details

property	value
status	complete
benchmark	Ex2_Luc03b_GM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n102.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.206022024155 seconds
cpu usage	0.202714479
max memory	9383936.0

stage attributes

key	value
output-size	11298
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!6220from : [o] --> o a!6220!6220fst : [o * o] --> o a!6220!6220len : [o] --> o add : [o * o] --> o cons : [o * o] --> o from : [o] --> o fst : [o * o] --> o len : [o] --> o mark : [o] --> o nil : [] --> o s : [o] --> o a!6220!6220fst(0, X) => nil a!6220!6220fst(s(X), cons(Y, Z)) => cons(mark(Y), fst(X, Z)) a!6220!6220from(X) => cons(mark(X), from(s(X))) a!6220!6220add(0, X) => mark(X) a!6220!6220add(s(X), Y) => s(add(X, Y)) a!6220!6220len(nil) => 0 a!6220!6220len(cons(X, Y)) => s(len(Y)) mark(fst(X, Y)) => a!6220!6220fst(mark(X), mark(Y)) mark(from(X)) => a!6220!6220from(mark(X)) mark(add(X, Y)) => a!6220!6220add(mark(X), mark(Y)) mark(len(X)) => a!6220!6220len(mark(X)) mark(0) => 0 mark(s(X)) => s(X) mark(nil) => nil mark(cons(X, Y)) => cons(mark(X), Y) a!6220!6220fst(X, Y) => fst(X, Y) a!6220!6220from(X) => from(X) a!6220!6220add(X, Y) => add(X, Y) a!6220!6220len(X) => len(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] a!6220!6220fst#(s(X), cons(Y, Z)) =#> mark#(Y) 1] a!6220!6220from#(X) =#> mark#(X) 2] a!6220!6220add#(0, X) =#> mark#(X) 3] mark#(fst(X, Y)) =#> a!6220!6220fst#(mark(X), mark(Y)) 4] mark#(fst(X, Y)) =#> mark#(X) 5] mark#(fst(X, Y)) =#> mark#(Y) 6] mark#(from(X)) =#> a!6220!6220from#(mark(X)) 7] mark#(from(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#(len(X)) =#> a!6220!6220len#(mark(X)) 12] mark#(len(X)) =#> mark#(X) 13] mark#(cons(X, Y)) =#> mark#(X) Rules R_0: a!6220!6220fst(0, X) => nil a!6220!6220fst(s(X), cons(Y, Z)) => cons(mark(Y), fst(X, Z)) a!6220!6220from(X) => cons(mark(X), from(s(X))) a!6220!6220add(0, X) => mark(X) a!6220!6220add(s(X), Y) => s(add(X, Y)) a!6220!6220len(nil) => 0 a!6220!6220len(cons(X, Y)) => s(len(Y)) mark(fst(X, Y)) => a!6220!6220fst(mark(X), mark(Y)) mark(from(X)) => a!6220!6220from(mark(X)) mark(add(X, Y)) => a!6220!6220add(mark(X), mark(Y)) mark(len(X)) => a!6220!6220len(mark(X)) mark(0) => 0 mark(s(X)) => s(X) mark(nil) => nil mark(cons(X, Y)) => cons(mark(X), Y) a!6220!6220fst(X, Y) => fst(X, Y) a!6220!6220from(X) => from(X) a!6220!6220add(X, Y) => add(X, Y) a!6220!6220len(X) => len(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 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 * 1 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 * 2 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 * 3 : 0 * 4 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 * 5 : 3, 4, 5, 6, 7, 8, 9, 10, 11, 12, 13 popout

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