details

property	value
status	complete
benchmark	Ex14_Luc06_iGM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n186.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	NTI-TC20-firstrun
configuration	Default 200
runtime (wallclock)	5.75644 seconds
cpu usage	18.9884
user time	18.1931
system time	0.795292
max virtual memory	3.539934E7
max residence set size	2377208.0

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] mark(h(a)) -> mark(h(a)) 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(h(a)) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = mark(h(a)) 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^#(h(_0)) -> mark^#(g(_0,_0)), mark^#(h(_0)) -> active^#(h(mark(_0))), active^#(g(a,_0)) -> mark^#(f(b,_0)), mark^#(g(_0,_1)) -> active^#(g(mark(_0),_1)), active^#(f(_0,_0)) -> mark^#(h(a)), mark^#(f(_0,_1)) -> active^#(f(mark(_0),_1)), mark^#(h(_0)) -> mark^#(_0), mark^#(g(_0,_1)) -> mark^#(_0), mark^#(f(_0,_1)) -> mark^#(_0)] TRS = {active(h(_0)) -> mark(g(_0,_0)), active(g(a,_0)) -> mark(f(b,_0)), active(f(_0,_0)) -> mark(h(a)), active(a) -> mark(b), mark(h(_0)) -> active(h(mark(_0))), mark(g(_0,_1)) -> active(g(mark(_0),_1)), mark(a) -> active(a), mark(f(_0,_1)) -> active(f(mark(_0),_1)), mark(b) -> active(b), h(mark(_0)) -> h(_0), h(active(_0)) -> h(_0), g(mark(_0),_1) -> g(_0,_1), g(_0,mark(_1)) -> g(_0,_1), g(active(_0),_1) -> g(_0,_1), g(_0,active(_1)) -> g(_0,_1), f(mark(_0),_1) -> f(_0,_1), f(_0,mark(_1)) -> f(_0,_1), f(active(_0),_1) -> f(_0,_1), f(_0,active(_1)) -> f(_0,_1)} ## Trying with homeomorphic embeddings... Failed! ## Trying with polynomial interpretations... Too many coefficients (17)! 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, 9 unfolded rules generated. # Iteration 1: nontermination not detected, 47 unfolded rules generated. # Iteration 2: nontermination not detected, 103 unfolded rules generated. # Iteration 3: nontermination not detected, 223 unfolded rules generated. # Iteration 4: nontermination not detected, 624 unfolded rules generated. # Iteration 5: nontermination not detected, 1333 unfolded rules generated. # Iteration 6: nontermination not detected, 2326 unfolded rules generated. # Iteration 7: nontermination not detected, 4119 unfolded rules generated. # Iteration 8: nontermination not detected, 6996 unfolded rules generated. # Iteration 9: nontermination detected, 315 unfolded rules generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = mark^#(h(_0)) -> active^#(h(mark(_0))) [trans] is in U_IR^0. D = active^#(h(_0)) -> mark^#(g(_0,_0)) is a dependency pair of IR. We build a composed triple from L0 and D. ==> L1 = [mark^#(h(_0)) -> active^#(h(mark(_0))), active^#(h(_1)) -> mark^#(g(_1,_1))] [comp] is in U_IR^1. Let p1 = [0, 0]. We unfold the first rule of L1 forwards at position p1 with the rule mark(a) -> active(a). ==> L2 = mark^#(h(a)) -> mark^#(g(active(a),active(a))) [trans] is in U_IR^2. D = mark^#(g(_0,_1)) -> active^#(g(mark(_0),_1)) is a dependency pair of IR. We build a composed triple from L2 and D. ==> L3 = [mark^#(h(a)) -> mark^#(g(active(a),active(a))), mark^#(g(_0,_1)) -> active^#(g(mark(_0),_1))] [comp] is in U_IR^3. Let p3 = [0]. We unfold the first rule of L3 forwards at position p3 with the rule g(active(_0),_1) -> g(_0,_1). ==> L4 = mark^#(h(a)) -> active^#(g(mark(a),active(a))) [trans] is in U_IR^4. D = active^#(g(a,_0)) -> mark^#(f(b,_0)) is a dependency pair of IR. We build a composed triple from L4 and D. ==> L5 = [mark^#(h(a)) -> active^#(g(mark(a),active(a))), active^#(g(a,_0)) -> mark^#(f(b,_0))] [comp] is in U_IR^5. Let p5 = [0]. We unfold the first rule of L5 forwards at position p5 with the rule g(mark(_0),_1) -> g(_0,_1). ==> L6 = [mark^#(h(a)) -> active^#(g(a,active(a))), active^#(g(a,_0)) -> mark^#(f(b,_0))] [comp] is in U_IR^6. Let p6 = [0, 1]. We unfold the first rule of L6 forwards at position p6 with the rule active(a) -> mark(b). ==> L7 = mark^#(h(a)) -> mark^#(f(b,mark(b))) [trans] is in U_IR^7. D = mark^#(f(_0,_1)) -> active^#(f(mark(_0),_1)) is a dependency pair of IR. We build a composed triple from L7 and D. ==> L8 = mark^#(h(a)) -> active^#(f(mark(b),mark(b))) [trans] is in U_IR^8. D = active^#(f(_0,_0)) -> mark^#(h(a)) is a dependency pair of IR. We build a composed triple from L8 and D. ==> L9 = mark^#(h(a)) -> mark^#(h(a)) [trans] 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 = 16095 Number of unfolded rules generated by all the parallel proofs = 83578 popout

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all output return to TRS Standard