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TRS Stand 20472 pair #381716215
details
property
value
status
complete
benchmark
PEANO_nokinds_GM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n046.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.665938854218 seconds
cpu usage
0.663634155
max memory
2.174976E7
stage attributes
key
value
output-size
30646
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 a!6220!6220U11 : [o * o] --> o a!6220!6220U21 : [o * o * o] --> o a!6220!6220and : [o * o] --> o a!6220!6220isNat : [o] --> o a!6220!6220plus : [o * o] --> o and : [o * o] --> o isNat : [o] --> o mark : [o] --> o plus : [o * o] --> o s : [o] --> o tt : [] --> o a!6220!6220U11(tt, X) => mark(X) a!6220!6220U21(tt, X, Y) => s(a!6220!6220plus(mark(Y), mark(X))) a!6220!6220and(tt, X) => mark(X) a!6220!6220isNat(0) => tt a!6220!6220isNat(plus(X, Y)) => a!6220!6220and(a!6220!6220isNat(X), isNat(Y)) a!6220!6220isNat(s(X)) => a!6220!6220isNat(X) a!6220!6220plus(X, 0) => a!6220!6220U11(a!6220!6220isNat(X), X) a!6220!6220plus(X, s(Y)) => a!6220!6220U21(a!6220!6220and(a!6220!6220isNat(Y), isNat(X)), Y, X) mark(U11(X, Y)) => a!6220!6220U11(mark(X), Y) mark(U21(X, Y, Z)) => a!6220!6220U21(mark(X), Y, Z) mark(plus(X, Y)) => a!6220!6220plus(mark(X), mark(Y)) mark(and(X, Y)) => a!6220!6220and(mark(X), Y) mark(isNat(X)) => a!6220!6220isNat(X) mark(tt) => tt mark(s(X)) => s(mark(X)) mark(0) => 0 a!6220!6220U11(X, Y) => U11(X, Y) a!6220!6220U21(X, Y, Z) => U21(X, Y, Z) a!6220!6220plus(X, Y) => plus(X, Y) a!6220!6220and(X, Y) => and(X, Y) a!6220!6220isNat(X) => isNat(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]): a!6220!6220U11(tt, X) >? mark(X) a!6220!6220U21(tt, X, Y) >? s(a!6220!6220plus(mark(Y), mark(X))) a!6220!6220and(tt, X) >? mark(X) a!6220!6220isNat(0) >? tt a!6220!6220isNat(plus(X, Y)) >? a!6220!6220and(a!6220!6220isNat(X), isNat(Y)) a!6220!6220isNat(s(X)) >? a!6220!6220isNat(X) a!6220!6220plus(X, 0) >? a!6220!6220U11(a!6220!6220isNat(X), X) a!6220!6220plus(X, s(Y)) >? a!6220!6220U21(a!6220!6220and(a!6220!6220isNat(Y), isNat(X)), Y, X) mark(U11(X, Y)) >? a!6220!6220U11(mark(X), Y) mark(U21(X, Y, Z)) >? a!6220!6220U21(mark(X), Y, Z) mark(plus(X, Y)) >? a!6220!6220plus(mark(X), mark(Y)) mark(and(X, Y)) >? a!6220!6220and(mark(X), Y) mark(isNat(X)) >? a!6220!6220isNat(X) mark(tt) >? tt mark(s(X)) >? s(mark(X)) mark(0) >? 0 a!6220!6220U11(X, Y) >? U11(X, Y) a!6220!6220U21(X, Y, Z) >? U21(X, Y, Z) a!6220!6220plus(X, Y) >? plus(X, Y) a!6220!6220and(X, Y) >? and(X, Y) a!6220!6220isNat(X) >? isNat(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) [[a!6220!6220U21(x_1, x_2, x_3)]] = a!6220!6220U21(x_2, x_3, x_1) [[a!6220!6220and(x_1, x_2)]] = a!6220!6220and(x_2, x_1) [[a!6220!6220plus(x_1, x_2)]] = a!6220!6220plus(x_2, x_1) [[and(x_1, x_2)]] = and(x_2, x_1) [[mark(x_1)]] = x_1 [[plus(x_1, x_2)]] = plus(x_2, x_1) [[tt]] = _|_ We choose Lex = {U21, a!6220!6220U21, a!6220!6220and, a!6220!6220plus, and, plus} and Mul = {U11, a!6220!6220U11, a!6220!6220isNat, isNat, s}, and the following precedence: U21 = a!6220!6220U21 = a!6220!6220plus = plus > U11 = a!6220!6220U11 > a!6220!6220isNat = isNat = s > a!6220!6220and = and Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: a!6220!6220U11(_|_, X) >= X a!6220!6220U21(_|_, X, Y) >= s(a!6220!6220plus(Y, X)) a!6220!6220and(_|_, X) >= X a!6220!6220isNat(_|_) >= _|_ a!6220!6220isNat(plus(X, Y)) > a!6220!6220and(a!6220!6220isNat(X), isNat(Y)) a!6220!6220isNat(s(X)) > a!6220!6220isNat(X)
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