details

property	value
status	complete
benchmark	OvConsOS_nosorts_noand_GM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n144.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	NTI_22
configuration	default
runtime (wallclock)	0.808251857758 seconds
cpu usage	2.200878691
max memory	4.88644608E8

stage attributes

key	value
output-size	5496
starexec-result	NO

output

/export/starexec/sandbox/solver/bin/starexec_run_default /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- NO ** BEGIN proof argument ** The following rule was generated while unfolding the analyzed TRS: [iteration = 4] a__U11(tt,zeros) -> a__U11(tt,zeros) Let l be the left-hand side and r be the right-hand side of this rule. Let p = epsilon, theta1 = {} and theta2 = {}. We have r|p = a__U11(tt,zeros) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = a__U11(tt,zeros) loops w.r.t. the analyzed TRS. ** END proof argument ** ** BEGIN proof description ** ## Searching for a generalized rewrite rule (a rule whose right-hand side contains a variable that does not occur in the left-hand side)... No generalized rewrite rule found! ## Applying the DP framework... ## 1 initial DP problem to solve. ## First, we try to decompose this problem into smaller problems. ## Round 1 [1 DP problem]: ## DP problem: Dependency pairs = [a__U11^#(tt,_0) -> a__U12^#(tt,_0), mark^#(U11(_0,_1)) -> a__U11^#(mark(_0),_1), a__length^#(cons(_0,_1)) -> a__U11^#(tt,_1), mark^#(length(_0)) -> a__length^#(mark(_0)), a__U12^#(tt,_0) -> a__length^#(mark(_0)), mark^#(U12(_0,_1)) -> a__U12^#(mark(_0),_1), a__U23^#(tt,_0,_1,_2) -> mark^#(_2), mark^#(U23(_0,_1,_2,_3)) -> a__U23^#(mark(_0),_1,_2,_3), a__U22^#(tt,_0,_1,_2) -> a__U23^#(tt,_0,_1,_2), mark^#(U22(_0,_1,_2,_3)) -> a__U22^#(mark(_0),_1,_2,_3), a__U21^#(tt,_0,_1,_2) -> a__U22^#(tt,_0,_1,_2), mark^#(U21(_0,_1,_2,_3)) -> a__U21^#(mark(_0),_1,_2,_3), a__take^#(s(_0),cons(_1,_2)) -> a__U21^#(tt,_2,_0,_1), mark^#(take(_0,_1)) -> a__take^#(mark(_0),mark(_1)), mark^#(U11(_0,_1)) -> mark^#(_0), mark^#(U12(_0,_1)) -> mark^#(_0), mark^#(length(_0)) -> mark^#(_0), mark^#(U21(_0,_1,_2,_3)) -> mark^#(_0), mark^#(U22(_0,_1,_2,_3)) -> mark^#(_0), mark^#(U23(_0,_1,_2,_3)) -> mark^#(_0), mark^#(take(_0,_1)) -> mark^#(_0), mark^#(take(_0,_1)) -> mark^#(_1), mark^#(cons(_0,_1)) -> mark^#(_0), mark^#(s(_0)) -> mark^#(_0), a__U12^#(tt,_0) -> mark^#(_0)] TRS = {a__zeros -> cons(0,zeros), a__U11(tt,_0) -> a__U12(tt,_0), a__U12(tt,_0) -> s(a__length(mark(_0))), a__U21(tt,_0,_1,_2) -> a__U22(tt,_0,_1,_2), a__U22(tt,_0,_1,_2) -> a__U23(tt,_0,_1,_2), a__U23(tt,_0,_1,_2) -> cons(mark(_2),take(_1,_0)), a__length(nil) -> 0, a__length(cons(_0,_1)) -> a__U11(tt,_1), a__take(0,_0) -> nil, a__take(s(_0),cons(_1,_2)) -> a__U21(tt,_2,_0,_1), mark(zeros) -> a__zeros, mark(U11(_0,_1)) -> a__U11(mark(_0),_1), mark(U12(_0,_1)) -> a__U12(mark(_0),_1), mark(length(_0)) -> a__length(mark(_0)), mark(U21(_0,_1,_2,_3)) -> a__U21(mark(_0),_1,_2,_3), mark(U22(_0,_1,_2,_3)) -> a__U22(mark(_0),_1,_2,_3), mark(U23(_0,_1,_2,_3)) -> a__U23(mark(_0),_1,_2,_3), mark(take(_0,_1)) -> a__take(mark(_0),mark(_1)), mark(cons(_0,_1)) -> cons(mark(_0),_1), mark(0) -> 0, mark(tt) -> tt, mark(s(_0)) -> s(mark(_0)), mark(nil) -> nil, a__zeros -> zeros, a__U11(_0,_1) -> U11(_0,_1), a__U12(_0,_1) -> U12(_0,_1), a__length(_0) -> length(_0), a__U21(_0,_1,_2,_3) -> U21(_0,_1,_2,_3), a__U22(_0,_1,_2,_3) -> U22(_0,_1,_2,_3), a__U23(_0,_1,_2,_3) -> U23(_0,_1,_2,_3), a__take(_0,_1) -> take(_0,_1)} ## Trying with homeomorphic embeddings... Failed! ## Trying with polynomial interpretations... This DP problem is too complex! Aborting! ## Trying with lexicographic path orders... Too many argument filtering possibilities (546750000)! Aborting! ## Trying with Knuth-Bendix orders... This DP problem is too complex! Aborting! Don't know whether this DP problem is finite. ## A DP problem could not be proved finite. ## Now, we try to prove that this problem is infinite. ## Trying to find a loop (forward=true, backward=false, max=-1) # max_depth=3, unfold_variables=false: # Iteration 0: no loop found, 25 unfolded rules generated. # Iteration 1: no loop found, 409 unfolded rules generated. # Iteration 2: no loop found, 2261 unfolded rules generated. # Iteration 3: no loop found, 4257 unfolded rules generated. # Iteration 4: success, found a loop, 3 unfolded rules generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = a__U11^#(tt,_0) -> a__U12^#(tt,_0) [trans] is in U_IR^0. D = a__U12^#(tt,_0) -> a__length^#(mark(_0)) is a dependency pair of IR. We build a composed triple from L0 and D. ==> L1 = a__U11^#(tt,_0) -> a__length^#(mark(_0)) [trans] is in U_IR^1. D = a__length^#(cons(_0,_1)) -> a__U11^#(tt,_1) is a dependency pair of IR. We build a composed triple from L1 and D. ==> L2 = [a__U11^#(tt,_0) -> a__length^#(mark(_0)), a__length^#(cons(_1,_2)) -> a__U11^#(tt,_2)] [comp] is in U_IR^2. Let p2 = [0]. We unfold the first rule of L2 forwards at position p2 with the rule mark(zeros) -> a__zeros. ==> L3 = [a__U11^#(tt,zeros) -> a__length^#(a__zeros), a__length^#(cons(_0,_1)) -> a__U11^#(tt,_1)] [comp] is in U_IR^3. Let p3 = [0]. We unfold the first rule of L3 forwards at position p3 with the rule a__zeros -> cons(0,zeros). ==> L4 = a__U11^#(tt,zeros) -> a__U11^#(tt,zeros) [trans] is in U_IR^4. This DP problem is infinite. Proof run on Linux version 3.10.0-1160.25.1.el7.x86_64 for amd64 using Java version 1.8.0_292 ** END proof description ** Total number of generated unfolded rules = 15872 popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to TRS Standard