details

property	value
status	complete
benchmark	Ex14_AEGL02_GM.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n085.star.cs.uiowa.edu
space	Transformed_CSR_04

run statistics

property	value
solver	Wanda
configuration	FirstOrder
runtime (wallclock)	0.301056861877 seconds
cpu usage	0.120540081
max memory	4726784.0

stage attributes

key	value
output-size	9737
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 a!6220!6220from : [o] --> o a!6220!6220length : [o] --> o a!6220!6220length1 : [o] --> o cons : [o * o] --> o from : [o] --> o length : [o] --> o length1 : [o] --> o mark : [o] --> o nil : [] --> o s : [o] --> o a!6220!6220from(X) => cons(mark(X), from(s(X))) a!6220!6220length(nil) => 0 a!6220!6220length(cons(X, Y)) => s(a!6220!6220length1(Y)) a!6220!6220length1(X) => a!6220!6220length(X) mark(from(X)) => a!6220!6220from(mark(X)) mark(length(X)) => a!6220!6220length(X) mark(length1(X)) => a!6220!6220length1(X) mark(cons(X, Y)) => cons(mark(X), Y) mark(s(X)) => s(mark(X)) mark(nil) => nil mark(0) => 0 a!6220!6220from(X) => from(X) a!6220!6220length(X) => length(X) a!6220!6220length1(X) => length1(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!6220from#(X) =#> mark#(X) 1] a!6220!6220length#(cons(X, Y)) =#> a!6220!6220length1#(Y) 2] a!6220!6220length1#(X) =#> a!6220!6220length#(X) 3] mark#(from(X)) =#> a!6220!6220from#(mark(X)) 4] mark#(from(X)) =#> mark#(X) 5] mark#(length(X)) =#> a!6220!6220length#(X) 6] mark#(length1(X)) =#> a!6220!6220length1#(X) 7] mark#(cons(X, Y)) =#> mark#(X) 8] mark#(s(X)) =#> mark#(X) Rules R_0: a!6220!6220from(X) => cons(mark(X), from(s(X))) a!6220!6220length(nil) => 0 a!6220!6220length(cons(X, Y)) => s(a!6220!6220length1(Y)) a!6220!6220length1(X) => a!6220!6220length(X) mark(from(X)) => a!6220!6220from(mark(X)) mark(length(X)) => a!6220!6220length(X) mark(length1(X)) => a!6220!6220length1(X) mark(cons(X, Y)) => cons(mark(X), Y) mark(s(X)) => s(mark(X)) mark(nil) => nil mark(0) => 0 a!6220!6220from(X) => from(X) a!6220!6220length(X) => length(X) a!6220!6220length1(X) => length1(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 * 1 : 2 * 2 : 1 * 3 : 0 * 4 : 3, 4, 5, 6, 7, 8 * 5 : 1 * 6 : 2 * 7 : 3, 4, 5, 6, 7, 8 * 8 : 3, 4, 5, 6, 7, 8 This graph has the following strongly connected components: P_1: a!6220!6220from#(X) =#> mark#(X) mark#(from(X)) =#> a!6220!6220from#(mark(X)) mark#(from(X)) =#> mark#(X) mark#(cons(X, Y)) =#> mark#(X) mark#(s(X)) =#> mark#(X) P_2: a!6220!6220length#(cons(X, Y)) =#> a!6220!6220length1#(Y) popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to TRS Stand 20472