details

property	value
status	complete
benchmark	nonTermF.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n007.star.cs.uiowa.edu
space	Applicative_05

run statistics

property	value
solver	NTI_22
configuration	default
runtime (wallclock)	0.454447984695 seconds
cpu usage	0.867355365
max memory	9.29792E7

stage attributes

key	value
output-size	4094
starexec-result	NO

output

/export/starexec/sandbox2/solver/bin/starexec_run_default /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- NO ** BEGIN proof argument ** The following rule was generated while unfolding the analyzed TRS: [iteration = 3] ap(ap(ap(foldr,ap(f,_0)),_1),ap(ap(cons,_0),nil)) -> ap(ap(ap(_0,ap(f,_0)),ap(ap(cons,_0),nil)),_1) Let l be the left-hand side and r be the right-hand side of this rule. Let p = epsilon, theta1 = {_0->foldr, _1->ap(ap(cons,foldr),nil)} and theta2 = {}. We have r|p = ap(ap(ap(_0,ap(f,_0)),ap(ap(cons,_0),nil)),_1) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = ap(ap(ap(foldr,ap(f,foldr)),ap(ap(cons,foldr),nil)),ap(ap(cons,foldr),nil)) 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... ## 1 initial DP problem to solve. ## First, we try to decompose this problem into smaller problems. ## Round 1 [1 DP problem]: ## DP problem: Dependency pairs = [ap^#(ap(f,_0),_0) -> ap^#(ap(_0,ap(f,_0)),ap(ap(cons,_0),nil)), ap^#(ap(f,_0),_0) -> ap^#(_0,ap(f,_0)), ap^#(ap(f,_0),_0) -> ap^#(ap(cons,_0),nil), ap^#(ap(ap(foldr,_0),_1),ap(ap(cons,_2),_3)) -> ap^#(ap(_0,_2),ap(ap(ap(foldr,_0),_1),_3)), ap^#(ap(ap(foldr,_0),_1),ap(ap(cons,_2),_3)) -> ap^#(_0,_2), ap^#(ap(ap(foldr,_0),_1),ap(ap(cons,_2),_3)) -> ap^#(ap(ap(foldr,_0),_1),_3), ap^#(ap(ap(foldr,_0),_1),ap(ap(cons,_2),_3)) -> ap^#(ap(foldr,_0),_1)] TRS = {ap(ap(f,_0),_0) -> ap(ap(_0,ap(f,_0)),ap(ap(cons,_0),nil)), ap(ap(ap(foldr,_0),_1),nil) -> _1, ap(ap(ap(foldr,_0),_1),ap(ap(cons,_2),_3)) -> ap(ap(_0,_2),ap(ap(ap(foldr,_0),_1),_3))} ## Trying with homeomorphic embeddings... Failed! ## Trying with polynomial interpretations... Failed! ## Trying with lexicographic path orders... Failed! ## Trying with Knuth-Bendix orders... Failed! Don't know whether this DP problem is finite. ## A DP problem could not be proved finite. ## Now, we try to prove that this problem is infinite. ## Trying to find a loop (forward=true, backward=false, max=-1) # max_depth=4, unfold_variables=false: # Iteration 0: no loop found, 7 unfolded rules generated. # Iteration 1: no loop found, 52 unfolded rules generated. # Iteration 2: no loop found, 71 unfolded rules generated. # Iteration 3: success, found a loop, 22 unfolded rules generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = ap^#(ap(ap(foldr,_0),_1),ap(ap(cons,_2),_3)) -> ap^#(ap(_0,_2),ap(ap(ap(foldr,_0),_1),_3)) [trans] is in U_IR^0. D = ap^#(ap(ap(foldr,_0),_1),ap(ap(cons,_2),_3)) -> ap^#(_0,_2) is a dependency pair of IR. We build a composed triple from L0 and D. ==> L1 = [ap^#(ap(ap(foldr,_0),_1),ap(ap(cons,_2),_3)) -> ap^#(ap(_0,_2),ap(ap(ap(foldr,_0),_1),_3)), ap^#(ap(ap(foldr,_4),_5),ap(ap(cons,_6),_7)) -> ap^#(_4,_6)] [comp] is in U_IR^1. Let p1 = [1]. We unfold the first rule of L1 forwards at position p1 with the rule ap(ap(ap(foldr,_0),_1),nil) -> _1. ==> L2 = [ap^#(ap(ap(foldr,_0),_1),ap(ap(cons,_2),nil)) -> ap^#(ap(_0,_2),_1), ap^#(ap(ap(foldr,_3),_4),ap(ap(cons,_5),_6)) -> ap^#(_3,_5)] [comp] is in U_IR^2. Let p2 = [0]. We unfold the first rule of L2 forwards at position p2 with the rule ap(ap(f,_0),_0) -> ap(ap(_0,ap(f,_0)),ap(ap(cons,_0),nil)). ==> L3 = [ap^#(ap(ap(foldr,ap(f,_0)),_1),ap(ap(cons,_0),nil)) -> ap^#(ap(ap(_0,ap(f,_0)),ap(ap(cons,_0),nil)),_1), ap^#(ap(ap(foldr,_2),_3),ap(ap(cons,_4),_5)) -> ap^#(_2,_4)] [comp] is in U_IR^3. This DP problem is infinite. Proof run on Linux version 3.10.0-1160.25.1.el7.x86_64 for amd64 using Java version 1.8.0_292 ** END proof description ** Total number of generated unfolded rules = 510 popout

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all output return to TRS Standard