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TRS Stand 20472 pair #381714567
details
property
value
status
complete
benchmark
Ex4_7_15_Bor03_Z.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
Transformed_CSR_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.049329996109 seconds
cpu usage
0.045658722
max memory
1900544.0
stage attributes
key
value
output-size
3610
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 activate : [o] --> o cons : [o * o] --> o f : [o] --> o n!6220!6220f : [o] --> o p : [o] --> o s : [o] --> o f(0) => cons(0, n!6220!6220f(s(0))) f(s(0)) => f(p(s(0))) p(s(0)) => 0 f(X) => n!6220!6220f(X) activate(n!6220!6220f(X)) => f(X) activate(X) => 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(0) >? cons(0, n!6220!6220f(s(0))) f(s(0)) >? f(p(s(0))) p(s(0)) >? 0 f(X) >? n!6220!6220f(X) activate(n!6220!6220f(X)) >? f(X) activate(X) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 activate = \y0.3 + 3y0 cons = \y0y1.y0 + y1 f = \y0.1 + y0 n!6220!6220f = \y0.y0 p = \y0.2y0 s = \y0.y0 Using this interpretation, the requirements translate to: [[f(0)]] = 1 > 0 = [[cons(0, n!6220!6220f(s(0)))]] [[f(s(0))]] = 1 >= 1 = [[f(p(s(0)))]] [[p(s(0))]] = 0 >= 0 = [[0]] [[f(_x0)]] = 1 + x0 > x0 = [[n!6220!6220f(_x0)]] [[activate(n!6220!6220f(_x0))]] = 3 + 3x0 > 1 + x0 = [[f(_x0)]] [[activate(_x0)]] = 3 + 3x0 > x0 = [[_x0]] We can thus remove the following rules: f(0) => cons(0, n!6220!6220f(s(0))) f(X) => n!6220!6220f(X) activate(n!6220!6220f(X)) => f(X) activate(X) => 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(s(0)) >? f(p(s(0))) p(s(0)) >? 0 We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 1 f = \y0.y0 p = \y0.y0 s = \y0.2y0 Using this interpretation, the requirements translate to: [[f(s(0))]] = 2 >= 2 = [[f(p(s(0)))]] [[p(s(0))]] = 2 > 1 = [[0]] We can thus remove the following rules: p(s(0)) => 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] f#(s(0)) =#> f#(p(s(0))) Rules R_0:
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