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TRS Standard pair #487070855
details
property
value
status
complete
benchmark
OvConsOS_nosorts-noand_Z.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n178.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
NTI-TC20-firstrun
configuration
Default 200
runtime (wallclock)
0.427643 seconds
cpu usage
1.11267
user time
1.03682
system time
0.075852
max virtual memory
113188.0
max residence set size
155368.0
stage attributes
key
value
starexec-result
NO
output
NO Prover = TRS(tech=GUIDED_UNF_TRIPLES, nb_unfoldings=unlimited, unfold_variables=false, max_nb_coefficients=12, max_nb_unfolded_rules=-1, strategy=LEFTMOST_NE) ** BEGIN proof argument ** The following rule was generated while unfolding the analyzed TRS: [iteration = 7] length(cons(_0,n__zeros)) -> length(cons(0,n__zeros)) Let l be the left-hand side and r be the right-hand side of this rule. Let p = epsilon, theta1 = {} and theta2 = {_0->0}. We have r|p = length(cons(0,n__zeros)) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = length(cons(_0,n__zeros)) 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... ## Round 1: ## DP problem: Dependency pairs = [U11^#(tt,_0) -> U12^#(tt,activate(_0)), length^#(cons(_0,_1)) -> U11^#(tt,activate(_1)), U12^#(tt,_0) -> length^#(activate(_0))] TRS = {zeros -> cons(0,n__zeros), U11(tt,_0) -> U12(tt,activate(_0)), U12(tt,_0) -> s(length(activate(_0))), U21(tt,_0,_1,_2) -> U22(tt,activate(_0),activate(_1),activate(_2)), U22(tt,_0,_1,_2) -> U23(tt,activate(_0),activate(_1),activate(_2)), U23(tt,_0,_1,_2) -> cons(activate(_2),n__take(activate(_1),activate(_0))), length(nil) -> 0, length(cons(_0,_1)) -> U11(tt,activate(_1)), take(0,_0) -> nil, take(s(_0),cons(_1,_2)) -> U21(tt,activate(_2),_0,_1), zeros -> n__zeros, take(_0,_1) -> n__take(_0,_1), activate(n__zeros) -> zeros, activate(n__take(_0,_1)) -> take(_0,_1), activate(_0) -> _0} ## Trying with homeomorphic embeddings... Failed! ## Trying with polynomial interpretations... Too many coefficients (80)! Aborting! ## Trying with lexicographic path orders... Too many argument filtering possibilities (243000)! Aborting! ## Trying to prove nontermination by unfolding the dependency pairs with the rules of the TRS # max_depth=3, unfold_variables=false: # Iteration 0: nontermination not detected, 3 unfolded rules generated. # Iteration 1: nontermination not detected, 5 unfolded rules generated. # Iteration 2: nontermination not detected, 21 unfolded rules generated. # Iteration 3: nontermination not detected, 46 unfolded rules generated. # Iteration 4: nontermination not detected, 57 unfolded rules generated. # Iteration 5: nontermination not detected, 16 unfolded rules generated. # Iteration 6: nontermination not detected, 36 unfolded rules generated. # Iteration 7: nontermination detected, 69 unfolded rules generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = length^#(cons(_0,_1)) -> U11^#(tt,activate(_1)) [trans] is in U_IR^0. D = U11^#(tt,_0) -> U12^#(tt,activate(_0)) is a dependency pair of IR. We build a composed triple from L0 and D. ==> L1 = [length^#(cons(_0,_1)) -> U11^#(tt,activate(_1)), U11^#(tt,_2) -> U12^#(tt,activate(_2))] [comp] is in U_IR^1. Let p1 = [1]. We unfold the first rule of L1 forwards at position p1 with the rule activate(n__zeros) -> zeros. ==> L2 = length^#(cons(_0,n__zeros)) -> U12^#(tt,activate(zeros)) [trans] is in U_IR^2. D = U12^#(tt,_0) -> length^#(activate(_0)) is a dependency pair of IR. We build a composed triple from L2 and D. ==> L3 = length^#(cons(_0,n__zeros)) -> length^#(activate(activate(zeros))) [trans] is in U_IR^3. We build a unit triple from L3. ==> L4 = length^#(cons(_0,n__zeros)) -> length^#(activate(activate(zeros))) [unit] is in U_IR^4. Let p4 = [0]. We unfold the rule of L4 forwards at position p4 with the rule activate(_0) -> _0. ==> L5 = length^#(cons(_0,n__zeros)) -> length^#(activate(zeros)) [unit] is in U_IR^5. Let p5 = [0]. We unfold the rule of L5 forwards at position p5 with the rule activate(_0) -> _0. ==> L6 = length^#(cons(_0,n__zeros)) -> length^#(zeros) [unit] is in U_IR^6. Let p6 = [0]. We unfold the rule of L6 forwards at position p6 with the rule zeros -> cons(0,n__zeros). ==> L7 = length^#(cons(_0,n__zeros)) -> length^#(cons(0,n__zeros)) [unit] is in U_IR^7. This DP problem is infinite. ** END proof description ** Proof stopped at iteration 7 Number of unfolded rules generated by this proof = 253 Number of unfolded rules generated by all the parallel proofs = 1608
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