details

property	value
status	complete
benchmark	LengthOfFiniteLists_nosorts-noand_FR.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	Transformed_CSR_04

run statistics

property	value
solver	NTI_22
configuration	default
runtime (wallclock)	0.572634935379 seconds
cpu usage	1.102412556
max memory	9.385984E7

stage attributes

key	value
output-size	4417
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 = 7] length(cons(_0,n__zeros)) -> length(cons(0,n__zeros)) Let l be the left-hand side and r be the right-hand side of this rule. Let p = epsilon, theta1 = {} and theta2 = {_0->0}. We have r|p = length(cons(0,n__zeros)) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = length(cons(_0,n__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... ## 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 = [U11^#(tt,_0) -> U12^#(tt,activate(_0)), length^#(cons(_0,_1)) -> U11^#(tt,activate(_1)), U12^#(tt,_0) -> length^#(activate(_0))] TRS = {zeros -> cons(0,n__zeros), U11(tt,_0) -> U12(tt,activate(_0)), U12(tt,_0) -> s(length(activate(_0))), length(nil) -> 0, length(cons(_0,_1)) -> U11(tt,activate(_1)), zeros -> n__zeros, activate(n__zeros) -> zeros, activate(_0) -> _0} ## Trying with homeomorphic embeddings... Failed! ## Trying with polynomial interpretations... This DP problem is too complex! Aborting! ## 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=3, unfold_variables=false: # Iteration 0: no loop found, 3 unfolded rules generated. # Iteration 1: no loop found, 5 unfolded rules generated. # Iteration 2: no loop found, 16 unfolded rules generated. # Iteration 3: no loop found, 30 unfolded rules generated. # Iteration 4: no loop found, 36 unfolded rules generated. # Iteration 5: no loop found, 5 unfolded rules generated. # Iteration 6: no loop found, 9 unfolded rules generated. # Iteration 7: success, found a loop, 1 unfolded rule generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = length^#(cons(_0,_1)) -> U11^#(tt,activate(_1)) [trans] is in U_IR^0. D = U11^#(tt,_0) -> U12^#(tt,activate(_0)) is a dependency pair of IR. We build a composed triple from L0 and D. ==> L1 = [length^#(cons(_0,_1)) -> U11^#(tt,activate(_1)), U11^#(tt,_2) -> U12^#(tt,activate(_2))] [comp] is in U_IR^1. Let p1 = [1]. We unfold the first rule of L1 forwards at position p1 with the rule activate(n__zeros) -> zeros. ==> L2 = length^#(cons(_0,n__zeros)) -> U12^#(tt,activate(zeros)) [trans] is in U_IR^2. D = U12^#(tt,_0) -> length^#(activate(_0)) is a dependency pair of IR. We build a composed triple from L2 and D. ==> L3 = length^#(cons(_0,n__zeros)) -> length^#(activate(activate(zeros))) [trans] is in U_IR^3. We build a unit triple from L3. ==> L4 = length^#(cons(_0,n__zeros)) -> length^#(activate(activate(zeros))) [unit] is in U_IR^4. Let p4 = [0]. We unfold the rule of L4 forwards at position p4 with the rule activate(_0) -> _0. ==> L5 = length^#(cons(_0,n__zeros)) -> length^#(activate(zeros)) [unit] is in U_IR^5. Let p5 = [0]. We unfold the rule of L5 forwards at position p5 with the rule activate(_0) -> _0. ==> L6 = length^#(cons(_0,n__zeros)) -> length^#(zeros) [unit] is in U_IR^6. Let p6 = [0]. We unfold the rule of L6 forwards at position p6 with the rule zeros -> cons(0,n__zeros). ==> L7 = length^#(cons(_0,n__zeros)) -> length^#(cons(0,n__zeros)) [unit] is in U_IR^7. 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 = 532 popout

output may be truncated. 'popout' for the full output.

job log

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actions

all output return to TRS Standard