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Higher Order Rewriting Union Beta pair #487093819
details
property
value
status
complete
benchmark
filter.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n146.star.cs.uiowa.edu
space
Mixed_HO_10
run statistics
property
value
solver
Wanda 2.2a
configuration
default
runtime (wallclock)
0.0535409 seconds
cpu usage
0.05362
user time
0.041572
system time
0.012048
max virtual memory
113188.0
max residence set size
2516.0
stage attributes
key
value
starexec-result
YES
output
YES We consider the system theBenchmark. Alphabet: 0 : [] --> nat bool : [nat] --> boolean cons : [nat * list] --> list consif : [boolean * nat * list] --> list false : [] --> boolean filter : [nat -> boolean * list] --> list nil : [] --> list rand : [nat] --> nat s : [nat] --> nat true : [] --> boolean Rules: rand(x) => x rand(s(x)) => rand(x) bool(0) => false bool(s(0)) => true filter(f, nil) => nil filter(f, cons(x, y)) => consif(f x, x, filter(f, y)) consif(true, x, y) => cons(x, y) consif(false, x, y) => y This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): rand(X) >? X rand(s(X)) >? rand(X) bool(0) >? false bool(s(0)) >? true filter(F, nil) >? nil filter(F, cons(X, Y)) >? consif(F X, X, filter(F, Y)) consif(true, X, Y) >? cons(X, Y) consif(false, X, Y) >? Y We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 3 bool = \y0.3 + 3y0 cons = \y0y1.2 + 2y0 + 2y1 consif = \y0y1y2.1 + y0 + 2y1 + 2y2 false = 1 filter = \G0y1.2y1 + G0(y1) + 2y1G0(y1) nil = 2 rand = \y0.3 + 2y0 s = \y0.3 + 3y0 true = 3 Using this interpretation, the requirements translate to: [[rand(_x0)]] = 3 + 2x0 > x0 = [[_x0]] [[rand(s(_x0))]] = 9 + 6x0 > 3 + 2x0 = [[rand(_x0)]] [[bool(0)]] = 12 > 1 = [[false]] [[bool(s(0))]] = 39 > 3 = [[true]] [[filter(_F0, nil)]] = 4 + 5F0(2) > 2 = [[nil]] [[filter(_F0, cons(_x1, _x2))]] = 4 + 4x1 + 4x2 + 4x1F0(2 + 2x1 + 2x2) + 4x2F0(2 + 2x1 + 2x2) + 5F0(2 + 2x1 + 2x2) > 1 + 3x1 + 4x2 + F0(x1) + 2F0(x2) + 4x2F0(x2) = [[consif(_F0 _x1, _x1, filter(_F0, _x2))]] [[consif(true, _x0, _x1)]] = 4 + 2x0 + 2x1 > 2 + 2x0 + 2x1 = [[cons(_x0, _x1)]] [[consif(false, _x0, _x1)]] = 2 + 2x0 + 2x1 > x1 = [[_x1]] We can thus remove the following rules: rand(X) => X rand(s(X)) => rand(X) bool(0) => false bool(s(0)) => true filter(F, nil) => nil filter(F, cons(X, Y)) => consif(F X, X, filter(F, Y)) consif(true, X, Y) => cons(X, Y) consif(false, X, Y) => Y All rules were succesfully removed. Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold. +++ Citations +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012.
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