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TRS Stand 20472 pair #381715195
details
property
value
status
complete
benchmark
Ex26_Luc03b_Z.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n016.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
1.19696903229 seconds
cpu usage
0.703313491
max memory
1.5384576E7
stage attributes
key
value
output-size
34960
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!6220add : [o * o] --> o n!6220!6220dbl : [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(s(X))) sqr(0) => 0 sqr(s(X)) => s(n!6220!6220add(sqr(activate(X)), dbl(activate(X)))) dbl(0) => 0 dbl(s(X)) => s(n!6220!6220s(n!6220!6220dbl(activate(X)))) add(0, X) => X add(s(X), Y) => s(n!6220!6220add(activate(X), Y)) first(0, X) => nil first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(activate(X), activate(Z))) terms(X) => n!6220!6220terms(X) add(X, Y) => n!6220!6220add(X, Y) s(X) => n!6220!6220s(X) dbl(X) => n!6220!6220dbl(X) first(X, Y) => n!6220!6220first(X, Y) activate(n!6220!6220terms(X)) => terms(X) activate(n!6220!6220add(X, Y)) => add(X, Y) activate(n!6220!6220s(X)) => s(X) activate(n!6220!6220dbl(X)) => dbl(X) activate(n!6220!6220first(X, Y)) => first(X, 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] terms#(X) =#> s#(X) 2] sqr#(s(X)) =#> s#(n!6220!6220add(sqr(activate(X)), dbl(activate(X)))) 3] sqr#(s(X)) =#> sqr#(activate(X)) 4] sqr#(s(X)) =#> activate#(X) 5] sqr#(s(X)) =#> dbl#(activate(X)) 6] sqr#(s(X)) =#> activate#(X) 7] dbl#(s(X)) =#> s#(n!6220!6220s(n!6220!6220dbl(activate(X)))) 8] dbl#(s(X)) =#> activate#(X) 9] add#(s(X), Y) =#> s#(n!6220!6220add(activate(X), Y)) 10] add#(s(X), Y) =#> activate#(X) 11] first#(s(X), cons(Y, Z)) =#> activate#(X) 12] first#(s(X), cons(Y, Z)) =#> activate#(Z) 13] activate#(n!6220!6220terms(X)) =#> terms#(X) 14] activate#(n!6220!6220add(X, Y)) =#> add#(X, Y) 15] activate#(n!6220!6220s(X)) =#> s#(X) 16] activate#(n!6220!6220dbl(X)) =#> dbl#(X) 17] activate#(n!6220!6220first(X, Y)) =#> first#(X, Y) Rules R_0: terms(X) => cons(recip(sqr(X)), n!6220!6220terms(s(X))) sqr(0) => 0 sqr(s(X)) => s(n!6220!6220add(sqr(activate(X)), dbl(activate(X)))) dbl(0) => 0 dbl(s(X)) => s(n!6220!6220s(n!6220!6220dbl(activate(X)))) add(0, X) => X add(s(X), Y) => s(n!6220!6220add(activate(X), Y)) first(0, X) => nil first(s(X), cons(Y, Z)) => cons(Y, n!6220!6220first(activate(X), activate(Z))) terms(X) => n!6220!6220terms(X) add(X, Y) => n!6220!6220add(X, Y) s(X) => n!6220!6220s(X) dbl(X) => n!6220!6220dbl(X) first(X, Y) => n!6220!6220first(X, Y) activate(n!6220!6220terms(X)) => terms(X) activate(n!6220!6220add(X, Y)) => add(X, Y) activate(n!6220!6220s(X)) => s(X) activate(n!6220!6220dbl(X)) => dbl(X) activate(n!6220!6220first(X, Y)) => first(X, 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).
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