details

property	value
status	complete
benchmark	secret4.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n053.star.cs.uiowa.edu
space	Secret_07_TRS

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.608064889908 seconds
cpu usage	0.605488652
max memory	2.4838144E7

stage attributes

key	value
output-size	25520
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. !plus : [o * o] --> o 1 : [] --> o a : [o * o * o * o] --> o app : [o * o] --> o cons : [o * o] --> o h : [] --> o nil : [] --> o s : [o] --> o sum : [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 app(nil, X) => X app(X, nil) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) sum(cons(X, nil)) => cons(X, nil) sum(cons(X, cons(Y, Z))) => sum(cons(a(X, Y, h, h), Z)) 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 app(nil, X) >? X app(X, nil) >? X app(cons(X, Y), Z) >? cons(X, app(Y, Z)) sum(cons(X, nil)) >? cons(X, nil) sum(cons(X, cons(Y, Z))) >? sum(cons(a(X, Y, h, h), Z)) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[1]] = _|_ [[cons(x_1, x_2)]] = cons(x_2, x_1) [[h]] = _|_ We choose Lex = {!plus, a, cons} and Mul = {app, nil, s, sum}, and the following precedence: sum > !plus > nil > app > cons > a > s Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: a(_|_, _|_, _|_, X) > s(X) a(X, Y, s(Z), _|_) >= a(X, Y, Z, s(_|_)) a(X, Y, s(Z), s(U)) >= a(X, Y, Z, a(X, Y, s(Z), U)) a(X, s(Y), _|_, Z) > a(X, Y, Z, Z) a(s(X), _|_, _|_, Y) > a(X, Y, _|_, Y) !plus(X, _|_) > X !plus(_|_, X) > X !plus(s(X), s(Y)) > s(s(!plus(X, Y))) !plus(!plus(X, Y), Z) >= !plus(X, !plus(Y, Z)) s(_|_) >= _|_ app(nil, X) >= X app(X, nil) > X app(cons(X, Y), Z) >= cons(X, app(Y, Z)) sum(cons(X, nil)) >= cons(X, nil) sum(cons(X, cons(Y, Z))) >= sum(cons(a(X, Y, _|_, _|_), Z)) With these choices, we have: 1] a(_|_, _|_, _|_, X) > s(X) because [2], by definition 2] a*(_|_, _|_, _|_, X) >= s(X) because a > s and [3], by (Copy) 3] a*(_|_, _|_, _|_, X) >= X because [4], by (Select) 4] X >= X by (Meta) 5] a(X, Y, s(Z), _|_) >= a(X, Y, Z, s(_|_)) because [6], by (Star) 6] a*(X, Y, s(Z), _|_) >= a(X, Y, Z, s(_|_)) 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) popout

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