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TRS Standard pair #516967592
details
property
value
status
complete
benchmark
nonterm.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n063.star.cs.uiowa.edu
space
AProVE_06
run statistics
property
value
solver
NTI_22
configuration
default
runtime (wallclock)
0.465870857239 seconds
cpu usage
0.948320858
max memory
1.70242048E8
stage attributes
key
value
output-size
5437
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 = 9] f(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f(s(s(s(s(s(s(s(s(id(_0))))))))),_1,_1) Let l be the left-hand side and r be the right-hand side of this rule. Let p = epsilon, theta1 = {} and theta2 = {_0->id(_0)}. We have r|p = f(s(s(s(s(s(s(s(s(id(_0))))))))),_1,_1) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = f(s(s(s(s(s(s(s(s(_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... ## 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^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(id(s(s(s(s(s(s(s(s(_0))))))))),_1,_1)] TRS = {f(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f(id(s(s(s(s(s(s(s(s(_0))))))))),_1,_1), id(s(_0)) -> s(id(_0)), id(0) -> 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. ## DP problem: Dependency pairs = [id^#(s(_0)) -> id^#(_0)] TRS = {f(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f(id(s(s(s(s(s(s(s(s(_0))))))))),_1,_1), id(s(_0)) -> s(id(_0)), id(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=10, 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, 1 unfolded rule generated. # Iteration 3: no loop found, 1 unfolded rule generated. # Iteration 4: no loop found, 1 unfolded rule generated. # Iteration 5: no loop found, 1 unfolded rule generated. # Iteration 6: no loop found, 1 unfolded rule generated. # Iteration 7: no loop found, 1 unfolded rule generated. # Iteration 8: no loop found, 1 unfolded rule generated. # Iteration 9: success, found a loop, 1 unfolded rule generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(id(s(s(s(s(s(s(s(s(_0))))))))),_1,_1) [trans] is in U_IR^0. We build a unit triple from L0. ==> L1 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(id(s(s(s(s(s(s(s(s(_0))))))))),_1,_1) [unit] is in U_IR^1. Let p1 = [0]. We unfold the rule of L1 forwards at position p1 with the rule id(s(_0)) -> s(id(_0)). ==> L2 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(id(s(s(s(s(s(s(s(_0))))))))),_1,_1) [unit] is in U_IR^2. Let p2 = [0, 0]. We unfold the rule of L2 forwards at position p2 with the rule id(s(_0)) -> s(id(_0)). ==> L3 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(s(id(s(s(s(s(s(s(_0))))))))),_1,_1) [unit] is in U_IR^3. Let p3 = [0, 0, 0]. We unfold the rule of L3 forwards at position p3 with the rule id(s(_0)) -> s(id(_0)). ==> L4 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(s(s(id(s(s(s(s(s(_0))))))))),_1,_1) [unit] is in U_IR^4. Let p4 = [0, 0, 0, 0]. We unfold the rule of L4 forwards at position p4 with the rule id(s(_0)) -> s(id(_0)). ==> L5 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(s(s(s(id(s(s(s(s(_0))))))))),_1,_1) [unit] is in U_IR^5. Let p5 = [0, 0, 0, 0, 0]. We unfold the rule of L5 forwards at position p5 with the rule id(s(_0)) -> s(id(_0)). ==> L6 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(s(s(s(s(id(s(s(s(_0))))))))),_1,_1) [unit] is in U_IR^6. Let p6 = [0, 0, 0, 0, 0, 0]. We unfold the rule of L6 forwards at position p6 with the rule id(s(_0)) -> s(id(_0)). ==> L7 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(s(s(s(s(s(id(s(s(_0))))))))),_1,_1) [unit] is in U_IR^7. Let p7 = [0, 0, 0, 0, 0, 0, 0]. We unfold the rule of L7 forwards at position p7 with the rule id(s(_0)) -> s(id(_0)). ==> L8 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(s(s(s(s(s(s(id(s(_0))))))))),_1,_1) [unit] is in U_IR^8. Let p8 = [0, 0, 0, 0, 0, 0, 0, 0]. We unfold the rule of L8 forwards at position p8 with the rule id(s(_0)) -> s(id(_0)). ==> L9 = f^#(s(s(s(s(s(s(s(s(_0)))))))),_1,_1) -> f^#(s(s(s(s(s(s(s(s(id(_0))))))))),_1,_1) [unit] is in U_IR^9. 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 = 19
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