40/1.68 3.40/1.68 f4(ar_0) -> Com_1(f11(ar_0)) 3.40/1.68 3.40/1.68 f11(ar_0) -> Com_1(f14(ar_0)) [ 1 >= ar_0 ] 3.40/1.68 3.40/1.68 f11(ar_0) -> Com_1(f14(ar_0)) [ ar_0 >= 2 ] 3.40/1.68 3.40/1.68 strictly and produces the following problem: 3.40/1.68 3.40/1.68 3: T: 3.40/1.68 3.40/1.68 (Comp: ?, Cost: 1) f4(ar_0) -> Com_1(f5(ar_0)) [ 0 >= b + 1 ] 3.40/1.68 3.40/1.68 (Comp: ?, Cost: 1) f4(ar_0) -> Com_1(f5(ar_0)) 3.40/1.68 3.40/1.68 (Comp: 1, Cost: 1) f0(ar_0) -> Com_1(f4(0)) 3.40/1.68 3.40/1.68 (Comp: 2, Cost: 1) f5(ar_0) -> Com_1(f11(ar_0)) [ ar_0 >= 3 ] 3.40/1.68 3.40/1.68 (Comp: 2, Cost: 1) f4(ar_0) -> Com_1(f11(ar_0)) 3.40/1.68 3.40/1.68 (Comp: ?, Cost: 1) f5(ar_0) -> Com_1(f4(ar_0 + 1)) [ 2 >= ar_0 ] 3.40/1.68 3.40/1.68 (Comp: 2, Cost: 1) f11(ar_0) -> Com_1(f14(ar_0)) [ 1 >= ar_0 ] 3.40/1.68 3.40/1.68 (Comp: 2, Cost: 1) f11(ar_0) -> Com_1(f14(ar_0)) [ ar_0 >= 2 ] 3.40/1.68 3.40/1.68 (Comp: 1, Cost: 0) koat_start(ar_0) -> Com_1(f0(ar_0)) [ 0 <= 0 ] 3.40/1.68 3.40/1.68 start location: koat_start 3.40/1.68 3.40/1.68 leaf cost: 0 3.40/1.68 3.40/1.68 3.40/1.68 3.40/1.68 A polynomial rank function with 3.40/1.68 3.40/1.68 Pol(f4) = -V_1 + 3 3.40/1.68 3.40/1.68 Pol(f5) = -V_1 + 3 3.40/1.68 3.40/1.68 Pol(f0) = 3 3.40/1.68 3.40/1.68 Pol(f11) = -V_1 3.40/1.68 3.40/1.68 Pol(f14) = -V_1 3.40/1.68 3.40/1.68 Pol(koat_start) = 3 3.40/1.68 3.40/1.68 orients all transitions weakly and the transition 3.40/1.68 3.40/1.68 f5(ar_0) -> Com_1(f4(ar_0 + 1)) [ 2 >= ar_0 ] 3.40/1.68 3.40/1.68 strictly and produces the following problem: 3.40/1.68 3.40/1.68 4: T: 3.40/1.68 3.40/1.68 (Comp: ?, Cost: 1) f4(ar_0) -> Com_1(f5(ar_0)) [ 0 >= b + 1 ] 3.40/1.68 3.40/1.68 (Comp: ?, Cost: 1) f4(ar_0) -> Com_1(f5(ar_0)) 3.40/1.68 3.40/1.68 (Comp: 1, Cost: 1) f0(ar_0) -> Com_1(f4(0)) 3.40/1.68 3.40/1.68 (Comp: 2, Cost: 1) f5(ar_0) -> Com_1(f11(ar_0)) [ ar_0 >= 3 ] 3.40/1.68 3.40/1.68 (Comp: 2, Cost: 1) f4(ar_0) -> Com_1(f11(ar_0)) 3.40/1.68 3.40/1.68 (Comp: 3, Cost: 1) f5(ar_0) -> Com_1(f4(ar_0 + 1)) [ 2 >= ar_0 ] 3.40/1.68 3.40/1.68 (Comp: 2, Cost: 1) f11(ar_0) -> Com_1(f14(ar_0)) [ 1 >= ar_0 ] 3.40/1.68 3.40/1.68 (Comp: 2, Cost: 1) f11(ar_0) -> Com_1(f14(ar_0)) [ ar_0 >= 2 ] 3.40/1.68 3.40/1.68 (Comp: 1, Cost: 0) koat_start(ar_0) -> Com_1(f0(ar_0)) [ 0 <= 0 ] 3.40/1.68 3.40/1.68 start location: koat_start 3.40/1.68 3.40/1.68 leaf cost: 0 3.40/1.68 3.40/1.68 3.40/1.68 3.40/1.68 Repeatedly propagating knowledge in problem 4 produces the following problem: 3.40/1.68 3.40/1.68 5: T: 3.40/1.68 3.40/1.68 (Comp: 4, Cost: 1) f4(ar_0) -> Com_1(f5(ar_0)) [ 0 >= b + 1 ] 3.40/1.68 3.40/1.68 (Comp: 4, Cost: 1) f4(ar_0) -> Com_1(f5(ar_0)) 3.40/1.68 3.40/1.68 (Comp: 1, Cost: 1) f0(ar_0) -> Com_1(f4(0)) 3.40/1.68 3.40/1.68 (Comp: 2, Cost: 1) f5(ar_0) -> Com_1(f11(ar_0)) [ ar_0 >= 3 ] 3.40/1.68 3.40/1.68 (Comp: 2, Cost: 1) f4(ar_0) -> Com_1(f11(ar_0)) 3.40/1.68 3.40/1.68 (Comp: 3, Cost: 1) f5(ar_0) -> Com_1(f4(ar_0 + 1)) [ 2 >= ar_0 ] 3.40/1.68 3.40/1.68 (Comp: 2, Cost: 1) f11(ar_0) -> Com_1(f14(ar_0)) [ 1 >= ar_0 ] 3.40/1.68 3.40/1.68 (Comp: 2, Cost: 1) f11(ar_0) -> Com_1(f14(ar_0)) [ ar_0 >= 2 ] 3.40/1.68 3.40/1.68 (Comp: 1, Cost: 0) koat_start(ar_0) -> Com_1(f0(ar_0)) [ 0 <= 0 ] 3.40/1.68 3.40/1.68 start location: koat_start 3.40/1.68 3.40/1.68 leaf cost: 0 3.40/1.68 3.40/1.68 3.40/1.68 3.40/1.68 Complexity upper bound 20 3.40/1.68 3.40/1.68 3.40/1.68 3.40/1.68 Time: 0.059 sec (SMT: 0.056 sec) 3.40/1.68 3.40/1.68 3.40/1.68 ---------------------------------------- 3.40/1.68 3.40/1.68 (2) 3.40/1.68 BOUNDS(1, 1) 3.55/1.70 EOF