details

property	value
status	complete
benchmark	LengthOfFiniteLists_nosorts_iGM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n179.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	NTI-TC20-firstrun
configuration	Default 200
runtime (wallclock)	21.2069 seconds
cpu usage	55.6941
user time	51.9727
system time	3.7214
max virtual memory	3.5396896E7
max residence set size	9395748.0

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 = 5] mark(length(zeros)) -> mark(length(zeros)) Let l be the left-hand side and r be the right-hand side of this rule. Let p = epsilon, theta1 = {} and theta2 = {}. We have r|p = mark(length(zeros)) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = mark(length(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 = [active^#(zeros) -> mark^#(cons(0,zeros)), mark^#(zeros) -> active^#(zeros), active^#(and(tt,_0)) -> mark^#(_0), mark^#(cons(_0,_1)) -> active^#(cons(mark(_0),_1)), active^#(length(cons(_0,_1))) -> mark^#(s(length(_1))), mark^#(s(_0)) -> active^#(s(mark(_0))), mark^#(length(_0)) -> active^#(length(mark(_0))), mark^#(and(_0,_1)) -> active^#(and(mark(_0),_1)), mark^#(cons(_0,_1)) -> mark^#(_0), mark^#(and(_0,_1)) -> mark^#(_0), mark^#(length(_0)) -> mark^#(_0), mark^#(s(_0)) -> mark^#(_0)] TRS = {active(zeros) -> mark(cons(0,zeros)), active(and(tt,_0)) -> mark(_0), active(length(nil)) -> mark(0), active(length(cons(_0,_1))) -> mark(s(length(_1))), mark(zeros) -> active(zeros), mark(cons(_0,_1)) -> active(cons(mark(_0),_1)), mark(0) -> active(0), mark(and(_0,_1)) -> active(and(mark(_0),_1)), mark(tt) -> active(tt), mark(length(_0)) -> active(length(mark(_0))), mark(nil) -> active(nil), mark(s(_0)) -> active(s(mark(_0))), cons(mark(_0),_1) -> cons(_0,_1), cons(_0,mark(_1)) -> cons(_0,_1), cons(active(_0),_1) -> cons(_0,_1), cons(_0,active(_1)) -> cons(_0,_1), and(mark(_0),_1) -> and(_0,_1), and(_0,mark(_1)) -> and(_0,_1), and(active(_0),_1) -> and(_0,_1), and(_0,active(_1)) -> and(_0,_1), length(mark(_0)) -> length(_0), length(active(_0)) -> length(_0), s(mark(_0)) -> s(_0), s(active(_0)) -> s(_0)} ## Trying with homeomorphic embeddings... Failed! ## Trying with polynomial interpretations... Too many coefficients (21)! 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, 12 unfolded rules generated. # Iteration 1: nontermination not detected, 94 unfolded rules generated. # Iteration 2: nontermination not detected, 567 unfolded rules generated. # Iteration 3: nontermination not detected, 4562 unfolded rules generated. # Iteration 4: nontermination not detected, 41028 unfolded rules generated. # Iteration 5: nontermination detected, 104825 unfolded rules generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = mark^#(length(_0)) -> active^#(length(mark(_0))) [trans] is in U_IR^0. D = active^#(length(cons(_0,_1))) -> mark^#(s(length(_1))) is a dependency pair of IR. We build a composed triple from L0 and D. ==> L1 = [mark^#(length(_0)) -> active^#(length(mark(_0))), active^#(length(cons(_1,_2))) -> mark^#(s(length(_2)))] [comp] is in U_IR^1. Let p1 = [0]. We unfold the second rule of L1 backwards at position p1 with the rule length(mark(_0)) -> length(_0). ==> L2 = [mark^#(length(_0)) -> active^#(length(mark(_0))), active^#(length(mark(cons(_1,_2)))) -> mark^#(s(length(_2)))] [comp] is in U_IR^2. Let p2 = [0, 0]. We unfold the first rule of L2 forwards at position p2 with the rule mark(zeros) -> active(zeros). ==> L3 = [mark^#(length(zeros)) -> active^#(length(active(zeros))), active^#(length(mark(cons(_0,_1)))) -> mark^#(s(length(_1)))] [comp] is in U_IR^3. Let p3 = [0, 0]. We unfold the first rule of L3 forwards at position p3 with the rule active(zeros) -> mark(cons(0,zeros)). ==> L4 = mark^#(length(zeros)) -> mark^#(s(length(zeros))) [trans] is in U_IR^4. D = mark^#(s(_0)) -> mark^#(_0) is a dependency pair of IR. We build a composed triple from L4 and D. ==> L5 = mark^#(length(zeros)) -> mark^#(length(zeros)) [trans] is in U_IR^5. This DP problem is infinite. ** END proof description ** Proof stopped at iteration 5 Number of unfolded rules generated by this proof = 151088 Number of unfolded rules generated by all the parallel proofs = 354485 popout

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