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TRS Stand 20472 pair #381716085
details
property
value
status
complete
benchmark
Ex2_Luc02a_FR.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n100.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.361246109009 seconds
cpu usage
0.347590158
max memory
1.0539008E7
stage attributes
key
value
output-size
20872
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 activate : [o] --> o add : [o * o] --> o cons : [o * o] --> o dbl : [o] --> o first : [o * o] --> o n!6220!6220first : [o * o] --> o n!6220!6220s : [o] --> o n!6220!6220terms : [o] --> o nil : [] --> o recip : [o] --> o s : [o] --> o sqr : [o] --> o terms : [o] --> o terms(X) => cons(recip(sqr(X)), n!6220!6220terms(n!6220!6220s(X))) sqr(0) => 0 sqr(s(X)) => s(add(sqr(X), dbl(X))) dbl(0) => 0 dbl(s(X)) => s(s(dbl(X))) add(0, X) => X add(s(X), Y) => s(add(X, Y)) first(0, X) => nil first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(X, activate(Z))) terms(X) => n!6220!6220terms(X) s(X) => n!6220!6220s(X) first(X, Y) => n!6220!6220first(X, Y) activate(n!6220!6220terms(X)) => terms(activate(X)) activate(n!6220!6220s(X)) => s(activate(X)) activate(n!6220!6220first(X, Y)) => first(activate(X), activate(Y)) 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] terms#(X) =#> sqr#(X) 1] sqr#(s(X)) =#> s#(add(sqr(X), dbl(X))) 2] sqr#(s(X)) =#> add#(sqr(X), dbl(X)) 3] sqr#(s(X)) =#> sqr#(X) 4] sqr#(s(X)) =#> dbl#(X) 5] dbl#(s(X)) =#> s#(s(dbl(X))) 6] dbl#(s(X)) =#> s#(dbl(X)) 7] dbl#(s(X)) =#> dbl#(X) 8] add#(s(X), Y) =#> s#(add(X, Y)) 9] add#(s(X), Y) =#> add#(X, Y) 10] first#(s(X), cons(Y, Z)) =#> activate#(Z) 11] activate#(n!6220!6220terms(X)) =#> terms#(activate(X)) 12] activate#(n!6220!6220terms(X)) =#> activate#(X) 13] activate#(n!6220!6220s(X)) =#> s#(activate(X)) 14] activate#(n!6220!6220s(X)) =#> activate#(X) 15] activate#(n!6220!6220first(X, Y)) =#> first#(activate(X), activate(Y)) 16] activate#(n!6220!6220first(X, Y)) =#> activate#(X) 17] activate#(n!6220!6220first(X, Y)) =#> activate#(Y) Rules R_0: terms(X) => cons(recip(sqr(X)), n!6220!6220terms(n!6220!6220s(X))) sqr(0) => 0 sqr(s(X)) => s(add(sqr(X), dbl(X))) dbl(0) => 0 dbl(s(X)) => s(s(dbl(X))) add(0, X) => X add(s(X), Y) => s(add(X, Y)) first(0, X) => nil first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(X, activate(Z))) terms(X) => n!6220!6220terms(X) s(X) => n!6220!6220s(X) first(X, Y) => n!6220!6220first(X, Y) activate(n!6220!6220terms(X)) => terms(activate(X)) activate(n!6220!6220s(X)) => s(activate(X)) activate(n!6220!6220first(X, Y)) => first(activate(X), activate(Y)) activate(X) => X 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, 2, 3, 4 * 1 : * 2 : 8, 9 * 3 : 1, 2, 3, 4 * 4 : 5, 6, 7 * 5 : * 6 :
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