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TRS Stand 20472 pair #381714974
details
property
value
status
complete
benchmark
Ex1_Luc04b_C.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n030.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.739539861679 seconds
cpu usage
0.722503109
max memory
1.728512E7
stage attributes
key
value
output-size
30823
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 cons : [o * o] --> o head : [o] --> o incr : [o] --> o mark : [o] --> o nats : [] --> o odds : [] --> o ok : [o] --> o pairs : [] --> o proper : [o] --> o s : [o] --> o tail : [o] --> o top : [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) active(cons(X, Y)) => cons(active(X), Y) active(incr(X)) => incr(active(X)) active(s(X)) => s(active(X)) active(head(X)) => head(active(X)) active(tail(X)) => tail(active(X)) cons(mark(X), Y) => mark(cons(X, Y)) incr(mark(X)) => mark(incr(X)) s(mark(X)) => mark(s(X)) head(mark(X)) => mark(head(X)) tail(mark(X)) => mark(tail(X)) proper(nats) => ok(nats) proper(cons(X, Y)) => cons(proper(X), proper(Y)) proper(0) => ok(0) proper(incr(X)) => incr(proper(X)) proper(pairs) => ok(pairs) proper(odds) => ok(odds) proper(s(X)) => s(proper(X)) proper(head(X)) => head(proper(X)) proper(tail(X)) => tail(proper(X)) cons(ok(X), ok(Y)) => ok(cons(X, Y)) incr(ok(X)) => ok(incr(X)) s(ok(X)) => ok(s(X)) head(ok(X)) => ok(head(X)) tail(ok(X)) => ok(tail(X)) top(mark(X)) => top(proper(X)) top(ok(X)) => top(active(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) active(cons(X, Y)) >? cons(active(X), Y) active(incr(X)) >? incr(active(X)) active(s(X)) >? s(active(X)) active(head(X)) >? head(active(X)) active(tail(X)) >? tail(active(X)) cons(mark(X), Y) >? mark(cons(X, Y)) incr(mark(X)) >? mark(incr(X)) s(mark(X)) >? mark(s(X)) head(mark(X)) >? mark(head(X)) tail(mark(X)) >? mark(tail(X)) proper(nats) >? ok(nats) proper(cons(X, Y)) >? cons(proper(X), proper(Y)) proper(0) >? ok(0) proper(incr(X)) >? incr(proper(X)) proper(pairs) >? ok(pairs) proper(odds) >? ok(odds) proper(s(X)) >? s(proper(X)) proper(head(X)) >? head(proper(X)) proper(tail(X)) >? tail(proper(X)) cons(ok(X), ok(Y)) >? ok(cons(X, Y)) incr(ok(X)) >? ok(incr(X)) s(ok(X)) >? ok(s(X)) head(ok(X)) >? ok(head(X)) tail(ok(X)) >? ok(tail(X)) top(mark(X)) >? top(proper(X)) top(ok(X)) >? top(active(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements:
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