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TRS Stand 20472 pair #381716850
details
property
value
status
complete
benchmark
LISTUTILITIES_nosorts_FR.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n049.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.26464009285 seconds
cpu usage
0.260599225
max memory
1.359872E7
stage attributes
key
value
output-size
8754
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. 0 : [] --> o U11 : [o * o * o * o] --> o U12 : [o * o] --> o activate : [o] --> o afterNth : [o * o] --> o and : [o * o] --> o cons : [o * o] --> o fst : [o] --> o head : [o] --> o n!6220!6220natsFrom : [o] --> o n!6220!6220s : [o] --> o natsFrom : [o] --> o nil : [] --> o pair : [o * o] --> o s : [o] --> o sel : [o * o] --> o snd : [o] --> o splitAt : [o * o] --> o tail : [o] --> o take : [o * o] --> o tt : [] --> o U11(tt, X, Y, Z) => U12(splitAt(activate(X), activate(Z)), activate(Y)) U12(pair(X, Y), Z) => pair(cons(activate(Z), X), Y) afterNth(X, Y) => snd(splitAt(X, Y)) and(tt, X) => activate(X) fst(pair(X, Y)) => X head(cons(X, Y)) => X natsFrom(X) => cons(X, n!6220!6220natsFrom(n!6220!6220s(X))) sel(X, Y) => head(afterNth(X, Y)) snd(pair(X, Y)) => Y splitAt(0, X) => pair(nil, X) splitAt(s(X), cons(Y, Z)) => U11(tt, X, Y, activate(Z)) tail(cons(X, Y)) => activate(Y) take(X, Y) => fst(splitAt(X, Y)) natsFrom(X) => n!6220!6220natsFrom(X) s(X) => n!6220!6220s(X) activate(n!6220!6220natsFrom(X)) => natsFrom(activate(X)) activate(n!6220!6220s(X)) => s(activate(X)) activate(X) => X 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] U11#(tt, X, Y, Z) =#> U12#(splitAt(activate(X), activate(Z)), activate(Y)) 1] U11#(tt, X, Y, Z) =#> splitAt#(activate(X), activate(Z)) 2] U11#(tt, X, Y, Z) =#> activate#(X) 3] U11#(tt, X, Y, Z) =#> activate#(Z) 4] U11#(tt, X, Y, Z) =#> activate#(Y) 5] U12#(pair(X, Y), Z) =#> activate#(Z) 6] afterNth#(X, Y) =#> snd#(splitAt(X, Y)) 7] afterNth#(X, Y) =#> splitAt#(X, Y) 8] and#(tt, X) =#> activate#(X) 9] sel#(X, Y) =#> head#(afterNth(X, Y)) 10] sel#(X, Y) =#> afterNth#(X, Y) 11] splitAt#(s(X), cons(Y, Z)) =#> U11#(tt, X, Y, activate(Z)) 12] splitAt#(s(X), cons(Y, Z)) =#> activate#(Z) 13] tail#(cons(X, Y)) =#> activate#(Y) 14] take#(X, Y) =#> fst#(splitAt(X, Y)) 15] take#(X, Y) =#> splitAt#(X, Y) 16] activate#(n!6220!6220natsFrom(X)) =#> natsFrom#(activate(X)) 17] activate#(n!6220!6220natsFrom(X)) =#> activate#(X) 18] activate#(n!6220!6220s(X)) =#> s#(activate(X)) 19] activate#(n!6220!6220s(X)) =#> activate#(X) Rules R_0: U11(tt, X, Y, Z) => U12(splitAt(activate(X), activate(Z)), activate(Y)) U12(pair(X, Y), Z) => pair(cons(activate(Z), X), Y) afterNth(X, Y) => snd(splitAt(X, Y)) and(tt, X) => activate(X) fst(pair(X, Y)) => X head(cons(X, Y)) => X natsFrom(X) => cons(X, n!6220!6220natsFrom(n!6220!6220s(X))) sel(X, Y) => head(afterNth(X, Y)) snd(pair(X, Y)) => Y splitAt(0, X) => pair(nil, X) splitAt(s(X), cons(Y, Z)) => U11(tt, X, Y, activate(Z)) tail(cons(X, Y)) => activate(Y) take(X, Y) => fst(splitAt(X, Y)) natsFrom(X) => n!6220!6220natsFrom(X) s(X) => n!6220!6220s(X) activate(n!6220!6220natsFrom(X)) => natsFrom(activate(X)) activate(n!6220!6220s(X)) => s(activate(X)) activate(X) => X
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