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HRS union beta 16688 pair #381734509
details
property
value
status
complete
benchmark
lambda3.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n095.star.cs.uiowa.edu
space
Mixed_HO_10
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
0.0414340496063 seconds
cpu usage
0.023199327
max memory
1617920.0
stage attributes
key
value
output-size
2238
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. Alphabet: fapp : [o * o] --> o h : [o] --> o lam : [o -> o] --> o Rules: fapp(lam(f), x) => f h(x) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). We use the dependency pair framework as described in [Kop12, Ch. 6/7], with dynamic dependency pairs. After applying [Kop12, Thm. 7.22] to denote collapsing dependency pairs in an extended form, we thus obtain the following dependency pair problem (P_0, R_0, minimal, formative): Dependency Pairs P_0: 0] fapp#(lam(F), X) =#> F(h(X)) Rules R_0: fapp(lam(F), X) => F h(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 will use the reduction pair processor [Kop12, Thm. 7.16]. As the system is abstraction-simple and the formative flag is set, it suffices to find a tagged reduction pair [Kop12, Def. 6.70]. Thus, we must orient: fapp#(lam(F), X) >? F(h(X)) fapp(lam(F), X) >= F h(X) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: fapp = \y0y1.3 + 3y0 fapp# = \y0y1.3 + y0 h = \y0.0 lam = \G0.3 + G0(0) Using this interpretation, the requirements translate to: [[fapp#(lam(_F0), _x1)]] = 6 + F0(0) > F0(0) = [[_F0(h(_x1))]] [[fapp(lam(_F0), _x1)]] = 12 + 3F0(0) >= F0(0) = [[_F0 h(_x1)]] By the observations in [Kop12, Sec. 6.6], this reduction pair suffices; we may thus replace a dependency pair problem (P_0, R_0) by ({}, R_0). 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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