details

property	value
status	complete
benchmark	Ex49_GM04.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	Strategy_removed_CSR_05

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.0837280750275 seconds
cpu usage	0.079197451
max memory	4448256.0

stage attributes

key	value
output-size	4823
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. 0 : [] --> o div : [o * o] --> o false : [] --> o geq : [o * o] --> o if : [o * o * o] --> o minus : [o * o] --> o s : [o] --> o true : [] --> o minus(0, X) => 0 minus(s(X), s(Y)) => minus(X, Y) geq(X, 0) => true geq(0, s(X)) => false geq(s(X), s(Y)) => geq(X, Y) div(0, s(X)) => 0 div(s(X), s(Y)) => if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0) if(true, X, Y) => X if(false, X, Y) => Y 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 : [] --> ic div : [ic * ic] --> ic false : [] --> ob geq : [ic * ic] --> ob if : [ob * ic * ic] --> ic minus : [ic * ic] --> ic s : [ic] --> ic true : [] --> ob 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#(s(X), s(Y)) =#> minus#(X, Y) 1] geq#(s(X), s(Y)) =#> geq#(X, Y) 2] div#(s(X), s(Y)) =#> if#(geq(X, Y), s(div(minus(X, Y), s(Y))), 0) 3] div#(s(X), s(Y)) =#> geq#(X, Y) 4] div#(s(X), s(Y)) =#> div#(minus(X, Y), s(Y)) 5] div#(s(X), s(Y)) =#> minus#(X, Y) Rules R_0: minus(0, X) => 0 minus(s(X), s(Y)) => minus(X, Y) geq(X, 0) => true geq(0, s(X)) => false geq(s(X), s(Y)) => geq(X, Y) div(0, s(X)) => 0 div(s(X), s(Y)) => if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0) if(true, X, Y) => X if(false, X, Y) => Y 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 : 1 * 2 : * 3 : 1 * 4 : * 5 : 0 This graph has the following strongly connected components: P_1: minus#(s(X), s(Y)) =#> minus#(X, Y) P_2: geq#(s(X), s(Y)) =#> geq#(X, Y) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f) and (P_2, R_0, m, f). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(geq#) = 1 Thus, we can orient the dependency pairs as follows: popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to TRS Stand 20472