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TRS Stand 20472 pair #381711700
details
property
value
status
complete
benchmark
011.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n010.star.cs.uiowa.edu
space
AotoYamada_05
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.951241970062 seconds
cpu usage
0.93951525
max memory
4.4335104E7
stage attributes
key
value
output-size
8283
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 curry : [] --> o inc : [] --> o map : [] --> o nil : [] --> o plus : [] --> o s : [] --> o app(app(plus, 0), X) => X app(app(plus, app(s, X)), Y) => app(s, app(app(plus, X), 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)) app(app(app(curry, X), Y), Z) => app(app(X, Y), Z) inc => app(map, app(app(curry, plus), 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(map, X), app(app(cons, Y), Z)) =#> app#(app(cons, app(X, Y)), app(app(map, X), Z)) 4] app#(app(map, X), app(app(cons, Y), Z)) =#> app#(cons, app(X, Y)) 5] app#(app(map, X), app(app(cons, Y), Z)) =#> app#(X, Y) 6] app#(app(map, X), app(app(cons, Y), Z)) =#> app#(app(map, X), Z) 7] app#(app(map, X), app(app(cons, Y), Z)) =#> app#(map, X) 8] app#(app(app(curry, X), Y), Z) =#> app#(app(X, Y), Z) 9] app#(app(app(curry, X), Y), Z) =#> app#(X, Y) 10] inc# =#> app#(map, app(app(curry, plus), app(s, 0))) 11] inc# =#> app#(app(curry, plus), app(s, 0)) 12] inc# =#> app#(curry, plus) 13] inc# =#> 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(map, X), nil) => nil app(app(map, X), app(app(cons, Y), Z)) => app(app(cons, app(X, Y)), app(app(map, X), Z)) app(app(app(curry, X), Y), Z) => app(app(X, Y), Z) inc => app(map, app(app(curry, plus), 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 * 2 : * 3 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 * 4 : * 5 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 * 6 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 * 7 : * 8 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 * 9 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 * 10 : * 11 : 0, 1, 2, 3, 4, 5, 6, 7, 8, 9 * 12 : * 13 : This graph has the following strongly connected components: P_1: app#(app(plus, app(s, X)), Y) =#> app#(app(plus, X), Y) app#(app(map, X), app(app(cons, Y), Z)) =#> app#(app(cons, app(X, Y)), app(app(map, X), Z)) app#(app(map, X), app(app(cons, Y), Z)) =#> app#(X, Y) app#(app(map, X), app(app(cons, Y), Z)) =#> app#(app(map, X), Z) app#(app(app(curry, X), Y), Z) =#> app#(app(X, Y), Z) app#(app(app(curry, X), Y), Z) =#> app#(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). 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(plus, 0), X) => X app(app(plus, app(s, X)), Y) => app(s, app(app(plus, X), Y))
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