details

property	value
status	complete
benchmark	014.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n071.star.cs.uiowa.edu
space	AotoYamada_05

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	1.51580190659 seconds
cpu usage	1.485275666
max memory	5.2916224E7

stage attributes

key	value
output-size	17605
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 app : [o * o] --> o cons : [] --> o double : [] --> o inc : [] --> o map : [] --> o nil : [] --> o plus : [] --> o s : [] --> o times : [] --> o app(app(plus, 0), X) => X app(app(plus, app(s, X)), Y) => app(s, app(app(plus, X), Y)) app(app(times, 0), X) => 0 app(app(times, app(s, X)), Y) => app(app(plus, app(app(times, X), Y)), Y) app(app(map, X), nil) => nil app(app(map, X), app(app(cons, Y), Z)) => app(app(cons, app(X, Y)), app(app(map, X), Z)) inc => app(map, app(plus, app(s, 0))) double => app(map, app(times, app(s, app(s, 0)))) 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] app#(app(plus, app(s, X)), Y) =#> app#(s, app(app(plus, X), Y)) 1] app#(app(plus, app(s, X)), Y) =#> app#(app(plus, X), Y) 2] app#(app(plus, app(s, X)), Y) =#> app#(plus, X) 3] app#(app(times, app(s, X)), Y) =#> app#(app(plus, app(app(times, X), Y)), Y) 4] app#(app(times, app(s, X)), Y) =#> app#(plus, app(app(times, X), Y)) 5] app#(app(times, app(s, X)), Y) =#> app#(app(times, X), Y) 6] app#(app(times, app(s, X)), Y) =#> app#(times, X) 7] app#(app(map, X), app(app(cons, Y), Z)) =#> app#(app(cons, app(X, Y)), app(app(map, X), Z)) 8] app#(app(map, X), app(app(cons, Y), Z)) =#> app#(cons, app(X, Y)) 9] app#(app(map, X), app(app(cons, Y), Z)) =#> app#(X, Y) 10] app#(app(map, X), app(app(cons, Y), Z)) =#> app#(app(map, X), Z) 11] app#(app(map, X), app(app(cons, Y), Z)) =#> app#(map, X) 12] inc# =#> app#(map, app(plus, app(s, 0))) 13] inc# =#> app#(plus, app(s, 0)) 14] inc# =#> app#(s, 0) 15] double# =#> app#(map, app(times, app(s, app(s, 0)))) 16] double# =#> app#(times, app(s, app(s, 0))) 17] double# =#> app#(s, app(s, 0)) 18] double# =#> app#(s, 0) Rules R_0: app(app(plus, 0), X) => X app(app(plus, app(s, X)), Y) => app(s, app(app(plus, X), Y)) app(app(times, 0), X) => 0 app(app(times, app(s, X)), Y) => app(app(plus, app(app(times, X), Y)), Y) app(app(map, X), nil) => nil app(app(map, X), app(app(cons, Y), Z)) => app(app(cons, app(X, Y)), app(app(map, X), Z)) inc => app(map, app(plus, app(s, 0))) double => app(map, app(times, app(s, app(s, 0)))) 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, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 2 : * 3 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 4 : * 5 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 6 : * 7 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 8 : * 9 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 10 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9, 10, 11 * 11 : * 12 : * 13 : * 14 : * 15 : * 16 : * 17 : * 18 : This graph has the following strongly connected components: P_1: app#(app(plus, app(s, X)), Y) =#> app#(app(plus, X), Y) app#(app(times, app(s, X)), Y) =#> app#(app(plus, app(app(times, X), Y)), Y) popout

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