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TRS Standard pair #516961862
details
property
value
status
complete
benchmark
Ex1_Luc04b_FR.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n139.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
NTI_22
configuration
default
runtime (wallclock)
1.2531208992 seconds
cpu usage
2.406937368
max memory
1.89300736E8
stage attributes
key
value
output-size
3603
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] activate(n__incr(n__nats)) -> activate(n__incr(n__nats)) Let l be the left-hand side and r be the right-hand side of this rule. Let p = epsilon, theta1 = {} and theta2 = {}. We have r|p = activate(n__incr(n__nats)) and theta2(theta1(l)) = theta1(r|p). Hence, the term theta1(l) = activate(n__incr(n__nats)) 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 = [odds^# -> incr^#(pairs), activate^#(n__odds) -> odds^#, incr^#(cons(_0,_1)) -> activate^#(_1), activate^#(n__incr(_0)) -> incr^#(activate(_0)), activate^#(n__incr(_0)) -> activate^#(_0)] TRS = {nats -> cons(0,n__incr(n__nats)), pairs -> cons(0,n__incr(n__odds)), odds -> incr(pairs), incr(cons(_0,_1)) -> cons(s(_0),n__incr(activate(_1))), head(cons(_0,_1)) -> _0, tail(cons(_0,_1)) -> activate(_1), incr(_0) -> n__incr(_0), nats -> n__nats, odds -> n__odds, activate(n__incr(_0)) -> incr(activate(_0)), activate(n__nats) -> nats, activate(n__odds) -> odds, 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, 5 unfolded rules generated. # Iteration 1: no loop found, 14 unfolded rules generated. # Iteration 2: no loop found, 19 unfolded rules generated. # Iteration 3: success, found a loop, 20 unfolded rules generated. Here is the successful unfolding. Let IR be the TRS under analysis. L0 = activate^#(n__incr(_0)) -> incr^#(activate(_0)) [trans] is in U_IR^0. D = incr^#(cons(_0,_1)) -> activate^#(_1) is a dependency pair of IR. We build a composed triple from L0 and D. ==> L1 = [activate^#(n__incr(_0)) -> incr^#(activate(_0)), incr^#(cons(_1,_2)) -> activate^#(_2)] [comp] is in U_IR^1. Let p1 = [0]. We unfold the first rule of L1 forwards at position p1 with the rule activate(n__nats) -> nats. ==> L2 = [activate^#(n__incr(n__nats)) -> incr^#(nats), incr^#(cons(_0,_1)) -> activate^#(_1)] [comp] is in U_IR^2. Let p2 = [0]. We unfold the first rule of L2 forwards at position p2 with the rule nats -> cons(0,n__incr(n__nats)). ==> L3 = activate^#(n__incr(n__nats)) -> activate^#(n__incr(n__nats)) [trans] 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 = 154
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