Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
TRS Stand 20472 pair #381709907
details
property
value
status
complete
benchmark
motivation.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n104.star.cs.uiowa.edu
space
Mixed_TRS
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0892288684845 seconds
cpu usage
0.057272876
max memory
2162688.0
stage attributes
key
value
output-size
4526
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. f : [o * o] --> o g : [o] --> o h : [o] --> o g(h(g(X))) => g(X) g(g(X)) => g(h(g(X))) h(h(X)) => h(f(h(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] g#(h(g(X))) =#> g#(X) 1] g#(g(X)) =#> g#(h(g(X))) 2] g#(g(X)) =#> h#(g(X)) 3] g#(g(X)) =#> g#(X) 4] h#(h(X)) =#> h#(f(h(X), X)) 5] h#(h(X)) =#> h#(X) Rules R_0: g(h(g(X))) => g(X) g(g(X)) => g(h(g(X))) h(h(X)) => h(f(h(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). 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, 2, 3 * 1 : 0 * 2 : * 3 : 0, 1, 2, 3 * 4 : * 5 : 4, 5 This graph has the following strongly connected components: P_1: g#(h(g(X))) =#> g#(X) g#(g(X)) =#> g#(h(g(X))) g#(g(X)) =#> g#(X) P_2: h#(h(X)) =#> h#(X) 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) and (P_2, R_0, m, f). Thus, the original system is terminating if each of (P_1, R_0, minimal, formative) and (P_2, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_2, R_0, minimal, formative). We apply the subterm criterion with the following projection function: nu(h#) = 1 Thus, we can orient the dependency pairs as follows: nu(h#(h(X))) = h(X) |> X = nu(h#(X)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_2, R_0, minimal, f) by ({}, R_0, minimal, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite. We consider the dependency pair problem (P_1, R_0, minimal, formative). We will use the reduction pair processor [Kop12, Thm. 7.16]. It suffices to find a standard reduction pair [Kop12, Def. 6.69]. Thus, we must orient: g#(h(g(X))) >? g#(X) g#(g(X)) >? g#(h(g(X))) g#(g(X)) >? g#(X) g(h(g(X))) >= g(X) g(g(X)) >= g(h(g(X))) h(h(X)) >= h(f(h(X), X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: f = \y0y1.0 g = \y0.1 + y0 g# = \y0.2y0 h = \y0.y0
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to TRS Stand 20472