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TRS Stand 20472 pair #381716340
details
property
value
status
complete
benchmark
Ex26_Luc03b_GM.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n192.star.cs.uiowa.edu
space
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.536817789078 seconds
cpu usage
0.518663094
max memory
1.4688256E7
stage attributes
key
value
output-size
27193
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 a!6220!6220add : [o * o] --> o a!6220!6220dbl : [o] --> o a!6220!6220first : [o * o] --> o a!6220!6220sqr : [o] --> o a!6220!6220terms : [o] --> o add : [o * o] --> o cons : [o * o] --> o dbl : [o] --> o first : [o * o] --> o mark : [o] --> o nil : [] --> o recip : [o] --> o s : [o] --> o sqr : [o] --> o terms : [o] --> o a!6220!6220terms(X) => cons(recip(a!6220!6220sqr(mark(X))), terms(s(X))) a!6220!6220sqr(0) => 0 a!6220!6220sqr(s(X)) => s(add(sqr(X), dbl(X))) a!6220!6220dbl(0) => 0 a!6220!6220dbl(s(X)) => s(s(dbl(X))) a!6220!6220add(0, X) => mark(X) a!6220!6220add(s(X), Y) => s(add(X, Y)) a!6220!6220first(0, X) => nil a!6220!6220first(s(X), cons(Y, Z)) => cons(mark(Y), first(X, Z)) mark(terms(X)) => a!6220!6220terms(mark(X)) mark(sqr(X)) => a!6220!6220sqr(mark(X)) mark(add(X, Y)) => a!6220!6220add(mark(X), mark(Y)) mark(dbl(X)) => a!6220!6220dbl(mark(X)) mark(first(X, Y)) => a!6220!6220first(mark(X), mark(Y)) mark(cons(X, Y)) => cons(mark(X), Y) mark(recip(X)) => recip(mark(X)) mark(s(X)) => s(X) mark(0) => 0 mark(nil) => nil a!6220!6220terms(X) => terms(X) a!6220!6220sqr(X) => sqr(X) a!6220!6220add(X, Y) => add(X, Y) a!6220!6220dbl(X) => dbl(X) a!6220!6220first(X, Y) => first(X, Y) 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] a!6220!6220terms#(X) =#> a!6220!6220sqr#(mark(X)) 1] a!6220!6220terms#(X) =#> mark#(X) 2] a!6220!6220add#(0, X) =#> mark#(X) 3] a!6220!6220first#(s(X), cons(Y, Z)) =#> mark#(Y) 4] mark#(terms(X)) =#> a!6220!6220terms#(mark(X)) 5] mark#(terms(X)) =#> mark#(X) 6] mark#(sqr(X)) =#> a!6220!6220sqr#(mark(X)) 7] mark#(sqr(X)) =#> mark#(X) 8] mark#(add(X, Y)) =#> a!6220!6220add#(mark(X), mark(Y)) 9] mark#(add(X, Y)) =#> mark#(X) 10] mark#(add(X, Y)) =#> mark#(Y) 11] mark#(dbl(X)) =#> a!6220!6220dbl#(mark(X)) 12] mark#(dbl(X)) =#> mark#(X) 13] mark#(first(X, Y)) =#> a!6220!6220first#(mark(X), mark(Y)) 14] mark#(first(X, Y)) =#> mark#(X) 15] mark#(first(X, Y)) =#> mark#(Y) 16] mark#(cons(X, Y)) =#> mark#(X) 17] mark#(recip(X)) =#> mark#(X) Rules R_0: a!6220!6220terms(X) => cons(recip(a!6220!6220sqr(mark(X))), terms(s(X))) a!6220!6220sqr(0) => 0 a!6220!6220sqr(s(X)) => s(add(sqr(X), dbl(X))) a!6220!6220dbl(0) => 0 a!6220!6220dbl(s(X)) => s(s(dbl(X))) a!6220!6220add(0, X) => mark(X) a!6220!6220add(s(X), Y) => s(add(X, Y)) a!6220!6220first(0, X) => nil a!6220!6220first(s(X), cons(Y, Z)) => cons(mark(Y), first(X, Z)) mark(terms(X)) => a!6220!6220terms(mark(X)) mark(sqr(X)) => a!6220!6220sqr(mark(X)) mark(add(X, Y)) => a!6220!6220add(mark(X), mark(Y)) mark(dbl(X)) => a!6220!6220dbl(mark(X)) mark(first(X, Y)) => a!6220!6220first(mark(X), mark(Y)) mark(cons(X, Y)) => cons(mark(X), Y) mark(recip(X)) => recip(mark(X)) mark(s(X)) => s(X) mark(0) => 0 mark(nil) => nil a!6220!6220terms(X) => terms(X)
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