details

property	value
status	complete
benchmark	#4.17.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)	139.245783091 seconds
cpu usage	140.546416408
max memory	5.5566336E8

stage attributes

key	value
output-size	3435
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 = 4] f(g(s(0)),s(0),_0) -> f(g(s(0)),_0,s(0)) Let l be the left-hand side and r be the right-hand side of this rule. Let p = epsilon, theta1 = {_0->s(0)} and theta2 = {}. We have r|p = f(g(s(0)),_0,s(0)) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = f(g(s(0)),s(0),s(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... ## 2 initial DP problems to solve. ## First, we try to decompose these problems into smaller problems. ## Round 1 [2 DP problems]: ## DP problem: Dependency pairs = [f^#(g(_0),s(0),_1) -> f^#(g(s(0)),_1,g(_0))] TRS = {f(g(_0),s(0),_1) -> f(g(s(0)),_1,g(_0)), g(s(_0)) -> s(g(_0)), g(0) -> 0} ## Trying with homeomorphic embeddings... Failed! ## Trying with polynomial interpretations... Failed! ## Trying with lexicographic path orders... Failed! ## Trying with Knuth-Bendix orders... Failed! Don't know whether this DP problem is finite. ## DP problem: Dependency pairs = [g^#(s(_0)) -> g^#(_0)] TRS = {f(g(_0),s(0),_1) -> f(g(s(0)),_1,g(_0)), g(s(_0)) -> s(g(_0)), g(0) -> 0} ## Trying with homeomorphic embeddings... Success! 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, 1 unfolded rule generated. # Iteration 1: no loop found, 1 unfolded rule generated. # Iteration 2: no loop found, 2 unfolded rules generated. # Iteration 3: no loop found, 6 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 = f^#(g(_0),s(0),_1) -> f^#(g(s(0)),_1,g(_0)) [trans] is in U_IR^0. We build a unit triple from L0. ==> L1 = f^#(g(_0),s(0),_1) -> f^#(g(s(0)),_1,g(_0)) [unit] is in U_IR^1. Let p1 = [0, 0]. The subterm at position p1 in the left-hand side of the rule of L1 unifies with the subterm at position p1 in the right-hand side of the rule of L1. ==> L2 = f^#(g(s(0)),s(0),_0) -> f^#(g(s(0)),_0,g(s(0))) [unit] is in U_IR^2. Let p2 = [2]. We unfold the rule of L2 forwards at position p2 with the rule g(s(_0)) -> s(g(_0)). ==> L3 = f^#(g(s(0)),s(0),_0) -> f^#(g(s(0)),_0,s(g(0))) [unit] is in U_IR^3. Let p3 = [2, 0]. We unfold the rule of L3 forwards at position p3 with the rule g(0) -> 0. ==> L4 = f^#(g(s(0)),s(0),_0) -> f^#(g(s(0)),_0,s(0)) [unit] 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 = 28 popout

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job log

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actions

all output return to TRS Standard