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TRS Stand 20472 pair #381710593
details
property
value
status
complete
benchmark
ExIntrod_GM04_GM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n012.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.396930217743 seconds
cpu usage
0.393779635
max memory
7553024.0
stage attributes
key
value
output-size
26449
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 a!6220!6220adx : [o] --> o a!6220!6220hd : [o] --> o a!6220!6220incr : [o] --> o a!6220!6220nats : [] --> o a!6220!6220tl : [o] --> o a!6220!6220zeros : [] --> 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 a!6220!6220nats => a!6220!6220adx(a!6220!6220zeros) a!6220!6220zeros => cons(0, zeros) a!6220!6220incr(cons(X, Y)) => cons(s(X), incr(Y)) a!6220!6220adx(cons(X, Y)) => a!6220!6220incr(cons(X, adx(Y))) a!6220!6220hd(cons(X, Y)) => mark(X) a!6220!6220tl(cons(X, Y)) => mark(Y) mark(nats) => a!6220!6220nats mark(adx(X)) => a!6220!6220adx(mark(X)) mark(zeros) => a!6220!6220zeros mark(incr(X)) => a!6220!6220incr(mark(X)) mark(hd(X)) => a!6220!6220hd(mark(X)) mark(tl(X)) => a!6220!6220tl(mark(X)) mark(cons(X, Y)) => cons(X, Y) mark(0) => 0 mark(s(X)) => s(X) a!6220!6220nats => nats a!6220!6220adx(X) => adx(X) a!6220!6220zeros => zeros a!6220!6220incr(X) => incr(X) a!6220!6220hd(X) => hd(X) a!6220!6220tl(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]): a!6220!6220nats >? a!6220!6220adx(a!6220!6220zeros) a!6220!6220zeros >? cons(0, zeros) a!6220!6220incr(cons(X, Y)) >? cons(s(X), incr(Y)) a!6220!6220adx(cons(X, Y)) >? a!6220!6220incr(cons(X, adx(Y))) a!6220!6220hd(cons(X, Y)) >? mark(X) a!6220!6220tl(cons(X, Y)) >? mark(Y) mark(nats) >? a!6220!6220nats mark(adx(X)) >? a!6220!6220adx(mark(X)) mark(zeros) >? a!6220!6220zeros mark(incr(X)) >? a!6220!6220incr(mark(X)) mark(hd(X)) >? a!6220!6220hd(mark(X)) mark(tl(X)) >? a!6220!6220tl(mark(X)) mark(cons(X, Y)) >? cons(X, Y) mark(0) >? 0 mark(s(X)) >? s(X) a!6220!6220nats >? nats a!6220!6220adx(X) >? adx(X) a!6220!6220zeros >? zeros a!6220!6220incr(X) >? incr(X) a!6220!6220hd(X) >? hd(X) a!6220!6220tl(X) >? tl(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 a!6220!6220adx = \y0.y0 a!6220!6220hd = \y0.y0 a!6220!6220incr = \y0.y0 a!6220!6220nats = 2 a!6220!6220tl = \y0.y0 a!6220!6220zeros = 0 adx = \y0.y0 cons = \y0y1.y1 + 2y0 hd = \y0.y0 incr = \y0.y0 mark = \y0.y0 nats = 2 s = \y0.y0 tl = \y0.y0 zeros = 0 Using this interpretation, the requirements translate to: [[a!6220!6220nats]] = 2 > 0 = [[a!6220!6220adx(a!6220!6220zeros)]]
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