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TRS Stand 20472 pair #381711730
details
property
value
status
complete
benchmark
PEANO_nokinds_FR.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n054.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.548007011414 seconds
cpu usage
0.545548659
max memory
1.78176E7
stage attributes
key
value
output-size
27064
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 U11 : [o * o] --> o U21 : [o * o * o] --> o activate : [o] --> o and : [o * o] --> o isNat : [o] --> o n!6220!62200 : [] --> o n!6220!6220isNat : [o] --> o n!6220!6220plus : [o * o] --> o n!6220!6220s : [o] --> o plus : [o * o] --> o s : [o] --> o tt : [] --> o U11(tt, X) => activate(X) U21(tt, X, Y) => s(plus(activate(Y), activate(X))) and(tt, X) => activate(X) isNat(n!6220!62200) => tt isNat(n!6220!6220plus(X, Y)) => and(isNat(activate(X)), n!6220!6220isNat(activate(Y))) isNat(n!6220!6220s(X)) => isNat(activate(X)) plus(X, 0) => U11(isNat(X), X) plus(X, s(Y)) => U21(and(isNat(Y), n!6220!6220isNat(X)), Y, X) 0 => n!6220!62200 plus(X, Y) => n!6220!6220plus(X, Y) isNat(X) => n!6220!6220isNat(X) s(X) => n!6220!6220s(X) activate(n!6220!62200) => 0 activate(n!6220!6220plus(X, Y)) => plus(activate(X), activate(Y)) activate(n!6220!6220isNat(X)) => isNat(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]): U11(tt, X) >? activate(X) U21(tt, X, Y) >? s(plus(activate(Y), activate(X))) and(tt, X) >? activate(X) isNat(n!6220!62200) >? tt isNat(n!6220!6220plus(X, Y)) >? and(isNat(activate(X)), n!6220!6220isNat(activate(Y))) isNat(n!6220!6220s(X)) >? isNat(activate(X)) plus(X, 0) >? U11(isNat(X), X) plus(X, s(Y)) >? U21(and(isNat(Y), n!6220!6220isNat(X)), Y, X) 0 >? n!6220!62200 plus(X, Y) >? n!6220!6220plus(X, Y) isNat(X) >? n!6220!6220isNat(X) s(X) >? n!6220!6220s(X) activate(n!6220!62200) >? 0 activate(n!6220!6220plus(X, Y)) >? plus(activate(X), activate(Y)) activate(n!6220!6220isNat(X)) >? isNat(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: [[0]] = _|_ [[U21(x_1, x_2, x_3)]] = U21(x_2, x_3, x_1) [[activate(x_1)]] = x_1 [[n!6220!62200]] = _|_ [[n!6220!6220plus(x_1, x_2)]] = n!6220!6220plus(x_2, x_1) [[plus(x_1, x_2)]] = plus(x_2, x_1) [[tt]] = _|_ We choose Lex = {U21, n!6220!6220plus, plus} and Mul = {U11, and, isNat, n!6220!6220isNat, n!6220!6220s, s}, and the following precedence: U21 = n!6220!6220plus = plus > U11 > and = isNat = n!6220!6220isNat > n!6220!6220s = s Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: U11(_|_, X) >= X U21(_|_, X, Y) >= s(plus(Y, X)) and(_|_, X) >= X isNat(_|_) >= _|_ isNat(n!6220!6220plus(X, Y)) >= and(isNat(X), n!6220!6220isNat(Y)) isNat(n!6220!6220s(X)) >= isNat(X) plus(X, _|_) > U11(isNat(X), X) plus(X, s(Y)) >= U21(and(isNat(Y), n!6220!6220isNat(X)), Y, X) _|_ >= _|_ plus(X, Y) >= n!6220!6220plus(X, Y) isNat(X) >= n!6220!6220isNat(X) s(X) >= n!6220!6220s(X) _|_ >= _|_ n!6220!6220plus(X, Y) >= plus(X, Y) n!6220!6220isNat(X) >= isNat(X) n!6220!6220s(X) >= s(X) X >= X
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