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TRS Standard pair #516964412
details
property
value
status
complete
benchmark
ExSec11_1_Luc02a.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n095.star.cs.uiowa.edu
space
Strategy_removed_mixed_05
run statistics
property
value
solver
NTI_22
configuration
default
runtime (wallclock)
0.503920078278 seconds
cpu usage
1.056816074
max memory
1.1546624E8
stage attributes
key
value
output-size
4933
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] terms(_0) -> terms(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 = terms(s(_0)) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = terms(_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... ## 6 initial DP problems to solve. ## First, we try to decompose these problems into smaller problems. ## Round 1 [6 DP problems]: ## DP problem: Dependency pairs = [half^#(s(s(_0))) -> half^#(_0)] TRS = {terms(_0) -> cons(recip(sqr(_0)),terms(s(_0))), sqr(0) -> 0, sqr(s(_0)) -> s(add(sqr(_0),dbl(_0))), dbl(0) -> 0, dbl(s(_0)) -> s(s(dbl(_0))), add(0,_0) -> _0, add(s(_0),_1) -> s(add(_0,_1)), first(0,_0) -> nil, first(s(_0),cons(_1,_2)) -> cons(_1,first(_0,_2)), half(0) -> 0, half(s(0)) -> 0, half(s(s(_0))) -> s(half(_0)), half(dbl(_0)) -> _0} ## Trying with homeomorphic embeddings... Success! This DP problem is finite. ## DP problem: Dependency pairs = [first^#(s(_0),cons(_1,_2)) -> first^#(_0,_2)] TRS = {terms(_0) -> cons(recip(sqr(_0)),terms(s(_0))), sqr(0) -> 0, sqr(s(_0)) -> s(add(sqr(_0),dbl(_0))), dbl(0) -> 0, dbl(s(_0)) -> s(s(dbl(_0))), add(0,_0) -> _0, add(s(_0),_1) -> s(add(_0,_1)), first(0,_0) -> nil, first(s(_0),cons(_1,_2)) -> cons(_1,first(_0,_2)), half(0) -> 0, half(s(0)) -> 0, half(s(s(_0))) -> s(half(_0)), half(dbl(_0)) -> _0} ## Trying with homeomorphic embeddings... Success! This DP problem is finite. ## DP problem: Dependency pairs = [terms^#(_0) -> terms^#(s(_0))] TRS = {terms(_0) -> cons(recip(sqr(_0)),terms(s(_0))), sqr(0) -> 0, sqr(s(_0)) -> s(add(sqr(_0),dbl(_0))), dbl(0) -> 0, dbl(s(_0)) -> s(s(dbl(_0))), add(0,_0) -> _0, add(s(_0),_1) -> s(add(_0,_1)), first(0,_0) -> nil, first(s(_0),cons(_1,_2)) -> cons(_1,first(_0,_2)), half(0) -> 0, half(s(0)) -> 0, half(s(s(_0))) -> s(half(_0)), half(dbl(_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. ## DP problem: Dependency pairs = [sqr^#(s(_0)) -> sqr^#(_0)] TRS = {terms(_0) -> cons(recip(sqr(_0)),terms(s(_0))), sqr(0) -> 0, sqr(s(_0)) -> s(add(sqr(_0),dbl(_0))), dbl(0) -> 0, dbl(s(_0)) -> s(s(dbl(_0))), add(0,_0) -> _0, add(s(_0),_1) -> s(add(_0,_1)), first(0,_0) -> nil, first(s(_0),cons(_1,_2)) -> cons(_1,first(_0,_2)), half(0) -> 0, half(s(0)) -> 0, half(s(s(_0))) -> s(half(_0)), half(dbl(_0)) -> _0} ## Trying with homeomorphic embeddings... Success! This DP problem is finite. ## DP problem: Dependency pairs = [dbl^#(s(_0)) -> dbl^#(_0)] TRS = {terms(_0) -> cons(recip(sqr(_0)),terms(s(_0))), sqr(0) -> 0, sqr(s(_0)) -> s(add(sqr(_0),dbl(_0))), dbl(0) -> 0, dbl(s(_0)) -> s(s(dbl(_0))), add(0,_0) -> _0, add(s(_0),_1) -> s(add(_0,_1)), first(0,_0) -> nil, first(s(_0),cons(_1,_2)) -> cons(_1,first(_0,_2)), half(0) -> 0, half(s(0)) -> 0, half(s(s(_0))) -> s(half(_0)), half(dbl(_0)) -> _0} ## Trying with homeomorphic embeddings... Success! This DP problem is finite. ## DP problem: Dependency pairs = [add^#(s(_0),_1) -> add^#(_0,_1)] TRS = {terms(_0) -> cons(recip(sqr(_0)),terms(s(_0))), sqr(0) -> 0, sqr(s(_0)) -> s(add(sqr(_0),dbl(_0))), dbl(0) -> 0, dbl(s(_0)) -> s(s(dbl(_0))), add(0,_0) -> _0, add(s(_0),_1) -> s(add(_0,_1)), first(0,_0) -> nil, first(s(_0),cons(_1,_2)) -> cons(_1,first(_0,_2)), half(0) -> 0, half(s(0)) -> 0, half(s(s(_0))) -> s(half(_0)), half(dbl(_0)) -> _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 = terms^#(_0) -> terms^#(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 = 3
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