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TRS Standard pair #516965907
details
property
value
status
complete
benchmark
ExIntrod_GM99.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
Strategy_removed_CSR_05
run statistics
property
value
solver
NTI_22
configuration
default
runtime (wallclock)
1.99026417732 seconds
cpu usage
3.346842283
max memory
3.95776E8
stage attributes
key
value
output-size
3281
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 = 0] from(_0) -> from(s(_0)) Let l be the left-hand side and r be the right-hand side of this rule. Let p = epsilon, theta1 = {} and theta2 = {_0->s(_0)}. We have r|p = from(s(_0)) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = from(_0) 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... ## 2 initial DP problems to solve. ## First, we try to decompose these problems into smaller problems. ## Round 1 [2 DP problems]: ## DP problem: Dependency pairs = [from^#(_0) -> from^#(s(_0))] TRS = {primes -> sieve(from(s(s(0)))), from(_0) -> cons(_0,from(s(_0))), head(cons(_0,_1)) -> _0, tail(cons(_0,_1)) -> _1, if(true,_0,_1) -> _0, if(false,_0,_1) -> _1, filter(s(s(_0)),cons(_1,_2)) -> if(divides(s(s(_0)),_1),filter(s(s(_0)),_2),cons(_1,filter(_0,sieve(_1)))), sieve(cons(_0,_1)) -> cons(_0,filter(_0,sieve(_1)))} ## 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. ## DP problem: Dependency pairs = [sieve^#(cons(_0,_1)) -> filter^#(_0,sieve(_1)), sieve^#(cons(_0,_1)) -> sieve^#(_1), filter^#(s(s(_0)),cons(_1,_2)) -> sieve^#(_1), filter^#(s(s(_0)),cons(_1,_2)) -> filter^#(s(s(_0)),_2), filter^#(s(s(_0)),cons(_1,_2)) -> filter^#(_0,sieve(_1))] TRS = {primes -> sieve(from(s(s(0)))), from(_0) -> cons(_0,from(s(_0))), head(cons(_0,_1)) -> _0, tail(cons(_0,_1)) -> _1, if(true,_0,_1) -> _0, if(false,_0,_1) -> _1, filter(s(s(_0)),cons(_1,_2)) -> if(divides(s(s(_0)),_1),filter(s(s(_0)),_2),cons(_1,filter(_0,sieve(_1)))), sieve(cons(_0,_1)) -> cons(_0,filter(_0,sieve(_1)))} ## 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. ## Some DP problems could not be proved finite. ## Now, we try to prove that one of these problems is infinite. ## Trying to find a loop (forward=true, backward=true, max=20) # max_depth=20, unfold_variables=false: # Iteration 0: success, found a loop, 1 unfolded rule generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = from^#(_0) -> from^#(s(_0)) [trans] is in U_IR^0. 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 = 8
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