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TRS Stand 20472 pair #381715745
details
property
value
status
complete
benchmark
ExIntrod_GM04_iGM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n054.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.503130912781 seconds
cpu usage
0.49219456
max memory
1.4544896E7
stage attributes
key
value
output-size
24553
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_FirstOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/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 adx : [o] --> o cons : [o * o] --> o hd : [o] --> o incr : [o] --> o mark : [o] --> o nats : [] --> o s : [o] --> o tl : [o] --> o zeros : [] --> o active(nats) => mark(adx(zeros)) active(zeros) => mark(cons(0, zeros)) active(incr(cons(X, Y))) => mark(cons(s(X), incr(Y))) active(adx(cons(X, Y))) => mark(incr(cons(X, adx(Y)))) active(hd(cons(X, Y))) => mark(X) active(tl(cons(X, Y))) => mark(Y) mark(nats) => active(nats) mark(adx(X)) => active(adx(mark(X))) mark(zeros) => active(zeros) mark(cons(X, Y)) => active(cons(X, Y)) mark(0) => active(0) mark(incr(X)) => active(incr(mark(X))) mark(s(X)) => active(s(X)) mark(hd(X)) => active(hd(mark(X))) mark(tl(X)) => active(tl(mark(X))) adx(mark(X)) => adx(X) adx(active(X)) => adx(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) hd(mark(X)) => hd(X) hd(active(X)) => hd(X) tl(mark(X)) => tl(X) tl(active(X)) => tl(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(adx(zeros)) active(zeros) >? mark(cons(0, zeros)) active(incr(cons(X, Y))) >? mark(cons(s(X), incr(Y))) active(adx(cons(X, Y))) >? mark(incr(cons(X, adx(Y)))) active(hd(cons(X, Y))) >? mark(X) active(tl(cons(X, Y))) >? mark(Y) mark(nats) >? active(nats) mark(adx(X)) >? active(adx(mark(X))) mark(zeros) >? active(zeros) mark(cons(X, Y)) >? active(cons(X, Y)) mark(0) >? active(0) mark(incr(X)) >? active(incr(mark(X))) mark(s(X)) >? active(s(X)) mark(hd(X)) >? active(hd(mark(X))) mark(tl(X)) >? active(tl(mark(X))) adx(mark(X)) >? adx(X) adx(active(X)) >? adx(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) hd(mark(X)) >? hd(X) hd(active(X)) >? hd(X) tl(mark(X)) >? tl(X) tl(active(X)) >? tl(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 active = \y0.y0 adx = \y0.2y0 cons = \y0y1.y0 + y1 hd = \y0.2y0 incr = \y0.y0 mark = \y0.y0 nats = 1 s = \y0.y0
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