details

property	value
status	complete
benchmark	11.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n027.star.cs.uiowa.edu
space	Der95

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.387051820755 seconds
cpu usage	0.383032164
max memory	1.60768E7

stage attributes

key	value
output-size	17662
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. !minus : [o * o] --> o !plus : [o * o] --> o !times : [o * o] --> o 0 : [] --> o 1 : [] --> o 2 : [] --> o D : [o] --> o constant : [] --> o div : [o * o] --> o ln : [o] --> o minus : [o] --> o pow : [o * o] --> o t : [] --> o D(t) => 1 D(constant) => 0 D(!plus(X, Y)) => !plus(D(X), D(Y)) D(!times(X, Y)) => !plus(!times(Y, D(X)), !times(X, D(Y))) D(!minus(X, Y)) => !minus(D(X), D(Y)) D(minus(X)) => minus(D(X)) D(div(X, Y)) => !minus(div(D(X), Y), div(!times(X, D(Y)), pow(Y, 2))) D(ln(X)) => div(D(X), X) D(pow(X, Y)) => !plus(!times(!times(Y, pow(X, !minus(Y, 1))), D(X)), !times(!times(pow(X, Y), ln(X)), D(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]): D(t) >? 1 D(constant) >? 0 D(!plus(X, Y)) >? !plus(D(X), D(Y)) D(!times(X, Y)) >? !plus(!times(Y, D(X)), !times(X, D(Y))) D(!minus(X, Y)) >? !minus(D(X), D(Y)) D(minus(X)) >? minus(D(X)) D(div(X, Y)) >? !minus(div(D(X), Y), div(!times(X, D(Y)), pow(Y, 2))) D(ln(X)) >? div(D(X), X) D(pow(X, Y)) >? !plus(!times(!times(Y, pow(X, !minus(Y, 1))), D(X)), !times(!times(pow(X, Y), ln(X)), D(Y))) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[1]] = _|_ [[2]] = _|_ We choose Lex = {} and Mul = {!minus, !plus, !times, D, constant, div, ln, minus, pow, t}, and the following precedence: D = ln = minus > !times > pow > constant > t > div > !minus > !plus Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: D(t) >= _|_ D(constant) >= _|_ D(!plus(X, Y)) >= !plus(D(X), D(Y)) D(!times(X, Y)) >= !plus(!times(Y, D(X)), !times(X, D(Y))) D(!minus(X, Y)) > !minus(D(X), D(Y)) D(minus(X)) >= minus(D(X)) D(div(X, Y)) >= !minus(div(D(X), Y), div(!times(X, D(Y)), pow(Y, _|_))) D(ln(X)) >= div(D(X), X) D(pow(X, Y)) > !plus(!times(!times(Y, pow(X, !minus(Y, _|_))), D(X)), !times(!times(pow(X, Y), ln(X)), D(Y))) With these choices, we have: 1] D(t) >= _|_ by (Bot) 2] D(constant) >= _|_ by (Bot) 3] D(!plus(X, Y)) >= !plus(D(X), D(Y)) because [4], by (Star) 4] D*(!plus(X, Y)) >= !plus(D(X), D(Y)) because D > !plus, [5] and [9], by (Copy) 5] D*(!plus(X, Y)) >= D(X) because D in Mul and [6], by (Stat) 6] !plus(X, Y) > X because [7], by definition 7] !plus*(X, Y) >= X because [8], by (Select) 8] X >= X by (Meta) 9] D*(!plus(X, Y)) >= D(Y) because D in Mul and [10], by (Stat) 10] !plus(X, Y) > Y because [11], by definition 11] !plus*(X, Y) >= Y because [12], by (Select) 12] Y >= Y by (Meta) 13] D(!times(X, Y)) >= !plus(!times(Y, D(X)), !times(X, D(Y))) because [14], by (Star) 14] D*(!times(X, Y)) >= !plus(!times(Y, D(X)), !times(X, D(Y))) because D > !plus, [15] and [22], by (Copy) 15] D*(!times(X, Y)) >= !times(Y, D(X)) because D > !times, [16] and [19], by (Copy) 16] D*(!times(X, Y)) >= Y because [17], by (Select) 17] !times(X, Y) >= Y because [18], by (Star) 18] !times*(X, Y) >= Y because [12], by (Select) 19] D*(!times(X, Y)) >= D(X) because D in Mul and [20], by (Stat) 20] !times(X, Y) > X because [21], by definition 21] !times*(X, Y) >= X because [8], by (Select) 22] D*(!times(X, Y)) >= !times(X, D(Y)) because D > !times, [23] and [25], by (Copy) popout

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