4.07/2.19	WORST_CASE(Omega(n^1), O(n^1))
4.11/2.20	proof of /export/starexec/sandbox/benchmark/theBenchmark.koat
4.11/2.20	# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty
4.11/2.20	
4.11/2.20	
4.11/2.20	The runtime complexity of the given CpxIntTrs could be proven to be BOUNDS(n^1, max(1, 1 + 2 * Arg_0)).
4.11/2.20	
4.11/2.20	(0) CpxIntTrs
4.11/2.20	(1) Koat2 Proof [FINISHED, 175 ms]
4.11/2.20	(2) BOUNDS(1, max(1, 1 + 2 * Arg_0))
4.11/2.20	(3) Loat Proof [FINISHED, 349 ms]
4.11/2.20	(4) BOUNDS(n^1, INF)
4.11/2.20	
4.11/2.20	
4.11/2.20	----------------------------------------
4.11/2.20	
4.11/2.20	(0)
4.11/2.20	Obligation:
4.11/2.20	Complexity Int TRS consisting of the following rules:
4.11/2.20	eval(A, B) -> Com_1(eval(A - 1, B)) :|: A >= 1 && B >= A
4.11/2.20	eval(A, B) -> Com_1(eval(B, B)) :|: A >= 1 && A >= B + 1
4.11/2.20	start(A, B) -> Com_1(eval(A, B)) :|: TRUE
4.11/2.20	
4.11/2.20	The start-symbols are:[start_2]
4.11/2.20	
4.11/2.20	
4.11/2.20	----------------------------------------
4.11/2.20	
4.11/2.20	(1) Koat2 Proof (FINISHED)
4.11/2.20	YES( ?, 1+2*max([0, Arg_0]) {O(n)})
4.11/2.20	
4.11/2.20	
4.11/2.20	
4.11/2.20	Initial Complexity Problem:
4.11/2.20	
4.11/2.20	  Start:  start  
4.11/2.20	
4.11/2.20	  Program_Vars:  Arg_0, Arg_1  
4.11/2.20	
4.11/2.20	  Temp_Vars:    
4.11/2.20	
4.11/2.20	  Locations:  eval, start  
4.11/2.20	
4.11/2.20	  Transitions:
4.11/2.20	
4.11/2.20	  eval(Arg_0,Arg_1) -> eval(Arg_0-1,Arg_1):|:1 <= Arg_0 && Arg_0 <= Arg_1
4.11/2.20	
4.11/2.20	  eval(Arg_0,Arg_1) -> eval(Arg_1,Arg_1):|:1 <= Arg_0 && Arg_1+1 <= Arg_0
4.11/2.20	
4.11/2.20	  start(Arg_0,Arg_1) -> eval(Arg_0,Arg_1):|:  
4.11/2.20	
4.11/2.20	
4.11/2.20	
4.11/2.20	Timebounds: 
4.11/2.20	
4.11/2.20	  Overall timebound: 1+2*max([0, Arg_0]) {O(n)}
4.11/2.20	
4.11/2.20	  0: eval->eval: max([0, Arg_0]) {O(n)}
4.11/2.20	
4.11/2.20	  1: eval->eval: max([0, Arg_0]) {O(n)}
4.11/2.20	
4.11/2.20	  2: start->eval: 1 {O(1)}
4.11/2.20	
4.11/2.20	
4.11/2.20	
4.11/2.20	Costbounds:
4.11/2.20	
4.11/2.20	  Overall costbound: 1+2*max([0, Arg_0]) {O(n)}
4.11/2.20	
4.11/2.20	  0: eval->eval: max([0, Arg_0]) {O(n)}
4.11/2.20	
4.11/2.20	  1: eval->eval: max([0, Arg_0]) {O(n)}
4.11/2.20	
4.11/2.20	  2: start->eval: 1 {O(1)}
4.11/2.20	
4.11/2.20	
4.11/2.20	
4.11/2.20	Sizebounds:
4.11/2.20	
4.11/2.20	`Lower:
4.11/2.20	
4.11/2.20	  0: eval->eval, Arg_0: 0 {O(1)}
4.11/2.20	
4.11/2.20	  0: eval->eval, Arg_1: 1 {O(1)}
4.11/2.20	
4.11/2.20	  1: eval->eval, Arg_0: Arg_1 {O(n)}
4.11/2.20	
4.11/2.20	  1: eval->eval, Arg_1: Arg_1 {O(n)}
4.11/2.20	
4.11/2.20	  2: start->eval, Arg_0: Arg_0 {O(n)}
4.11/2.20	
4.11/2.20	  2: start->eval, Arg_1: Arg_1 {O(n)}
4.11/2.20	
4.11/2.20	`Upper:
4.11/2.20	
4.11/2.20	  0: eval->eval, Arg_0: max([Arg_1, Arg_0]) {O(n)}
4.11/2.20	
4.11/2.20	  0: eval->eval, Arg_1: Arg_1 {O(n)}
4.11/2.20	
4.11/2.20	  1: eval->eval, Arg_0: Arg_1 {O(n)}
4.11/2.20	
4.11/2.20	  1: eval->eval, Arg_1: Arg_1 {O(n)}
4.11/2.20	
4.11/2.20	  2: start->eval, Arg_0: Arg_0 {O(n)}
4.11/2.20	
4.11/2.20	  2: start->eval, Arg_1: Arg_1 {O(n)}
4.11/2.20	
4.11/2.20	
4.11/2.20	----------------------------------------
4.11/2.20	
4.11/2.20	(2)
4.11/2.20	BOUNDS(1, max(1, 1 + 2 * Arg_0))
4.11/2.20	
4.11/2.20	----------------------------------------
4.11/2.20	
4.11/2.20	(3) Loat Proof (FINISHED)
4.11/2.20	
4.11/2.20	
4.11/2.20	### Pre-processing the ITS problem ###
4.11/2.20	
4.11/2.20	
4.11/2.20	
4.11/2.20	Initial linear ITS problem
4.11/2.20	
4.11/2.20	   Start location: start
4.11/2.20	
4.11/2.20	      0: eval -> eval : A'=-1+A, [ A>=1 && B>=A ], cost: 1
4.11/2.20	
4.11/2.20	      1: eval -> eval : A'=B, [ A>=1 && A>=1+B ], cost: 1
4.11/2.20	
4.11/2.20	      2: start -> eval : [], cost: 1
4.11/2.20	
4.11/2.20	
4.11/2.20	
4.11/2.20	### Simplification by acceleration and chaining ###
4.11/2.20	
4.11/2.20	
4.11/2.20	
4.11/2.20	Accelerating simple loops of location 0.
