details

property	value
status	complete
benchmark	JFP_Ex51.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n088.star.cs.uiowa.edu
space	AProVE_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.47412109375 seconds
cpu usage	0.46937188
max memory	1.3848576E7

stage attributes

key	value
output-size	18377
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 div : [o * o] --> o div!6220active : [o * o] --> o false : [] --> o ge : [o * o] --> o ge!6220active : [o * o] --> o if : [o * o * o] --> o if!6220active : [o * o * o] --> o mark : [o] --> o minus : [o * o] --> o minus!6220active : [o * o] --> o s : [o] --> o true : [] --> o minus!6220active(0, X) => 0 mark(0) => 0 minus!6220active(s(X), s(Y)) => minus!6220active(X, Y) mark(s(X)) => s(mark(X)) ge!6220active(X, 0) => true mark(minus(X, Y)) => minus!6220active(X, Y) ge!6220active(0, s(X)) => false mark(ge(X, Y)) => ge!6220active(X, Y) ge!6220active(s(X), s(Y)) => ge!6220active(X, Y) mark(div(X, Y)) => div!6220active(mark(X), Y) div!6220active(0, s(X)) => 0 mark(if(X, Y, Z)) => if!6220active(mark(X), Y, Z) div!6220active(s(X), s(Y)) => if!6220active(ge!6220active(X, Y), s(div(minus(X, Y), s(Y))), 0) if!6220active(true, X, Y) => mark(X) minus!6220active(X, Y) => minus(X, Y) if!6220active(false, X, Y) => mark(Y) ge!6220active(X, Y) => ge(X, Y) if!6220active(X, Y, Z) => if(X, Y, Z) div!6220active(X, Y) => div(X, Y) 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!6220active#(s(X), s(Y)) =#> minus!6220active#(X, Y) 1] mark#(s(X)) =#> mark#(X) 2] mark#(minus(X, Y)) =#> minus!6220active#(X, Y) 3] mark#(ge(X, Y)) =#> ge!6220active#(X, Y) 4] ge!6220active#(s(X), s(Y)) =#> ge!6220active#(X, Y) 5] mark#(div(X, Y)) =#> div!6220active#(mark(X), Y) 6] mark#(div(X, Y)) =#> mark#(X) 7] mark#(if(X, Y, Z)) =#> if!6220active#(mark(X), Y, Z) 8] mark#(if(X, Y, Z)) =#> mark#(X) 9] div!6220active#(s(X), s(Y)) =#> if!6220active#(ge!6220active(X, Y), s(div(minus(X, Y), s(Y))), 0) 10] div!6220active#(s(X), s(Y)) =#> ge!6220active#(X, Y) 11] if!6220active#(true, X, Y) =#> mark#(X) 12] if!6220active#(false, X, Y) =#> mark#(Y) Rules R_0: minus!6220active(0, X) => 0 mark(0) => 0 minus!6220active(s(X), s(Y)) => minus!6220active(X, Y) mark(s(X)) => s(mark(X)) ge!6220active(X, 0) => true mark(minus(X, Y)) => minus!6220active(X, Y) ge!6220active(0, s(X)) => false mark(ge(X, Y)) => ge!6220active(X, Y) ge!6220active(s(X), s(Y)) => ge!6220active(X, Y) mark(div(X, Y)) => div!6220active(mark(X), Y) div!6220active(0, s(X)) => 0 mark(if(X, Y, Z)) => if!6220active(mark(X), Y, Z) div!6220active(s(X), s(Y)) => if!6220active(ge!6220active(X, Y), s(div(minus(X, Y), s(Y))), 0) if!6220active(true, X, Y) => mark(X) minus!6220active(X, Y) => minus(X, Y) if!6220active(false, X, Y) => mark(Y) ge!6220active(X, Y) => ge(X, Y) if!6220active(X, Y, Z) => if(X, Y, Z) div!6220active(X, Y) => div(X, 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, 5, 6, 7, 8 * 2 : 0 * 3 : 4 * 4 : 4 * 5 : 9, 10 * 6 : 1, 2, 3, 5, 6, 7, 8 popout

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