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TRS Stand 20472 pair #381711872
details
property
value
status
complete
benchmark
Ex1_Luc04b_GM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n112.star.cs.uiowa.edu
space
Mixed_TRS
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.459130048752 seconds
cpu usage
0.449177118
max memory
1.0194944E7
stage attributes
key
value
output-size
30828
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!6220head : [o] --> o a!6220!6220incr : [o] --> o a!6220!6220nats : [] --> o a!6220!6220odds : [] --> o a!6220!6220pairs : [] --> o a!6220!6220tail : [o] --> o cons : [o * o] --> o head : [o] --> o incr : [o] --> o mark : [o] --> o nats : [] --> o nil : [] --> o odds : [] --> o pairs : [] --> o s : [o] --> o tail : [o] --> o a!6220!6220nats => cons(0, incr(nats)) a!6220!6220pairs => cons(0, incr(odds)) a!6220!6220odds => a!6220!6220incr(a!6220!6220pairs) a!6220!6220incr(cons(X, Y)) => cons(s(mark(X)), incr(Y)) a!6220!6220head(cons(X, Y)) => mark(X) a!6220!6220tail(cons(X, Y)) => mark(Y) mark(nats) => a!6220!6220nats mark(pairs) => a!6220!6220pairs mark(odds) => a!6220!6220odds mark(incr(X)) => a!6220!6220incr(mark(X)) mark(head(X)) => a!6220!6220head(mark(X)) mark(tail(X)) => a!6220!6220tail(mark(X)) mark(0) => 0 mark(s(X)) => s(mark(X)) mark(nil) => nil mark(cons(X, Y)) => cons(mark(X), Y) a!6220!6220nats => nats a!6220!6220pairs => pairs a!6220!6220odds => odds a!6220!6220incr(X) => incr(X) a!6220!6220head(X) => head(X) a!6220!6220tail(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]): a!6220!6220nats >? cons(0, incr(nats)) a!6220!6220pairs >? cons(0, incr(odds)) a!6220!6220odds >? a!6220!6220incr(a!6220!6220pairs) a!6220!6220incr(cons(X, Y)) >? cons(s(mark(X)), incr(Y)) a!6220!6220head(cons(X, Y)) >? mark(X) a!6220!6220tail(cons(X, Y)) >? mark(Y) mark(nats) >? a!6220!6220nats mark(pairs) >? a!6220!6220pairs mark(odds) >? a!6220!6220odds mark(incr(X)) >? a!6220!6220incr(mark(X)) mark(head(X)) >? a!6220!6220head(mark(X)) mark(tail(X)) >? a!6220!6220tail(mark(X)) mark(0) >? 0 mark(s(X)) >? s(mark(X)) mark(nil) >? nil mark(cons(X, Y)) >? cons(mark(X), Y) a!6220!6220nats >? nats a!6220!6220pairs >? pairs a!6220!6220odds >? odds a!6220!6220incr(X) >? incr(X) a!6220!6220head(X) >? head(X) a!6220!6220tail(X) >? tail(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 a!6220!6220head = \y0.y0 a!6220!6220incr = \y0.2y0 a!6220!6220nats = 0 a!6220!6220odds = 0 a!6220!6220pairs = 0 a!6220!6220tail = \y0.2 + y0 cons = \y0y1.2y0 + 2y1 head = \y0.y0 incr = \y0.2y0 mark = \y0.y0 nats = 0 nil = 0 odds = 0 pairs = 0 s = \y0.y0 tail = \y0.2 + y0
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