details

property	value
status	complete
benchmark	Ex24_Luc06_iGM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n185.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	NTI-TC20-firstrun
configuration	Default 200
runtime (wallclock)	0.545152 seconds
cpu usage	1.21254
user time	1.11239
system time	0.100144
max virtual memory	113188.0
max residence set size	180160.0

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] active(f(b,mark(c),c)) -> active(f(b,mark(c),c)) 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 = active(f(b,mark(c),c)) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = active(f(b,mark(c),c)) 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^#(f(b,_0,c)) -> mark^#(f(_0,c,_0)), mark^#(f(_0,_1,_2)) -> active^#(f(_0,mark(_1),_2)), mark^#(f(_0,_1,_2)) -> mark^#(_1)] TRS = {active(f(b,_0,c)) -> mark(f(_0,c,_0)), active(c) -> mark(b), mark(f(_0,_1,_2)) -> active(f(_0,mark(_1),_2)), mark(b) -> active(b), mark(c) -> active(c), f(mark(_0),_1,_2) -> f(_0,_1,_2), f(_0,mark(_1),_2) -> f(_0,_1,_2), f(_0,_1,mark(_2)) -> f(_0,_1,_2), f(active(_0),_1,_2) -> f(_0,_1,_2), f(_0,active(_1),_2) -> f(_0,_1,_2), f(_0,_1,active(_2)) -> f(_0,_1,_2)} ## Trying with homeomorphic embeddings... Failed! ## Trying with polynomial interpretations... Too many coefficients (15)! Aborting! ## Trying with lexicographic path orders... Failed! ## 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, 8 unfolded rules generated. # Iteration 2: nontermination not detected, 48 unfolded rules generated. # Iteration 3: nontermination not detected, 134 unfolded rules generated. # Iteration 4: nontermination not detected, 279 unfolded rules generated. # Iteration 5: nontermination detected, 26 unfolded rules generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = active^#(f(b,_0,c)) -> mark^#(f(_0,c,_0)) [trans] is in U_IR^0. D = mark^#(f(_0,_1,_2)) -> active^#(f(_0,mark(_1),_2)) is a dependency pair of IR. We build a composed triple from L0 and D. ==> L1 = [active^#(f(b,_0,c)) -> mark^#(f(_0,c,_0)), mark^#(f(_1,_2,_3)) -> active^#(f(_1,mark(_2),_3))] [comp] is in U_IR^1. Let p1 = [0]. We unfold the first rule of L1 forwards at position p1 with the rule f(_0,_1,mark(_2)) -> f(_0,_1,_2). ==> L2 = [active^#(f(b,mark(_0),c)) -> mark^#(f(mark(_0),c,_0)), mark^#(f(_1,_2,_3)) -> active^#(f(_1,mark(_2),_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 mark(c) -> active(c). ==> L3 = [active^#(f(b,mark(c),c)) -> mark^#(f(active(c),c,c)), mark^#(f(_0,_1,_2)) -> active^#(f(_0,mark(_1),_2))] [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(c) -> mark(b). ==> L4 = [active^#(f(b,mark(c),c)) -> mark^#(f(mark(b),c,c)), mark^#(f(_0,_1,_2)) -> active^#(f(_0,mark(_1),_2))] [comp] is in U_IR^4. Let p4 = [0]. We unfold the first rule of L4 forwards at position p4 with the rule f(mark(_0),_1,_2) -> f(_0,_1,_2). ==> L5 = active^#(f(b,mark(c),c)) -> active^#(f(b,mark(c),c)) [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 = 498 Number of unfolded rules generated by all the parallel proofs = 2247 popout

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all output return to TRS Standard