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TRS Relative pair #487081932
details
property
value
status
complete
benchmark
rtL-evo.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n151.star.cs.uiowa.edu
space
Relative_05
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
1.77296 seconds
cpu usage
4.24284
user time
4.06032
system time
0.182524
max virtual memory
3.6970892E7
max residence set size
285612.0
stage attributes
key
value
starexec-result
NO
output
NO proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination of the given RelTRS could be disproven: (0) RelTRS (1) RelTRSLoopFinderProof [COMPLETE, 91 ms] (2) NO ---------------------------------------- (0) Obligation: Relative term rewrite system: The relative TRS consists of the following R rules: top(U(x, y)) -> top(check(D(x, y))) D(x, B) -> U(x, B) F(x, U(O(y), z)) -> U(x, F(y, z)) F(x, U(N(y), z)) -> U(x, F(y, z)) D(O(x), F(y, z)) -> F(x, D(y, z)) D(N(x), F(y, z)) -> F(x, D(y, z)) F(x, U(E, y)) -> U(x, F(E, y)) D(E, F(x, y)) -> F(E, D(x, y)) The relative TRS consists of the following S rules: E -> N(E) check(O(x)) -> O(x) check(U(x, y)) -> U(check(x), y) check(U(x, y)) -> U(x, check(y)) check(D(x, y)) -> D(check(x), y) check(D(x, y)) -> D(x, check(y)) check(F(x, y)) -> F(check(x), y) check(F(x, y)) -> F(x, check(y)) check(O(x)) -> O(check(x)) check(N(x)) -> N(check(x)) ---------------------------------------- (1) RelTRSLoopFinderProof (COMPLETE) The following loop was found: ---------- Loop: ---------- top(check(D(x, B))) -> top(D(check(x), B)) with rule check(D(x', y)) -> D(check(x'), y) at position [0] and matcher [x' / x, y / B] top(D(check(x), B)) -> top(U(check(x), B)) with rule D(x', B) -> U(x', B) at position [0] and matcher [x' / check(x)] top(U(check(x), B)) -> top(check(D(check(x), B))) with rule top(U(x', y)) -> top(check(D(x', y))) at position [] and matcher [x' / check(x), y / B] Now an instance of the first term with Matcher [x / check(x)] occurs in the last term at position []. Context: [] Therefore, the relative TRS problem does not terminate. ---------------------------------------- (2) NO
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