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TRS Stand 20472 pair #381714694
details
property
value
status
complete
benchmark
ExSec4_2_DLMMU04_FR.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n086.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.224623203278 seconds
cpu usage
0.217427867
max memory
1.060864E7
stage attributes
key
value
output-size
12184
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 activate : [o] --> o afterNth : [o * o] --> o cons : [o * o] --> o fst : [o] --> o head : [o] --> o n!6220!6220natsFrom : [o] --> o n!6220!6220s : [o] --> o natsFrom : [o] --> o nil : [] --> o pair : [o * o] --> o s : [o] --> o sel : [o * o] --> o snd : [o] --> o splitAt : [o * o] --> o tail : [o] --> o take : [o * o] --> o u : [o * o * o * o] --> o natsFrom(X) => cons(X, n!6220!6220natsFrom(n!6220!6220s(X))) fst(pair(X, Y)) => X snd(pair(X, Y)) => Y splitAt(0, X) => pair(nil, X) splitAt(s(X), cons(Y, Z)) => u(splitAt(X, activate(Z)), X, Y, activate(Z)) u(pair(X, Y), Z, U, V) => pair(cons(activate(U), X), Y) head(cons(X, Y)) => X tail(cons(X, Y)) => activate(Y) sel(X, Y) => head(afterNth(X, Y)) take(X, Y) => fst(splitAt(X, Y)) afterNth(X, Y) => snd(splitAt(X, Y)) natsFrom(X) => n!6220!6220natsFrom(X) s(X) => n!6220!6220s(X) activate(n!6220!6220natsFrom(X)) => natsFrom(activate(X)) activate(n!6220!6220s(X)) => s(activate(X)) activate(X) => 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]): natsFrom(X) >? cons(X, n!6220!6220natsFrom(n!6220!6220s(X))) fst(pair(X, Y)) >? X snd(pair(X, Y)) >? Y splitAt(0, X) >? pair(nil, X) splitAt(s(X), cons(Y, Z)) >? u(splitAt(X, activate(Z)), X, Y, activate(Z)) u(pair(X, Y), Z, U, V) >? pair(cons(activate(U), X), Y) head(cons(X, Y)) >? X tail(cons(X, Y)) >? activate(Y) sel(X, Y) >? head(afterNth(X, Y)) take(X, Y) >? fst(splitAt(X, Y)) afterNth(X, Y) >? snd(splitAt(X, Y)) natsFrom(X) >? n!6220!6220natsFrom(X) s(X) >? n!6220!6220s(X) activate(n!6220!6220natsFrom(X)) >? natsFrom(activate(X)) activate(n!6220!6220s(X)) >? s(activate(X)) activate(X) >? X about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[fst(x_1)]] = x_1 [[nil]] = _|_ [[snd(x_1)]] = x_1 We choose Lex = {afterNth, splitAt} and Mul = {0, activate, cons, head, n!6220!6220natsFrom, n!6220!6220s, natsFrom, pair, s, sel, tail, take, u}, and the following precedence: tail > sel > take > afterNth = splitAt > activate = u > natsFrom > n!6220!6220natsFrom > n!6220!6220s = s > 0 > cons > pair > head Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: natsFrom(X) >= cons(X, n!6220!6220natsFrom(n!6220!6220s(X))) pair(X, Y) > X pair(X, Y) >= Y splitAt(0, X) > pair(_|_, X) splitAt(s(X), cons(Y, Z)) > u(splitAt(X, activate(Z)), X, Y, activate(Z)) u(pair(X, Y), Z, U, V) >= pair(cons(activate(U), X), Y) head(cons(X, Y)) >= X tail(cons(X, Y)) >= activate(Y) sel(X, Y) >= head(afterNth(X, Y)) take(X, Y) >= splitAt(X, Y) afterNth(X, Y) >= splitAt(X, Y) natsFrom(X) > n!6220!6220natsFrom(X) s(X) >= n!6220!6220s(X) activate(n!6220!6220natsFrom(X)) > natsFrom(activate(X)) activate(n!6220!6220s(X)) > s(activate(X)) activate(X) >= X With these choices, we have:
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