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TRS Stand 20472 pair #381716660
details
property
value
status
complete
benchmark
ExSec11_1_Luc02a_Z.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n021.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.294969797134 seconds
cpu usage
0.289534408
max memory
8507392.0
stage attributes
key
value
output-size
16225
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 activate : [o] --> o add : [o * o] --> o cons : [o * o] --> o dbl : [o] --> o first : [o * o] --> o half : [o] --> o n!6220!6220first : [o * o] --> o n!6220!6220terms : [o] --> o nil : [] --> o recip : [o] --> o s : [o] --> o sqr : [o] --> o terms : [o] --> o terms(X) => cons(recip(sqr(X)), n!6220!6220terms(s(X))) sqr(0) => 0 sqr(s(X)) => s(add(sqr(X), dbl(X))) dbl(0) => 0 dbl(s(X)) => s(s(dbl(X))) add(0, X) => X add(s(X), Y) => s(add(X, Y)) first(0, X) => nil first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(X, activate(Z))) half(0) => 0 half(s(0)) => 0 half(s(s(X))) => s(half(X)) half(dbl(X)) => X terms(X) => n!6220!6220terms(X) first(X, Y) => n!6220!6220first(X, Y) activate(n!6220!6220terms(X)) => terms(X) activate(n!6220!6220first(X, Y)) => first(X, Y) 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]): terms(X) >? cons(recip(sqr(X)), n!6220!6220terms(s(X))) sqr(0) >? 0 sqr(s(X)) >? s(add(sqr(X), dbl(X))) dbl(0) >? 0 dbl(s(X)) >? s(s(dbl(X))) add(0, X) >? X add(s(X), Y) >? s(add(X, Y)) first(0, X) >? nil first(s(X), cons(Y, Z)) >? cons(Y, n!6220!6220first(X, activate(Z))) half(0) >? 0 half(s(0)) >? 0 half(s(s(X))) >? s(half(X)) half(dbl(X)) >? X terms(X) >? n!6220!6220terms(X) first(X, Y) >? n!6220!6220first(X, Y) activate(n!6220!6220terms(X)) >? terms(X) activate(n!6220!6220first(X, Y)) >? first(X, Y) activate(X) >? X about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[n!6220!6220terms(x_1)]] = x_1 [[nil]] = _|_ We choose Lex = {} and Mul = {activate, add, cons, dbl, first, half, n!6220!6220first, recip, s, sqr, terms}, and the following precedence: activate = first > n!6220!6220first > terms > cons > recip > dbl = sqr > add > half = s Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: terms(X) >= cons(recip(sqr(X)), s(X)) sqr(_|_) >= _|_ sqr(s(X)) > s(add(sqr(X), dbl(X))) dbl(_|_) >= _|_ dbl(s(X)) >= s(s(dbl(X))) add(_|_, X) > X add(s(X), Y) > s(add(X, Y)) first(_|_, X) >= _|_ first(s(X), cons(Y, Z)) >= cons(Y, n!6220!6220first(X, activate(Z))) half(_|_) >= _|_ half(s(_|_)) >= _|_ half(s(s(X))) >= s(half(X)) half(dbl(X)) > X terms(X) >= X first(X, Y) >= n!6220!6220first(X, Y) activate(X) > terms(X) activate(n!6220!6220first(X, Y)) >= first(X, Y) activate(X) >= X
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