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TRS Stand 20472 pair #381712516
details
property
value
status
complete
benchmark
polo2.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n091.star.cs.uiowa.edu
space
Rubio_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.345877885818 seconds
cpu usage
0.342694762
max memory
1.3316096E7
stage attributes
key
value
output-size
17338
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. a : [] --> o div : [o * o] --> o dx : [o] --> o exp : [o * o] --> o ln : [o] --> o minus : [o * o] --> o neg : [o] --> o one : [] --> o plus : [o * o] --> o times : [o * o] --> o two : [] --> o zero : [] --> o dx(X) => one dx(a) => zero dx(plus(X, Y)) => plus(dx(X), dx(Y)) dx(times(X, Y)) => plus(times(Y, dx(X)), times(X, dx(Y))) dx(minus(X, Y)) => minus(dx(X), dx(Y)) dx(neg(X)) => neg(dx(X)) dx(div(X, Y)) => minus(div(dx(X), Y), times(X, div(dx(Y), exp(Y, two)))) dx(ln(X)) => div(dx(X), X) dx(exp(X, Y)) => plus(times(Y, times(exp(X, minus(Y, one)), dx(X))), times(exp(X, Y), times(ln(X), dx(Y)))) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): dx(X) >? one dx(a) >? zero dx(plus(X, Y)) >? plus(dx(X), dx(Y)) dx(times(X, Y)) >? plus(times(Y, dx(X)), times(X, dx(Y))) dx(minus(X, Y)) >? minus(dx(X), dx(Y)) dx(neg(X)) >? neg(dx(X)) dx(div(X, Y)) >? minus(div(dx(X), Y), times(X, div(dx(Y), exp(Y, two)))) dx(ln(X)) >? div(dx(X), X) dx(exp(X, Y)) >? plus(times(Y, times(exp(X, minus(Y, one)), dx(X))), times(exp(X, Y), times(ln(X), dx(Y)))) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[neg(x_1)]] = x_1 [[one]] = _|_ [[two]] = _|_ [[zero]] = _|_ We choose Lex = {} and Mul = {a, div, dx, exp, ln, minus, plus, times}, and the following precedence: a > dx = exp > plus > minus > ln > times > div Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: dx(X) > _|_ dx(a) >= _|_ dx(plus(X, Y)) > plus(dx(X), dx(Y)) dx(times(X, Y)) > plus(times(Y, dx(X)), times(X, dx(Y))) dx(minus(X, Y)) >= minus(dx(X), dx(Y)) dx(X) >= dx(X) dx(div(X, Y)) >= minus(div(dx(X), Y), times(X, div(dx(Y), exp(Y, _|_)))) dx(ln(X)) >= div(dx(X), X) dx(exp(X, Y)) > plus(times(Y, times(exp(X, minus(Y, _|_)), dx(X))), times(exp(X, Y), times(ln(X), dx(Y)))) With these choices, we have: 1] dx(X) > _|_ because [2], by definition 2] dx*(X) >= _|_ by (Bot) 3] dx(a) >= _|_ by (Bot) 4] dx(plus(X, Y)) > plus(dx(X), dx(Y)) because [5], by definition 5] dx*(plus(X, Y)) >= plus(dx(X), dx(Y)) because dx > plus, [6] and [10], by (Copy) 6] dx*(plus(X, Y)) >= dx(X) because dx in Mul and [7], by (Stat) 7] plus(X, Y) > X because [8], by definition 8] plus*(X, Y) >= X because [9], by (Select) 9] X >= X by (Meta) 10] dx*(plus(X, Y)) >= dx(Y) because dx in Mul and [11], by (Stat) 11] plus(X, Y) > Y because [12], by definition 12] plus*(X, Y) >= Y because [13], by (Select) 13] Y >= Y by (Meta) 14] dx(times(X, Y)) > plus(times(Y, dx(X)), times(X, dx(Y))) because [15], by definition 15] dx*(times(X, Y)) >= plus(times(Y, dx(X)), times(X, dx(Y))) because dx > plus, [16] and [23], by (Copy) 16] dx*(times(X, Y)) >= times(Y, dx(X)) because dx > times, [17] and [20], by (Copy) 17] dx*(times(X, Y)) >= Y because [18], by (Select) 18] times(X, Y) >= Y because [19], by (Star) 19] times*(X, Y) >= Y because [13], by (Select) 20] dx*(times(X, Y)) >= dx(X) because dx in Mul and [21], by (Stat) 21] times(X, Y) > X because [22], by definition 22] times*(X, Y) >= X because [9], by (Select)
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