details

property	value
status	complete
benchmark	#4.18.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	Strategy_removed_AG01

run statistics

property	value
solver	NTI_22
configuration	default
runtime (wallclock)	0.264547109604 seconds
cpu usage	0.260160047
max memory	4.0210432E7

stage attributes

key	value
output-size	2881
starexec-result	NO

output

/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO ** BEGIN proof argument ** The following rule was generated while unfolding the analyzed TRS: [iteration = 2] a(b(a(b(a(b(_0)))))) -> a(b(a(b(a(b(a(a(b(_0))))))))) Let l be the left-hand side and r be the right-hand side of this rule. Let p = epsilon, theta1 = {} and theta2 = {_0->a(a(b(_0)))}. We have r|p = a(b(a(b(a(b(a(a(b(_0))))))))) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = a(b(a(b(a(b(_0)))))) 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^#(b(a(b(_0)))) -> a^#(b(a(a(b(_0))))), a^#(b(a(b(_0)))) -> a^#(a(b(_0))), a^#(b(a(b(_0)))) -> a^#(b(_0))] TRS = {a(b(a(b(_0)))) -> b(a(b(a(a(b(_0))))))} ## Trying with homeomorphic embeddings... Failed! ## Trying with polynomial interpretations... This DP problem is too complex! Aborting! ## Trying with lexicographic path orders... Failed! ## Trying with Knuth-Bendix orders... Failed! 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=6, unfold_variables=false: # Iteration 0: no loop found, 3 unfolded rules generated. # Iteration 1: no loop found, 9 unfolded rules generated. # Iteration 2: success, found a loop, 1 unfolded rule generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = a^#(b(a(b(_0)))) -> a^#(b(a(a(b(_0))))) [trans] is in U_IR^0. D = a^#(b(a(b(_0)))) -> a^#(a(b(_0))) is a dependency pair of IR. We build a composed triple from L0 and D. ==> L1 = [a^#(b(a(b(_0)))) -> a^#(b(a(a(b(_0))))), a^#(b(a(b(_1)))) -> a^#(a(b(_1)))] [comp] is in U_IR^1. Let p1 = [0, 0, 0]. We unfold the first rule of L1 forwards at position p1 with the rule a(b(a(b(_0)))) -> b(a(b(a(a(b(_0)))))). ==> L2 = [a^#(b(a(b(a(b(_0)))))) -> a^#(b(a(b(a(b(a(a(b(_0))))))))), a^#(b(a(b(_1)))) -> a^#(a(b(_1)))] [comp] is in U_IR^2. 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 = 37 popout

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job log

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actions

all output return to TRS Standard