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	n074.star.cs.uiowa.edu
space	AProVE_06

run statistics

property	value
solver	NTI-TC20-firstrun
configuration	Default 200
runtime (wallclock)	1.12527 seconds
cpu usage	3.82457
user time	3.62919
system time	0.195388
max virtual memory	3.5396896E7
max residence set size	598084.0

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 = 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... ## Round 1: ## 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... Too many coefficients (14)! Aborting! ## Trying with lexicographic path orders... Failed! ## Trying to prove nontermination by unfolding the dependency pairs with the rules of the TRS # max_depth=20, unfold_variables=false: # Iteration 0: nontermination not detected, 1 unfolded rule generated. # Iteration 1: nontermination not detected, 1 unfolded rule generated. # Iteration 2: nontermination not detected, 1 unfolded rule generated. # Iteration 3: nontermination not detected, 1 unfolded rule generated. # Iteration 4: nontermination not detected, 1 unfolded rule generated. # Iteration 5: nontermination not detected, 1 unfolded rule generated. # Iteration 6: nontermination not detected, 1 unfolded rule generated. # Iteration 7: nontermination not detected, 1 unfolded rule generated. # Iteration 8: nontermination not detected, 1 unfolded rule generated. # Iteration 9: nontermination detected, 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. ** END proof description ** Proof stopped at iteration 9 Number of unfolded rules generated by this proof = 10 Number of unfolded rules generated by all the parallel proofs = 2378 popout

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