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Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270309
details
property
value
status
complete
benchmark
ordrec.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n167.star.cs.uiowa.edu
space
Mixed_HO_10
run statistics
property
value
solver
Wanda 2.1c
configuration
default
runtime (wallclock)
0.542765 seconds
cpu usage
0.53304
user time
0.476071
system time
0.056969
max virtual memory
113176.0
max residence set size
21688.0
stage attributes
key
value
starexec-result
YES
output
0.00/0.53 YES 0.00/0.54 We consider the system theBenchmark. 0.00/0.54 0.00/0.54 Alphabet: 0.00/0.54 0.00/0.54 0 : [] --> ord 0.00/0.54 lim : [nat -> ord] --> ord 0.00/0.54 rec : [ord * a * ord -> a -> a * (nat -> ord) -> (nat -> a) -> a] --> a 0.00/0.54 s : [ord] --> ord 0.00/0.54 0.00/0.54 Rules: 0.00/0.54 0.00/0.54 rec(0, x, f, g) => x 0.00/0.54 rec(s(x), y, f, g) => f x rec(x, y, f, g) 0.00/0.54 rec(lim(f), x, g, h) => h f (/\y.rec(f y, x, g, h)) 0.00/0.54 0.00/0.54 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 0.00/0.54 0.00/0.54 We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [Kop13]). 0.00/0.54 0.00/0.54 We thus obtain the following dependency pair problem (P_0, R_0, static, formative): 0.00/0.54 0.00/0.54 Dependency Pairs P_0: 0.00/0.54 0.00/0.54 0] rec#(s(X), Y, F, G) =#> rec#(X, Y, F, G) 0.00/0.54 1] rec#(lim(F), X, G, H) =#> rec#(F Y, X, G, H) 0.00/0.54 0.00/0.54 Rules R_0: 0.00/0.54 0.00/0.54 rec(0, X, F, G) => X 0.00/0.54 rec(s(X), Y, F, G) => F X rec(X, Y, F, G) 0.00/0.54 rec(lim(F), X, G, H) => H F (/\x.rec(F x, X, G, H)) 0.00/0.54 0.00/0.54 Thus, the original system is terminating if (P_0, R_0, static, formative) is finite. 0.00/0.54 0.00/0.54 We consider the dependency pair problem (P_0, R_0, static, formative). 0.00/0.54 0.00/0.54 We apply the accessible subterm criterion with the following projection function: 0.00/0.54 0.00/0.54 nu(rec#) = 1 0.00/0.54 0.00/0.54 Thus, we can orient the dependency pairs as follows: 0.00/0.54 0.00/0.54 nu(rec#(s(X), Y, F, G)) = s(X) [>] X = nu(rec#(X, Y, F, G)) 0.00/0.54 nu(rec#(lim(F), X, G, H)) = lim(F) [>] F Y = nu(rec#(F Y, X, G, H)) 0.00/0.54 0.00/0.54 By [Kop13, Thm. 6], we may replace a dependency pair problem (P_0, R_0, static, f) by ({}, R_0, static, f). By the empty set processor [Kop12, Thm. 7.15] this problem may be immediately removed. 0.00/0.54 0.00/0.54 As all dependency pair problems were succesfully simplified with sound (and complete) processors until nothing remained, we conclude termination. 0.00/0.54 0.00/0.54 0.00/0.54 +++ Citations +++ 0.00/0.54 0.00/0.54 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 0.00/0.54 [Kop13] C. Kop. Static Dependency Pairs with Accessibility. Unpublished manuscript, http://cl-informatik.uibk.ac.at/users/kop/static.pdf, 2013. 0.00/0.54 [KusIsoSakBla09] K. Kusakari, Y. Isogai, M. Sakai, and F. Blanqui. Static Dependency Pair Method Based On Strong Computability for Higher-Order Rewrite Systems. In volume 92(10) of IEICE Transactions on Information and Systems. 2007--2015, 2009. 0.00/0.54 EOF
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