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TRS Stand 20472 pair #381711246
details
property
value
status
complete
benchmark
p266.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n027.star.cs.uiowa.edu
space
Rubio_04
run statistics
property
value
solver
Wanda
configuration
FirstOrder
runtime (wallclock)
0.0527470111847 seconds
cpu usage
0.045733615
max memory
1806336.0
stage attributes
key
value
output-size
3164
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. a : [o] --> o b : [o] --> o f : [o] --> o g : [o] --> o f(f(X)) => f(a(b(f(X)))) f(a(g(X))) => b(X) b(X) => a(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(f(X)) >? f(a(b(f(X)))) f(a(g(X))) >? b(X) b(X) >? a(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: a = \y0.y0 b = \y0.y0 f = \y0.y0 g = \y0.3 + 3y0 Using this interpretation, the requirements translate to: [[f(f(_x0))]] = x0 >= x0 = [[f(a(b(f(_x0))))]] [[f(a(g(_x0)))]] = 3 + 3x0 > x0 = [[b(_x0)]] [[b(_x0)]] = x0 >= x0 = [[a(_x0)]] We can thus remove the following rules: f(a(g(X))) => b(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#(f(X)) =#> f#(a(b(f(X)))) 1] f#(f(X)) =#> b#(f(X)) 2] f#(f(X)) =#> f#(X) Rules R_0: f(f(X)) => f(a(b(f(X)))) b(X) => a(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 : * 1 : * 2 : 0, 1, 2 This graph has the following strongly connected components: P_1: f#(f(X)) =#> f#(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). 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 apply the subterm criterion with the following projection function: nu(f#) = 1 Thus, we can orient the dependency pairs as follows: nu(f#(f(X))) = f(X) |> X = nu(f#(X)) By [Kop12, Thm. 7.35], we may replace a dependency pair problem (P_1, R_0, minimal, f) by ({}, R_0, minimal, f). 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. +++ Citations +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012.
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