details

property	value
status	complete
benchmark	secret2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n113.star.cs.uiowa.edu
space	Secret_07_TRS

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.570471048355 seconds
cpu usage	0.564111936
max memory	2.3171072E7

stage attributes

key	value
output-size	24546
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 !times : [o * o] --> o 1 : [] --> o a : [o * o * o * o] --> o h : [] --> o s : [o] --> o a(h, h, h, X) => s(X) a(X, Y, s(Z), h) => a(X, Y, Z, s(h)) a(X, Y, s(Z), s(U)) => a(X, Y, Z, a(X, Y, s(Z), U)) a(X, s(Y), h, Z) => a(X, Y, Z, Z) a(s(X), h, h, Y) => a(X, Y, h, Y) !plus(X, h) => X !plus(h, X) => X !plus(s(X), s(Y)) => s(s(!plus(X, Y))) !plus(!plus(X, Y), Z) => !plus(X, !plus(Y, Z)) s(h) => 1 !times(h, X) => h !times(X, h) => h !times(s(X), s(Y)) => s(!plus(!plus(!times(X, Y), X), 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]): a(h, h, h, X) >? s(X) a(X, Y, s(Z), h) >? a(X, Y, Z, s(h)) a(X, Y, s(Z), s(U)) >? a(X, Y, Z, a(X, Y, s(Z), U)) a(X, s(Y), h, Z) >? a(X, Y, Z, Z) a(s(X), h, h, Y) >? a(X, Y, h, Y) !plus(X, h) >? X !plus(h, X) >? X !plus(s(X), s(Y)) >? s(s(!plus(X, Y))) !plus(!plus(X, Y), Z) >? !plus(X, !plus(Y, Z)) s(h) >? 1 !times(h, X) >? h !times(X, h) >? h !times(s(X), s(Y)) >? s(!plus(!plus(!times(X, Y), X), Y)) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[1]] = _|_ We choose Lex = {!plus, a} and Mul = {!times, h, s}, and the following precedence: a > !times > !plus > h = s Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: a(h, h, h, X) >= s(X) a(X, Y, s(Z), h) > a(X, Y, Z, s(h)) a(X, Y, s(Z), s(U)) >= a(X, Y, Z, a(X, Y, s(Z), U)) a(X, s(Y), h, Z) > a(X, Y, Z, Z) a(s(X), h, h, Y) >= a(X, Y, h, Y) !plus(X, h) > X !plus(h, X) > X !plus(s(X), s(Y)) > s(s(!plus(X, Y))) !plus(!plus(X, Y), Z) >= !plus(X, !plus(Y, Z)) s(h) >= _|_ !times(h, X) >= h !times(X, h) >= h !times(s(X), s(Y)) >= s(!plus(!plus(!times(X, Y), X), Y)) With these choices, we have: 1] a(h, h, h, X) >= s(X) because [2], by (Star) 2] a*(h, h, h, X) >= s(X) because a > s and [3], by (Copy) 3] a*(h, h, h, X) >= X because [4], by (Select) 4] X >= X by (Meta) 5] a(X, Y, s(Z), h) > a(X, Y, Z, s(h)) because [6], by definition 6] a*(X, Y, s(Z), h) >= a(X, Y, Z, s(h)) because [7], [8], [9], [12], [13], [14] and [16], by (Stat) 7] X >= X by (Meta) 8] Y >= Y by (Meta) 9] s(Z) > Z because [10], by definition 10] s*(Z) >= Z because [11], by (Select) 11] Z >= Z by (Meta) 12] a*(X, Y, s(Z), h) >= X because [7], by (Select) 13] a*(X, Y, s(Z), h) >= Y because [8], by (Select) 14] a*(X, Y, s(Z), h) >= Z because [15], by (Select) 15] s(Z) >= Z because [10], by (Star) 16] a*(X, Y, s(Z), h) >= s(h) because a > s and [17], by (Copy) 17] a*(X, Y, s(Z), h) >= h because [18], by (Select) 18] s(Z) >= h because [19], by (Star) 19] s*(Z) >= h because s = h and s in Mul, by (Stat) 20] a(X, Y, s(Z), s(U)) >= a(X, Y, Z, a(X, Y, s(Z), U)) because [21], by (Star) popout

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all output return to TRS Stand 20472