details

property	value
status	complete
benchmark	012.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n036.star.cs.uiowa.edu
space	AotoYamada_05

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	1.13167500496 seconds
cpu usage	1.128666254
max memory	4.4896256E7

stage attributes

key	value
output-size	13330
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. and : [] --> o app : [o * o] --> o cons : [] --> o false : [] --> o forall : [] --> o forsome : [] --> o nil : [] --> o or : [] --> o true : [] --> o app(app(and, true), true) => true app(app(and, X), false) => false app(app(and, false), X) => false app(app(or, true), X) => true app(app(or, X), true) => true app(app(or, false), false) => false app(app(forall, X), nil) => true app(app(forall, X), app(app(cons, Y), Z)) => app(app(and, app(X, Y)), app(app(forall, X), Z)) app(app(forsome, X), nil) => false app(app(forsome, X), app(app(cons, Y), Z)) => app(app(or, app(X, Y)), app(app(forsome, X), Z)) 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(forall, X), app(app(cons, Y), Z)) =#> app#(app(and, app(X, Y)), app(app(forall, X), Z)) 1] app#(app(forall, X), app(app(cons, Y), Z)) =#> app#(and, app(X, Y)) 2] app#(app(forall, X), app(app(cons, Y), Z)) =#> app#(X, Y) 3] app#(app(forall, X), app(app(cons, Y), Z)) =#> app#(app(forall, X), Z) 4] app#(app(forall, X), app(app(cons, Y), Z)) =#> app#(forall, X) 5] app#(app(forsome, X), app(app(cons, Y), Z)) =#> app#(app(or, app(X, Y)), app(app(forsome, X), Z)) 6] app#(app(forsome, X), app(app(cons, Y), Z)) =#> app#(or, app(X, Y)) 7] app#(app(forsome, X), app(app(cons, Y), Z)) =#> app#(X, Y) 8] app#(app(forsome, X), app(app(cons, Y), Z)) =#> app#(app(forsome, X), Z) 9] app#(app(forsome, X), app(app(cons, Y), Z)) =#> app#(forsome, X) Rules R_0: app(app(and, true), true) => true app(app(and, X), false) => false app(app(and, false), X) => false app(app(or, true), X) => true app(app(or, X), true) => true app(app(or, false), false) => false app(app(forall, X), nil) => true app(app(forall, X), app(app(cons, Y), Z)) => app(app(and, app(X, Y)), app(app(forall, X), Z)) app(app(forsome, X), nil) => false app(app(forsome, X), app(app(cons, Y), Z)) => app(app(or, app(X, Y)), app(app(forsome, X), Z)) 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, 2, 3, 4, 5, 6, 7, 8, 9 * 1 : * 2 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 * 3 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 * 4 : * 5 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 * 6 : * 7 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 * 8 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 * 9 : This graph has the following strongly connected components: P_1: app#(app(forall, X), app(app(cons, Y), Z)) =#> app#(app(and, app(X, Y)), app(app(forall, X), Z)) app#(app(forall, X), app(app(cons, Y), Z)) =#> app#(X, Y) app#(app(forall, X), app(app(cons, Y), Z)) =#> app#(app(forall, X), Z) app#(app(forsome, X), app(app(cons, Y), Z)) =#> app#(app(or, app(X, Y)), app(app(forsome, X), Z)) app#(app(forsome, X), app(app(cons, Y), Z)) =#> app#(X, Y) app#(app(forsome, X), app(app(cons, Y), Z)) =#> app#(app(forsome, X), Z) 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). Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_1, R_0, minimal, formative). The formative rules of (P_1, R_0) are R_1 ::= app(app(forall, X), app(app(cons, Y), Z)) => app(app(and, app(X, Y)), app(app(forall, X), Z)) app(app(forsome, X), app(app(cons, Y), Z)) => app(app(or, app(X, Y)), app(app(forsome, X), Z)) popout

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