details

property	value
status	complete
benchmark	#4.13.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n045.star.cs.uiowa.edu
space	Strategy_removed_AG01

run statistics

property	value
solver	NTI_22
configuration	default
runtime (wallclock)	0.344091892242 seconds
cpu usage	0.617529575
max memory	7.602176E7

stage attributes

key	value
output-size	3583
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 = 5] f(g(0,_0,_0),g(2,1,1),g(0,_0,_0)) -> f(g(0,_0,_0),g(2,_0,_0),g(0,_0,_0)) Let l be the left-hand side and r be the right-hand side of this rule. Let p = epsilon, theta1 = {_0->1} and theta2 = {}. We have r|p = f(g(0,_0,_0),g(2,_0,_0),g(0,_0,_0)) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = f(g(0,1,1),g(2,1,1),g(0,1,1)) 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 = [f^#(0,1,_0) -> f^#(_0,_0,_0)] TRS = {f(0,1,_0) -> f(_0,_0,_0), f(_0,_1,_2) -> 2, 0 -> 2, 1 -> 2, g(_0,_0,_1) -> _1, g(_0,_1,_1) -> _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=true, max=20) # max_depth=20, unfold_variables=false: # Iteration 0: no loop found, 1 unfolded rule generated. # Iteration 1: no loop found, 1 unfolded rule generated. # Iteration 2: no loop found, 5 unfolded rules generated. # Iteration 3: no loop found, 24 unfolded rules generated. # Iteration 4: no loop found, 138 unfolded rules generated. # Iteration 5: success, found a loop, 1044 unfolded rules generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = f^#(0,1,_0) -> f^#(_0,_0,_0) [trans] is in U_IR^0. We build a unit triple from L0. ==> L1 = f^#(0,1,_0) -> f^#(_0,_0,_0) [unit] is in U_IR^1. Let p1 = [0]. We unfold the rule of L1 backwards at position p1 with the rule g(_0,_1,_1) -> _0. ==> L2 = f^#(g(0,_0,_0),1,g(0,_0,_0)) -> f^#(g(0,_0,_0),g(0,_0,_0),g(0,_0,_0)) [unit] is in U_IR^2. Let p2 = [1]. We unfold the rule of L2 backwards at position p2 with the rule g(_0,_0,_1) -> _1. ==> L3 = f^#(g(0,_0,_0),g(_1,_1,1),g(0,_0,_0)) -> f^#(g(0,_0,_0),g(0,_0,_0),g(0,_0,_0)) [unit] is in U_IR^3. Let p3 = [1, 0]. We unfold the rule of L3 forwards at position p3 with the rule 0 -> 2. ==> L4 = f^#(g(0,_0,_0),g(2,2,1),g(0,_0,_0)) -> f^#(g(0,_0,_0),g(2,_0,_0),g(0,_0,_0)) [unit] is in U_IR^4. Let p4 = [1, 1]. We unfold the rule of L4 backwards at position p4 with the rule 1 -> 2. ==> L5 = f^#(g(0,_0,_0),g(2,1,1),g(0,_0,_0)) -> f^#(g(0,_0,_0),g(2,_0,_0),g(0,_0,_0)) [unit] is in U_IR^5. 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 = 1223 popout

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

job log

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actions

all output return to TRS Standard