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TRS Stand 20472 pair #381715605
details
property
value
status
complete
benchmark
Ex1_2_AEL03_FR.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n023.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.318480968475 seconds
cpu usage
0.307244672
max memory
1.2775424E7
stage attributes
key
value
output-size
15313
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 2ndsneg : [o * o] --> o 2ndspos : [o * o] --> o activate : [o] --> o cons : [o * o] --> o from : [o] --> o n!6220!6220cons : [o * o] --> o n!6220!6220from : [o] --> o n!6220!6220s : [o] --> o negrecip : [o] --> o pi : [o] --> o plus : [o * o] --> o posrecip : [o] --> o rcons : [o * o] --> o rnil : [] --> o s : [o] --> o square : [o] --> o times : [o * o] --> o from(X) => cons(X, n!6220!6220from(n!6220!6220s(X))) 2ndspos(0, X) => rnil 2ndspos(s(X), cons(Y, n!6220!6220cons(Z, U))) => rcons(posrecip(activate(Z)), 2ndsneg(X, activate(U))) 2ndsneg(0, X) => rnil 2ndsneg(s(X), cons(Y, n!6220!6220cons(Z, U))) => rcons(negrecip(activate(Z)), 2ndspos(X, activate(U))) pi(X) => 2ndspos(X, from(0)) plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) times(0, X) => 0 times(s(X), Y) => plus(Y, times(X, Y)) square(X) => times(X, X) from(X) => n!6220!6220from(X) s(X) => n!6220!6220s(X) cons(X, Y) => n!6220!6220cons(X, Y) activate(n!6220!6220from(X)) => from(activate(X)) activate(n!6220!6220s(X)) => s(activate(X)) activate(n!6220!6220cons(X, Y)) => cons(activate(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]): from(X) >? cons(X, n!6220!6220from(n!6220!6220s(X))) 2ndspos(0, X) >? rnil 2ndspos(s(X), cons(Y, n!6220!6220cons(Z, U))) >? rcons(posrecip(activate(Z)), 2ndsneg(X, activate(U))) 2ndsneg(0, X) >? rnil 2ndsneg(s(X), cons(Y, n!6220!6220cons(Z, U))) >? rcons(negrecip(activate(Z)), 2ndspos(X, activate(U))) pi(X) >? 2ndspos(X, from(0)) plus(0, X) >? X plus(s(X), Y) >? s(plus(X, Y)) times(0, X) >? 0 times(s(X), Y) >? plus(Y, times(X, Y)) square(X) >? times(X, X) from(X) >? n!6220!6220from(X) s(X) >? n!6220!6220s(X) cons(X, Y) >? n!6220!6220cons(X, Y) activate(n!6220!6220from(X)) >? from(activate(X)) activate(n!6220!6220s(X)) >? s(activate(X)) activate(n!6220!6220cons(X, Y)) >? cons(activate(X), Y) activate(X) >? X about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[posrecip(x_1)]] = x_1 [[rnil]] = _|_ We choose Lex = {2ndsneg, 2ndspos} and Mul = {activate, cons, from, n!6220!6220cons, n!6220!6220from, n!6220!6220s, negrecip, pi, plus, rcons, s, square, times}, and the following precedence: pi > square > times > plus > 2ndsneg = 2ndspos > negrecip > rcons > activate > from > n!6220!6220s = s > n!6220!6220from > cons > n!6220!6220cons Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: from(X) > cons(X, n!6220!6220from(n!6220!6220s(X))) 2ndspos(_|_, X) >= _|_ 2ndspos(s(X), cons(Y, n!6220!6220cons(Z, U))) >= rcons(activate(Z), 2ndsneg(X, activate(U))) 2ndsneg(_|_, X) >= _|_ 2ndsneg(s(X), cons(Y, n!6220!6220cons(Z, U))) >= rcons(negrecip(activate(Z)), 2ndspos(X, activate(U))) pi(X) > 2ndspos(X, from(_|_)) plus(_|_, X) >= X plus(s(X), Y) >= s(plus(X, Y)) times(_|_, X) >= _|_ times(s(X), Y) > plus(Y, times(X, Y)) square(X) >= times(X, X) from(X) >= n!6220!6220from(X) s(X) >= n!6220!6220s(X) cons(X, Y) > n!6220!6220cons(X, Y)
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