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TRS Stand 20472 pair #381716440
details
property
value
status
complete
benchmark
Ex1_Luc04b_iGM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n037.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
1.3694229126 seconds
cpu usage
0.859329055
max memory
1.8784256E7
stage attributes
key
value
output-size
37474
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. We are asked to determine termination of the following first-order TRS. 0 : [] --> o active : [o] --> o cons : [o * o] --> o head : [o] --> o incr : [o] --> o mark : [o] --> o nats : [] --> o odds : [] --> o pairs : [] --> o s : [o] --> o tail : [o] --> o active(nats) => mark(cons(0, incr(nats))) active(pairs) => mark(cons(0, incr(odds))) active(odds) => mark(incr(pairs)) active(incr(cons(X, Y))) => mark(cons(s(X), incr(Y))) active(head(cons(X, Y))) => mark(X) active(tail(cons(X, Y))) => mark(Y) mark(nats) => active(nats) mark(cons(X, Y)) => active(cons(mark(X), Y)) mark(0) => active(0) mark(incr(X)) => active(incr(mark(X))) mark(pairs) => active(pairs) mark(odds) => active(odds) mark(s(X)) => active(s(mark(X))) mark(head(X)) => active(head(mark(X))) mark(tail(X)) => active(tail(mark(X))) cons(mark(X), Y) => cons(X, Y) cons(X, mark(Y)) => cons(X, Y) cons(active(X), Y) => cons(X, Y) cons(X, active(Y)) => cons(X, Y) incr(mark(X)) => incr(X) incr(active(X)) => incr(X) s(mark(X)) => s(X) s(active(X)) => s(X) head(mark(X)) => head(X) head(active(X)) => head(X) tail(mark(X)) => tail(X) tail(active(X)) => tail(X) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): active(nats) >? mark(cons(0, incr(nats))) active(pairs) >? mark(cons(0, incr(odds))) active(odds) >? mark(incr(pairs)) active(incr(cons(X, Y))) >? mark(cons(s(X), incr(Y))) active(head(cons(X, Y))) >? mark(X) active(tail(cons(X, Y))) >? mark(Y) mark(nats) >? active(nats) mark(cons(X, Y)) >? active(cons(mark(X), Y)) mark(0) >? active(0) mark(incr(X)) >? active(incr(mark(X))) mark(pairs) >? active(pairs) mark(odds) >? active(odds) mark(s(X)) >? active(s(mark(X))) mark(head(X)) >? active(head(mark(X))) mark(tail(X)) >? active(tail(mark(X))) cons(mark(X), Y) >? cons(X, Y) cons(X, mark(Y)) >? cons(X, Y) cons(active(X), Y) >? cons(X, Y) cons(X, active(Y)) >? cons(X, Y) incr(mark(X)) >? incr(X) incr(active(X)) >? incr(X) s(mark(X)) >? s(X) s(active(X)) >? s(X) head(mark(X)) >? head(X) head(active(X)) >? head(X) tail(mark(X)) >? tail(X) tail(active(X)) >? tail(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 active = \y0.y0 cons = \y0y1.y0 + y1 head = \y0.y0 incr = \y0.y0 mark = \y0.y0 nats = 0 odds = 0 pairs = 0 s = \y0.y0 tail = \y0.1 + y0 Using this interpretation, the requirements translate to:
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