details

property	value
status	complete
benchmark	2.17.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n041.star.cs.uiowa.edu
space	SK90

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.0922451019287 seconds
cpu usage	0.072350012
max memory	2756608.0

stage attributes

key	value
output-size	4457
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. !plus : [o * o] --> o 0 : [] --> o s : [o] --> o sum : [o] --> o sum1 : [o] --> o sum(0) => 0 sum(s(X)) => !plus(sum(X), s(X)) sum1(0) => 0 sum1(s(X)) => s(!plus(sum1(X), !plus(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]): sum(0) >? 0 sum(s(X)) >? !plus(sum(X), s(X)) sum1(0) >? 0 sum1(s(X)) >? s(!plus(sum1(X), !plus(X, X))) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {} and Mul = {!plus, 0, s, sum, sum1}, and the following precedence: sum > 0 > sum1 > !plus > s With these choices, we have: 1] sum(0) >= 0 because [2], by (Star) 2] sum*(0) >= 0 because sum > 0, by (Copy) 3] sum(s(X)) > !plus(sum(X), s(X)) because [4], by definition 4] sum*(s(X)) >= !plus(sum(X), s(X)) because sum > !plus, [5] and [9], by (Copy) 5] sum*(s(X)) >= sum(X) because sum in Mul and [6], by (Stat) 6] s(X) > X because [7], by definition 7] s*(X) >= X because [8], by (Select) 8] X >= X by (Meta) 9] sum*(s(X)) >= s(X) because sum > s and [10], by (Copy) 10] sum*(s(X)) >= X because [11], by (Select) 11] s(X) >= X because [7], by (Star) 12] sum1(0) > 0 because [13], by definition 13] sum1*(0) >= 0 because [14], by (Select) 14] 0 >= 0 by (Fun) 15] sum1(s(X)) >= s(!plus(sum1(X), !plus(X, X))) because [16], by (Star) 16] sum1*(s(X)) >= s(!plus(sum1(X), !plus(X, X))) because sum1 > s and [17], by (Copy) 17] sum1*(s(X)) >= !plus(sum1(X), !plus(X, X)) because sum1 > !plus, [18] and [19], by (Copy) 18] sum1*(s(X)) >= sum1(X) because sum1 in Mul and [6], by (Stat) 19] sum1*(s(X)) >= !plus(X, X) because sum1 > !plus, [20] and [20], by (Copy) 20] sum1*(s(X)) >= X because [11], by (Select) We can thus remove the following rules: sum(s(X)) => !plus(sum(X), s(X)) sum1(0) => 0 We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): sum(0) >? 0 sum1(s(X)) >? s(!plus(sum1(X), !plus(X, X))) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ We choose Lex = {} and Mul = {!plus, s, sum, sum1}, and the following precedence: sum > sum1 > s > !plus Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: sum(_|_) >= _|_ sum1(s(X)) > s(!plus(sum1(X), !plus(X, X))) With these choices, we have: 1] sum(_|_) >= _|_ by (Bot) 2] sum1(s(X)) > s(!plus(sum1(X), !plus(X, X))) because [3], by definition 3] sum1*(s(X)) >= s(!plus(sum1(X), !plus(X, X))) because sum1 > s and [4], by (Copy) 4] sum1*(s(X)) >= !plus(sum1(X), !plus(X, X)) because sum1 > !plus, [5] and [9], by (Copy) 5] sum1*(s(X)) >= sum1(X) because sum1 in Mul and [6], by (Stat) 6] s(X) > X because [7], by definition 7] s*(X) >= X because [8], by (Select) popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to TRS Stand 20472