4.11/2.20	
4.11/2.20	   Accelerating the following rules:
4.11/2.20	
4.11/2.20	      0: eval -> eval : A'=-1+A, [ A>=1 && B>=A ], cost: 1
4.11/2.20	
4.11/2.20	      1: eval -> eval : A'=B, [ A>=1 && A>=1+B ], cost: 1
4.11/2.20	
4.11/2.20	
4.11/2.20	
4.11/2.20	   Accelerated rule 0 with metering function A, yielding the new rule 3.
4.11/2.20	
4.11/2.20	   Found no metering function for rule 1.
4.11/2.20	
4.11/2.20	   Removing the simple loops: 0.
4.11/2.20	
4.11/2.20	
4.11/2.20	
4.11/2.20	Accelerated all simple loops using metering functions (where possible):
4.11/2.20	
4.11/2.20	   Start location: start
4.11/2.20	
4.11/2.20	      1: eval -> eval : A'=B, [ A>=1 && A>=1+B ], cost: 1
4.11/2.20	
4.11/2.20	      3: eval -> eval : A'=0, [ A>=1 && B>=A ], cost: A
4.11/2.20	
4.11/2.20	      2: start -> eval : [], cost: 1
4.11/2.20	
4.11/2.20	
4.11/2.20	
4.11/2.20	Chained accelerated rules (with incoming rules):
4.11/2.20	
4.11/2.20	   Start location: start
4.11/2.20	
4.11/2.20	      2: start -> eval : [], cost: 1
4.11/2.20	
4.11/2.20	      4: start -> eval : A'=B, [ A>=1 && A>=1+B ], cost: 2
4.11/2.20	
4.11/2.20	      5: start -> eval : A'=0, [ A>=1 && B>=A ], cost: 1+A
4.11/2.20	
4.11/2.20	
4.11/2.20	
4.11/2.20	Removed unreachable locations (and leaf rules with constant cost):
4.11/2.20	
4.11/2.20	   Start location: start
4.11/2.20	
4.11/2.20	      5: start -> eval : A'=0, [ A>=1 && B>=A ], cost: 1+A
4.11/2.20	
4.11/2.20	
4.11/2.20	
4.11/2.20	### Computing asymptotic complexity ###
4.11/2.20	
4.11/2.20	
4.11/2.20	
4.11/2.20	Fully simplified ITS problem
4.11/2.20	
4.11/2.20	   Start location: start
4.11/2.20	
4.11/2.20	      5: start -> eval : A'=0, [ A>=1 && B>=A ], cost: 1+A
4.11/2.20	
4.11/2.20	
4.11/2.20	
4.11/2.20	Computing asymptotic complexity for rule 5
4.11/2.20	
4.11/2.20	   Solved the limit problem by the following transformations:
4.11/2.20	
4.11/2.20	      Created initial limit problem:
4.11/2.20	
4.11/2.20	      A (+/+!), 1-A+B (+/+!), 1+A (+) [not solved]
4.11/2.20	
4.11/2.20	
4.11/2.20	
4.11/2.20	      removing all constraints (solved by SMT)
4.11/2.20	
4.11/2.20	      resulting limit problem: [solved]
4.11/2.20	
4.11/2.20	
4.11/2.20	
4.11/2.20	      applying transformation rule (C) using substitution {A==n,B==n}
4.11/2.20	
4.11/2.20	      resulting limit problem:
4.11/2.20	
4.11/2.20	      [solved]
4.11/2.20	
4.11/2.20	
4.11/2.20	
4.11/2.20	   Solution:
4.11/2.20	
4.11/2.20	      A / n
4.11/2.20	
4.11/2.20	      B / n
4.11/2.20	
4.11/2.20	   Resulting cost 1+n has complexity: Poly(n^1)
4.11/2.20	
4.11/2.20	
4.11/2.20	
4.11/2.20	Found new complexity Poly(n^1).
4.11/2.20	
4.11/2.20	
4.11/2.20	
4.11/2.20	Obtained the following overall complexity (w.r.t. the length of the input n):
4.11/2.20	
4.11/2.20	   Complexity:  Poly(n^1)
4.11/2.20	
4.11/2.20	   Cpx degree:  1
4.11/2.20	
4.11/2.20	   Solved cost: 1+n
4.11/2.20	
4.11/2.20	   Rule cost:   1+A
4.11/2.20	
4.11/2.20	   Rule guard:  [ A>=1 && B>=A ]
4.11/2.20	
4.11/2.20	
4.11/2.20	
4.11/2.20	WORST_CASE(Omega(n^1),?)
4.11/2.20	
4.11/2.20	
4.11/2.20	----------------------------------------
4.11/2.20	
4.11/2.20	(4)
4.11/2.20	BOUNDS(n^1, INF)
4.12/2.23	EOF