details

property	value
status	complete
benchmark	08.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n045.star.cs.uiowa.edu
space	Der95

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.128805160522 seconds
cpu usage	0.119784895
max memory	4956160.0

stage attributes

key	value
output-size	7931
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. !minus : [o * o] --> o !plus : [o * o] --> o !times : [o * o] --> o 0 : [] --> o 1 : [] --> o D : [o] --> o constant : [] --> 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)) 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)) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[1]] = _|_ We choose Lex = {} and Mul = {!minus, !plus, !times, D, constant, t}, and the following precedence: !times = D > !plus > t > constant > !minus 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)) 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 [21], by (Copy) 15] D*(!times(X, Y)) >= !times(Y, D(X)) because D = !times, D in Mul, [16] and [18], by (Stat) 16] !times(X, Y) > Y because [17], by definition 17] !times*(X, Y) >= Y because [12], by (Select) 18] !times(X, Y) > D(X) because [19], by definition 19] !times*(X, Y) >= D(X) because !times = D, !times in Mul and [20], by (Stat) 20] X >= X by (Meta) 21] D*(!times(X, Y)) >= !times(X, D(Y)) because D = !times, D in Mul, [22] and [24], by (Stat) 22] !times(X, Y) > X because [23], by definition 23] !times*(X, Y) >= X because [20], by (Select) 24] !times(X, Y) > D(Y) because [25], by definition 25] !times*(X, Y) >= D(Y) because !times = D, !times in Mul and [26], by (Stat) 26] Y >= Y by (Meta) 27] D(!minus(X, Y)) > !minus(D(X), D(Y)) because [28], by definition 28] D*(!minus(X, Y)) >= !minus(D(X), D(Y)) because D > !minus, [29] and [32], by (Copy) 29] D*(!minus(X, Y)) >= D(X) because D in Mul and [30], by (Stat) 30] !minus(X, Y) > X because [31], by definition 31] !minus*(X, Y) >= X because [20], by (Select) 32] D*(!minus(X, Y)) >= D(Y) because D in Mul and [33], by (Stat) 33] !minus(X, Y) > Y because [34], by definition 34] !minus*(X, Y) >= Y because [26], by (Select) We can thus remove the following rules: D(!minus(X, Y)) => !minus(D(X), D(Y)) popout

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