Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
TRS Stand 20472 pair #381711874
details
property
value
status
complete
benchmark
logarquot.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n020.star.cs.uiowa.edu
space
Rubio_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.15411901474 seconds
cpu usage
0.138704221
max memory
5414912.0
stage attributes
key
value
output-size
5950
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 log : [o] --> o min : [o * o] --> o quot : [o * o] --> o s : [o] --> o min(X, 0) => X min(s(X), s(Y)) => min(X, Y) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(min(X, Y), s(Y))) log(s(0)) => 0 log(s(s(X))) => s(log(s(quot(X, s(s(0)))))) 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] min#(s(X), s(Y)) =#> min#(X, Y) 1] quot#(s(X), s(Y)) =#> quot#(min(X, Y), s(Y)) 2] quot#(s(X), s(Y)) =#> min#(X, Y) 3] log#(s(s(X))) =#> log#(s(quot(X, s(s(0))))) 4] log#(s(s(X))) =#> quot#(X, s(s(0))) Rules R_0: min(X, 0) => X min(s(X), s(Y)) => min(X, Y) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(min(X, Y), s(Y))) log(s(0)) => 0 log(s(s(X))) => s(log(s(quot(X, s(s(0)))))) 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 : 0 * 1 : 1, 2 * 2 : 0 * 3 : 3, 4 * 4 : 1, 2 This graph has the following strongly connected components: P_1: min#(s(X), s(Y)) =#> min#(X, Y) P_2: quot#(s(X), s(Y)) =#> quot#(min(X, Y), s(Y)) P_3: log#(s(s(X))) =#> log#(s(quot(X, s(s(0))))) By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f), (P_2, R_0, m, f) and (P_3, R_0, m, f). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative), (P_2, R_0, minimal, formative) and (P_3, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_3, R_0, minimal, formative). We will use the reduction pair processor with usable rules [Kop12, Thm. 7.44]. The usable rules of (P_3, R_0) are: min(X, 0) => X min(s(X), s(Y)) => min(X, Y) quot(0, s(X)) => 0 quot(s(X), s(Y)) => s(quot(min(X, Y), s(Y))) It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: log#(s(s(X))) >? log#(s(quot(X, s(s(0))))) min(X, 0) >= X min(s(X), s(Y)) >= min(X, Y) quot(0, s(X)) >= 0 quot(s(X), s(Y)) >= s(quot(min(X, Y), s(Y))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 log# = \y0.2y0 min = \y0y1.y0 quot = \y0y1.y0
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to TRS Stand 20472