details

property	value
status	complete
benchmark	gcd.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n051.star.cs.uiowa.edu
space	Rubio_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.168771982193 seconds
cpu usage	0.166245341
max memory	6230016.0

stage attributes

key	value
output-size	10184
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 false : [] --> o gcd : [o * o] --> o if : [o * o * o] --> o le : [o * o] --> o minus : [o * o] --> o pred : [o] --> o s : [o] --> o true : [] --> o minus(X, s(Y)) => pred(minus(X, Y)) minus(X, 0) => X pred(s(X)) => X le(s(X), s(Y)) => le(X, Y) le(s(X), 0) => false le(0, X) => true gcd(0, X) => 0 gcd(s(X), 0) => s(X) gcd(s(X), s(Y)) => if(le(Y, X), s(X), s(Y)) if(true, s(X), s(Y)) => gcd(minus(X, Y), s(Y)) if(false, s(X), s(Y)) => gcd(minus(Y, X), s(X)) As the system is orthogonal, it is terminating if it is innermost terminating by [Gra95]. Then, by [FuhGieParSchSwi11], it suffices to prove (innermost) termination of the typed system, with sort annotations chosen to respect the rules, as follows: 0 : [] --> hd false : [] --> cc gcd : [hd * hd] --> hd if : [cc * hd * hd] --> hd le : [hd * hd] --> cc minus : [hd * hd] --> hd pred : [hd] --> hd s : [hd] --> hd true : [] --> cc We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs in [Kop12, Ch. 7.8]). We thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] minus#(X, s(Y)) =#> pred#(minus(X, Y)) 1] minus#(X, s(Y)) =#> minus#(X, Y) 2] le#(s(X), s(Y)) =#> le#(X, Y) 3] gcd#(s(X), s(Y)) =#> if#(le(Y, X), s(X), s(Y)) 4] gcd#(s(X), s(Y)) =#> le#(Y, X) 5] if#(true, s(X), s(Y)) =#> gcd#(minus(X, Y), s(Y)) 6] if#(true, s(X), s(Y)) =#> minus#(X, Y) 7] if#(false, s(X), s(Y)) =#> gcd#(minus(Y, X), s(X)) 8] if#(false, s(X), s(Y)) =#> minus#(Y, X) Rules R_0: minus(X, s(Y)) => pred(minus(X, Y)) minus(X, 0) => X pred(s(X)) => X le(s(X), s(Y)) => le(X, Y) le(s(X), 0) => false le(0, X) => true gcd(0, X) => 0 gcd(s(X), 0) => s(X) gcd(s(X), s(Y)) => if(le(Y, X), s(X), s(Y)) if(true, s(X), s(Y)) => gcd(minus(X, Y), s(Y)) if(false, s(X), s(Y)) => gcd(minus(Y, X), s(X)) Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_0, R_0, minimal, formative). We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows: * 0 : * 1 : 0, 1 * 2 : 2 * 3 : 5, 6, 7, 8 * 4 : 2 * 5 : 3, 4 * 6 : 0, 1 * 7 : 3, 4 * 8 : 0, 1 This graph has the following strongly connected components: P_1: minus#(X, s(Y)) =#> minus#(X, Y) P_2: le#(s(X), s(Y)) =#> le#(X, Y) popout

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