details

property	value
status	complete
benchmark	Ex4_Zan97_FR.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n022.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.691685914993 seconds
cpu usage	0.127737784
max memory	4173824.0

stage attributes

key	value
output-size	9606
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. 0 : [] --> o activate : [o] --> o cons : [o * o] --> o f : [o] --> o g : [o] --> o n!6220!6220f : [o] --> o n!6220!6220g : [o] --> o s : [o] --> o sel : [o * o] --> o f(X) => cons(X, n!6220!6220f(n!6220!6220g(X))) g(0) => s(0) g(s(X)) => s(s(g(X))) sel(0, cons(X, Y)) => X sel(s(X), cons(Y, Z)) => sel(X, activate(Z)) f(X) => n!6220!6220f(X) g(X) => n!6220!6220g(X) activate(n!6220!6220f(X)) => f(activate(X)) activate(n!6220!6220g(X)) => g(activate(X)) activate(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]): f(X) >? cons(X, n!6220!6220f(n!6220!6220g(X))) g(0) >? s(0) g(s(X)) >? s(s(g(X))) sel(0, cons(X, Y)) >? X sel(s(X), cons(Y, Z)) >? sel(X, activate(Z)) f(X) >? n!6220!6220f(X) g(X) >? n!6220!6220g(X) activate(n!6220!6220f(X)) >? f(activate(X)) activate(n!6220!6220g(X)) >? g(activate(X)) activate(X) >? X about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ We choose Lex = {sel} and Mul = {activate, cons, f, g, n!6220!6220f, n!6220!6220g, s}, and the following precedence: sel > activate > f > cons > g > n!6220!6220f > n!6220!6220g > s Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: f(X) > cons(X, n!6220!6220f(n!6220!6220g(X))) g(_|_) >= s(_|_) g(s(X)) >= s(s(g(X))) sel(_|_, cons(X, Y)) >= X sel(s(X), cons(Y, Z)) >= sel(X, activate(Z)) f(X) >= n!6220!6220f(X) g(X) >= n!6220!6220g(X) activate(n!6220!6220f(X)) >= f(activate(X)) activate(n!6220!6220g(X)) > g(activate(X)) activate(X) >= X With these choices, we have: 1] f(X) > cons(X, n!6220!6220f(n!6220!6220g(X))) because [2], by definition 2] f*(X) >= cons(X, n!6220!6220f(n!6220!6220g(X))) because f > cons, [3] and [5], by (Copy) 3] f*(X) >= X because [4], by (Select) 4] X >= X by (Meta) 5] f*(X) >= n!6220!6220f(n!6220!6220g(X)) because f > n!6220!6220f and [6], by (Copy) 6] f*(X) >= n!6220!6220g(X) because f > n!6220!6220g and [3], by (Copy) 7] g(_|_) >= s(_|_) because [8], by (Star) 8] g*(_|_) >= s(_|_) because g > s and [9], by (Copy) 9] g*(_|_) >= _|_ by (Bot) 10] g(s(X)) >= s(s(g(X))) because [11], by (Star) 11] g*(s(X)) >= s(s(g(X))) because g > s and [12], by (Copy) 12] g*(s(X)) >= s(g(X)) because g > s and [13], by (Copy) 13] g*(s(X)) >= g(X) because g in Mul and [14], by (Stat) 14] s(X) > X because [15], by definition 15] s*(X) >= X because [4], by (Select) 16] sel(_|_, cons(X, Y)) >= X because [17], by (Star) 17] sel*(_|_, cons(X, Y)) >= X because [18], by (Select) 18] cons(X, Y) >= X because [19], by (Star) 19] cons*(X, Y) >= X because [4], by (Select) 20] sel(s(X), cons(Y, Z)) >= sel(X, activate(Z)) because [21], by (Star) 21] sel*(s(X), cons(Y, Z)) >= sel(X, activate(Z)) because [14], [22] and [24], by (Stat) 22] sel*(s(X), cons(Y, Z)) >= X because [23], by (Select) 23] s(X) >= X because [15], by (Star) 24] sel*(s(X), cons(Y, Z)) >= activate(Z) because sel > activate and [25], by (Copy) popout

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all output return to TRS Stand 20472