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Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270297
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
n031.star.cs.uiowa.edu
space
Mixed_HO_10
run statistics
property
value
solver
Wanda 2.1c
configuration
default
runtime (wallclock)
0.036406 seconds
cpu usage
0.036552
user time
0.025098
system time
0.011454
max virtual memory
113176.0
max residence set size
2400.0
stage attributes
key
value
starexec-result
YES
output
0.00/0.03 YES 0.00/0.03 We consider the system theBenchmark. 0.00/0.03 0.00/0.03 Alphabet: 0.00/0.03 0.00/0.03 fapp : [o * o] --> o 0.00/0.03 h : [o] --> o 0.00/0.03 lam : [o -> o] --> o 0.00/0.03 0.00/0.03 Rules: 0.00/0.03 0.00/0.03 fapp(lam(f), x) => f h(x) 0.00/0.03 0.00/0.03 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 0.00/0.03 0.00/0.03 We use the dependency pair framework as described in [Kop12, Ch. 6/7], with dynamic dependency pairs. 0.00/0.03 0.00/0.03 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): 0.00/0.03 0.00/0.03 Dependency Pairs P_0: 0.00/0.03 0.00/0.03 0] fapp#(lam(F), X) =#> F(h(X)) 0.00/0.03 0.00/0.03 Rules R_0: 0.00/0.03 0.00/0.03 fapp(lam(F), X) => F h(X) 0.00/0.03 0.00/0.03 Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite. 0.00/0.03 0.00/0.03 We consider the dependency pair problem (P_0, R_0, minimal, formative). 0.00/0.03 0.00/0.03 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: 0.00/0.03 0.00/0.03 fapp#(lam(F), X) >? F(h(X)) 0.00/0.03 fapp(lam(F), X) >= F h(X) 0.00/0.03 0.00/0.03 We orient these requirements with a polynomial interpretation in the natural numbers. 0.00/0.03 0.00/0.03 The following interpretation satisfies the requirements: 0.00/0.03 0.00/0.03 fapp = \y0y1.3 + 3y0 0.00/0.03 fapp# = \y0y1.3 + y0 0.00/0.03 h = \y0.0 0.00/0.03 lam = \G0.3 + G0(0) 0.00/0.03 0.00/0.03 Using this interpretation, the requirements translate to: 0.00/0.03 0.00/0.03 [[fapp#(lam(_F0), _x1)]] = 6 + F0(0) > F0(0) = [[_F0(h(_x1))]] 0.00/0.03 [[fapp(lam(_F0), _x1)]] = 12 + 3F0(0) >= F0(0) = [[_F0 h(_x1)]] 0.00/0.03 0.00/0.03 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. 0.00/0.03 0.00/0.03 As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. 0.00/0.03 0.00/0.03 0.00/0.03 +++ Citations +++ 0.00/0.03 0.00/0.03 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 0.00/0.03 EOF
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