details

property	value
status	complete
benchmark	perfect2.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	Mixed_TRS

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.246085166931 seconds
cpu usage	0.144988893
max memory	7180288.0

stage attributes

key	value
output-size	7149
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 f : [o * o * o * o] --> o false : [] --> o if : [o * o * o] --> o le : [o * o] --> o minus : [o * o] --> o perfectp : [o] --> o s : [o] --> o true : [] --> o minus(0, X) => 0 minus(s(X), 0) => s(X) minus(s(X), s(Y)) => minus(X, Y) le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) if(true, X, Y) => X if(false, X, Y) => Y perfectp(0) => false perfectp(s(X)) => f(X, s(0), s(X), s(X)) f(0, X, 0, Y) => true f(0, X, s(Y), Z) => false f(s(X), 0, Y, Z) => f(X, Z, minus(Y, s(X)), Z) f(s(X), s(Y), Z, U) => if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 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] le#(s(X), s(Y)) =#> le#(X, Y) 2] perfectp#(s(X)) =#> f#(X, s(0), s(X), s(X)) 3] f#(s(X), 0, Y, Z) =#> f#(X, Z, minus(Y, s(X)), Z) 4] f#(s(X), 0, Y, Z) =#> minus#(Y, s(X)) 5] f#(s(X), s(Y), Z, U) =#> if#(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 6] f#(s(X), s(Y), Z, U) =#> le#(X, Y) 7] f#(s(X), s(Y), Z, U) =#> f#(s(X), minus(Y, X), Z, U) 8] f#(s(X), s(Y), Z, U) =#> minus#(Y, X) 9] f#(s(X), s(Y), Z, U) =#> f#(X, U, Z, U) Rules R_0: minus(0, X) => 0 minus(s(X), 0) => s(X) minus(s(X), s(Y)) => minus(X, Y) le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) if(true, X, Y) => X if(false, X, Y) => Y perfectp(0) => false perfectp(s(X)) => f(X, s(0), s(X), s(X)) f(0, X, 0, Y) => true f(0, X, s(Y), Z) => false f(s(X), 0, Y, Z) => f(X, Z, minus(Y, s(X)), Z) f(s(X), s(Y), Z, U) => if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 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 : 5, 6, 7, 8, 9 * 3 : 3, 4, 5, 6, 7, 8, 9 * 4 : 0 * 5 : * 6 : 1 * 7 : 3, 4, 5, 6, 7, 8, 9 * 8 : 0 * 9 : 3, 4, 5, 6, 7, 8, 9 This graph has the following strongly connected components: P_1: minus#(s(X), s(Y)) =#> minus#(X, Y) P_2: le#(s(X), s(Y)) =#> le#(X, Y) P_3: f#(s(X), 0, Y, Z) =#> f#(X, Z, minus(Y, s(X)), Z) popout

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