Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
TRS Stand 20472 pair #381714095
details
property
value
status
complete
benchmark
#3.54.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n092.star.cs.uiowa.edu
space
AG01
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.156655073166 seconds
cpu usage
0.064334257
max memory
2002944.0
stage attributes
key
value
output-size
3637
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. f : [o] --> o f!450 : [o * o * o] --> o g : [o] --> o h : [o] --> o s : [o] --> o f(g(X)) => g(f(f(X))) f(h(X)) => h(g(X)) f!450(s(X), Y, Y) => f!450(Y, X, s(X)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): f(g(X)) >? g(f(f(X))) f(h(X)) >? h(g(X)) f!450(s(X), Y, Y) >? f!450(Y, X, s(X)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: f = \y0.y0 f!450 = \y0y1y2.y2 + 2y1 + 3y0 g = \y0.y0 h = \y0.y0 s = \y0.2 + y0 Using this interpretation, the requirements translate to: [[f(g(_x0))]] = x0 >= x0 = [[g(f(f(_x0)))]] [[f(h(_x0))]] = x0 >= x0 = [[h(g(_x0))]] [[f!450(s(_x0), _x1, _x1)]] = 6 + 3x0 + 3x1 > 2 + 3x0 + 3x1 = [[f!450(_x1, _x0, s(_x0))]] We can thus remove the following rules: f!450(s(X), Y, Y) => f!450(Y, X, s(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] f#(g(X)) =#> f#(f(X)) 1] f#(g(X)) =#> f#(X) Rules R_0: f(g(X)) => g(f(f(X))) f(h(X)) => h(g(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). The formative rules of (P_0, R_0) are R_1 ::= f(g(X)) => g(f(f(X))) By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_0, R_0, minimal, formative) by (P_0, R_1, minimal, formative). Thus, the original system is terminating if (P_0, R_1, minimal, formative) is finite. We consider the dependency pair problem (P_0, R_1, 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: f#(g(X)) >? f#(f(X)) f#(g(X)) >? f#(X) f(g(X)) >= g(f(f(X))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: f = \y0.y0 f# = \y0.3y0 g = \y0.3 + 3y0 Using this interpretation, the requirements translate to: [[f#(g(_x0))]] = 9 + 9x0 > 3x0 = [[f#(f(_x0))]] [[f#(g(_x0))]] = 9 + 9x0 > 3x0 = [[f#(_x0)]] [[f(g(_x0))]] = 3 + 3x0 >= 3 + 3x0 = [[g(f(f(_x0)))]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_0, R_1) by ({}, R_1). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination.
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to TRS Stand 20472