details

property	value
status	complete
benchmark	Ex3_2_Luc97.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n069.star.cs.uiowa.edu
space	Strategy_removed_CSR_05

run statistics

property	value
solver	NTI_22
configuration	default
runtime (wallclock)	0.314986944199 seconds
cpu usage	0.655705493
max memory	6.5908736E7

stage attributes

key	value
output-size	4093
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... ## 5 initial DP problems to solve. ## First, we try to decompose these problems into smaller problems. ## Round 1 [5 DP problems]: ## DP problem: Dependency pairs = [from^#(_0) -> from^#(s(_0))] TRS = {dbl(0) -> 0, dbl(s(_0)) -> s(s(dbl(_0))), dbls(nil) -> nil, dbls(cons(_0,_1)) -> cons(dbl(_0),dbls(_1)), sel(0,cons(_0,_1)) -> _0, sel(s(_0),cons(_1,_2)) -> sel(_0,_2), indx(nil,_0) -> nil, indx(cons(_0,_1),_2) -> cons(sel(_0,_2),indx(_1,_2)), from(_0) -> cons(_0,from(s(_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. ## DP problem: Dependency pairs = [indx^#(cons(_0,_1),_2) -> indx^#(_1,_2)] TRS = {dbl(0) -> 0, dbl(s(_0)) -> s(s(dbl(_0))), dbls(nil) -> nil, dbls(cons(_0,_1)) -> cons(dbl(_0),dbls(_1)), sel(0,cons(_0,_1)) -> _0, sel(s(_0),cons(_1,_2)) -> sel(_0,_2), indx(nil,_0) -> nil, indx(cons(_0,_1),_2) -> cons(sel(_0,_2),indx(_1,_2)), from(_0) -> cons(_0,from(s(_0)))} ## Trying with homeomorphic embeddings... Success! This DP problem is finite. ## DP problem: Dependency pairs = [sel^#(s(_0),cons(_1,_2)) -> sel^#(_0,_2)] TRS = {dbl(0) -> 0, dbl(s(_0)) -> s(s(dbl(_0))), dbls(nil) -> nil, dbls(cons(_0,_1)) -> cons(dbl(_0),dbls(_1)), sel(0,cons(_0,_1)) -> _0, sel(s(_0),cons(_1,_2)) -> sel(_0,_2), indx(nil,_0) -> nil, indx(cons(_0,_1),_2) -> cons(sel(_0,_2),indx(_1,_2)), from(_0) -> cons(_0,from(s(_0)))} ## Trying with homeomorphic embeddings... Success! This DP problem is finite. ## DP problem: Dependency pairs = [dbls^#(cons(_0,_1)) -> dbls^#(_1)] TRS = {dbl(0) -> 0, dbl(s(_0)) -> s(s(dbl(_0))), dbls(nil) -> nil, dbls(cons(_0,_1)) -> cons(dbl(_0),dbls(_1)), sel(0,cons(_0,_1)) -> _0, sel(s(_0),cons(_1,_2)) -> sel(_0,_2), indx(nil,_0) -> nil, indx(cons(_0,_1),_2) -> cons(sel(_0,_2),indx(_1,_2)), from(_0) -> cons(_0,from(s(_0)))} ## Trying with homeomorphic embeddings... Success! This DP problem is finite. ## DP problem: Dependency pairs = [dbl^#(s(_0)) -> dbl^#(_0)] TRS = {dbl(0) -> 0, dbl(s(_0)) -> s(s(dbl(_0))), dbls(nil) -> nil, dbls(cons(_0,_1)) -> cons(dbl(_0),dbls(_1)), sel(0,cons(_0,_1)) -> _0, sel(s(_0),cons(_1,_2)) -> sel(_0,_2), indx(nil,_0) -> nil, indx(cons(_0,_1),_2) -> cons(sel(_0,_2),indx(_1,_2)), from(_0) -> cons(_0,from(s(_0)))} ## Trying with homeomorphic embeddings... Success! 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=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 = 2 popout

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

job log

popout

actions

all output return to TRS Standard