details

property	value
status	complete
benchmark	#3.5b.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n192.star.cs.uiowa.edu
space	AG01

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.154128074646 seconds
cpu usage	0.150013329
max memory	6533120.0

stage attributes

key	value
output-size	9346
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 if!6220minus : [o * o * o] --> o if!6220mod : [o * o * o] --> o le : [o * o] --> o minus : [o * o] --> o mod : [o * o] --> o s : [o] --> o true : [] --> o le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) minus(0, X) => 0 minus(s(X), Y) => if!6220minus(le(s(X), Y), s(X), Y) if!6220minus(true, s(X), Y) => 0 if!6220minus(false, s(X), Y) => s(minus(X, Y)) mod(0, X) => 0 mod(s(X), 0) => 0 mod(s(X), s(Y)) => if!6220mod(le(Y, X), s(X), s(Y)) if!6220mod(true, s(X), s(Y)) => mod(minus(X, Y), s(Y)) if!6220mod(false, s(X), s(Y)) => 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 : [] --> td false : [] --> sc if!6220minus : [sc * td * td] --> td if!6220mod : [sc * td * td] --> td le : [td * td] --> sc minus : [td * td] --> td mod : [td * td] --> td s : [td] --> td true : [] --> sc 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] le#(s(X), s(Y)) =#> le#(X, Y) 1] minus#(s(X), Y) =#> if!6220minus#(le(s(X), Y), s(X), Y) 2] minus#(s(X), Y) =#> le#(s(X), Y) 3] if!6220minus#(false, s(X), Y) =#> minus#(X, Y) 4] mod#(s(X), s(Y)) =#> if!6220mod#(le(Y, X), s(X), s(Y)) 5] mod#(s(X), s(Y)) =#> le#(Y, X) 6] if!6220mod#(true, s(X), s(Y)) =#> mod#(minus(X, Y), s(Y)) 7] if!6220mod#(true, s(X), s(Y)) =#> minus#(X, Y) Rules R_0: le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) minus(0, X) => 0 minus(s(X), Y) => if!6220minus(le(s(X), Y), s(X), Y) if!6220minus(true, s(X), Y) => 0 if!6220minus(false, s(X), Y) => s(minus(X, Y)) mod(0, X) => 0 mod(s(X), 0) => 0 mod(s(X), s(Y)) => if!6220mod(le(Y, X), s(X), s(Y)) if!6220mod(true, s(X), s(Y)) => mod(minus(X, Y), s(Y)) if!6220mod(false, s(X), s(Y)) => 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 : 0 * 1 : 3 * 2 : 0 * 3 : 1, 2 * 4 : 6, 7 * 5 : 0 * 6 : 4, 5 * 7 : 1, 2 This graph has the following strongly connected components: P_1: le#(s(X), s(Y)) =#> le#(X, Y) P_2: minus#(s(X), Y) =#> if!6220minus#(le(s(X), Y), s(X), Y) popout

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