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TRS Standard pair #516963252
details
property
value
status
complete
benchmark
Ex4_4_Luc96b_FR.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n035.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
NTI_22
configuration
default
runtime (wallclock)
11.2204270363 seconds
cpu usage
11.531986672
max memory
5.42314496E8
stage attributes
key
value
output-size
3837
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] activate(n__f(n__g(g(_0)),_1)) -> activate(n__f(n__g(g(_0)),activate(_1))) Let l be the left-hand side and r be the right-hand side of this rule. Let p = epsilon, theta1 = {} and theta2 = {_1->activate(_1)}. We have r|p = activate(n__f(n__g(g(_0)),activate(_1))) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = activate(n__f(n__g(g(_0)),_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^#(g(_0),_1) -> f^#(_0,n__f(n__g(_0),activate(_1))), activate^#(n__f(_0,_1)) -> f^#(activate(_0),_1), activate^#(n__f(_0,_1)) -> activate^#(_0), activate^#(n__g(_0)) -> activate^#(_0), f^#(g(_0),_1) -> activate^#(_1)] TRS = {f(g(_0),_1) -> f(_0,n__f(n__g(_0),activate(_1))), f(_0,_1) -> n__f(_0,_1), g(_0) -> n__g(_0), activate(n__f(_0,_1)) -> f(activate(_0),_1), activate(n__g(_0)) -> g(activate(_0)), activate(_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. ## 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, 5 unfolded rules generated. # Iteration 1: no loop found, 18 unfolded rules generated. # Iteration 2: no loop found, 31 unfolded rules generated. # Iteration 3: no loop found, 34 unfolded rules generated. # Iteration 4: success, found a loop, 10 unfolded rules generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = activate^#(n__f(_0,_1)) -> f^#(activate(_0),_1) [trans] is in U_IR^0. D = f^#(g(_0),_1) -> f^#(_0,n__f(n__g(_0),activate(_1))) is a dependency pair of IR. We build a composed triple from L0 and D. ==> L1 = [activate^#(n__f(_0,_1)) -> f^#(activate(_0),_1), f^#(g(_2),_3) -> f^#(_2,n__f(n__g(_2),activate(_3)))] [comp] is in U_IR^1. Let p1 = [0]. We unfold the first rule of L1 forwards at position p1 with the rule activate(n__g(_0)) -> g(activate(_0)). ==> L2 = [activate^#(n__f(n__g(_0),_1)) -> f^#(g(activate(_0)),_1), f^#(g(_2),_3) -> f^#(_2,n__f(n__g(_2),activate(_3)))] [comp] is in U_IR^2. Let p2 = [0, 0]. We unfold the first rule of L2 forwards at position p2 with the rule activate(_0) -> _0. ==> L3 = activate^#(n__f(n__g(_0),_1)) -> f^#(_0,n__f(n__g(_0),activate(_1))) [trans] is in U_IR^3. D = f^#(g(_0),_1) -> activate^#(_1) is a dependency pair of IR. We build a composed triple from L3 and D. ==> L4 = activate^#(n__f(n__g(g(_0)),_1)) -> activate^#(n__f(n__g(g(_0)),activate(_1))) [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 = 438
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