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Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270273
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
n181.star.cs.uiowa.edu
space
Mixed_HO_10
run statistics
property
value
solver
Wanda 2.1c
configuration
default
runtime (wallclock)
0.0543251 seconds
cpu usage
0.054365
user time
0.04486
system time
0.009505
max virtual memory
113176.0
max residence set size
2548.0
stage attributes
key
value
starexec-result
YES
output
0.00/0.05 YES 0.00/0.05 We consider the system theBenchmark. 0.00/0.05 0.00/0.05 Alphabet: 0.00/0.05 0.00/0.05 0 : [] --> nat 0.00/0.05 bool : [nat] --> boolean 0.00/0.05 cons : [nat * list] --> list 0.00/0.05 consif : [boolean * nat * list] --> list 0.00/0.05 false : [] --> boolean 0.00/0.05 filter : [nat -> boolean * list] --> list 0.00/0.05 nil : [] --> list 0.00/0.05 rand : [nat] --> nat 0.00/0.05 s : [nat] --> nat 0.00/0.05 true : [] --> boolean 0.00/0.05 0.00/0.05 Rules: 0.00/0.05 0.00/0.05 rand(x) => x 0.00/0.05 rand(s(x)) => rand(x) 0.00/0.05 bool(0) => false 0.00/0.05 bool(s(0)) => true 0.00/0.05 filter(f, nil) => nil 0.00/0.05 filter(f, cons(x, y)) => consif(f x, x, filter(f, y)) 0.00/0.05 consif(true, x, y) => cons(x, y) 0.00/0.05 consif(false, x, y) => y 0.00/0.05 0.00/0.05 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 0.00/0.05 0.00/0.05 We use rule removal, following [Kop12, Theorem 2.23]. 0.00/0.05 0.00/0.05 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.00/0.05 0.00/0.05 rand(X) >? X 0.00/0.05 rand(s(X)) >? rand(X) 0.00/0.05 bool(0) >? false 0.00/0.05 bool(s(0)) >? true 0.00/0.05 filter(F, nil) >? nil 0.00/0.05 filter(F, cons(X, Y)) >? consif(F X, X, filter(F, Y)) 0.00/0.05 consif(true, X, Y) >? cons(X, Y) 0.00/0.05 consif(false, X, Y) >? Y 0.00/0.05 0.00/0.05 We orient these requirements with a polynomial interpretation in the natural numbers. 0.00/0.05 0.00/0.05 The following interpretation satisfies the requirements: 0.00/0.05 0.00/0.05 0 = 3 0.00/0.05 bool = \y0.3 + 3y0 0.00/0.05 cons = \y0y1.2 + 2y0 + 2y1 0.00/0.05 consif = \y0y1y2.1 + y0 + 2y1 + 2y2 0.00/0.05 false = 1 0.00/0.05 filter = \G0y1.2y1 + G0(y1) + 2y1G0(y1) 0.00/0.05 nil = 2 0.00/0.05 rand = \y0.3 + 2y0 0.00/0.05 s = \y0.3 + 3y0 0.00/0.05 true = 3 0.00/0.05 0.00/0.05 Using this interpretation, the requirements translate to: 0.00/0.05 0.00/0.05 [[rand(_x0)]] = 3 + 2x0 > x0 = [[_x0]] 0.00/0.05 [[rand(s(_x0))]] = 9 + 6x0 > 3 + 2x0 = [[rand(_x0)]] 0.00/0.05 [[bool(0)]] = 12 > 1 = [[false]] 0.00/0.05 [[bool(s(0))]] = 39 > 3 = [[true]] 0.00/0.05 [[filter(_F0, nil)]] = 4 + 5F0(2) > 2 = [[nil]] 0.00/0.05 [[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))]] 0.00/0.05 [[consif(true, _x0, _x1)]] = 4 + 2x0 + 2x1 > 2 + 2x0 + 2x1 = [[cons(_x0, _x1)]] 0.00/0.05 [[consif(false, _x0, _x1)]] = 2 + 2x0 + 2x1 > x1 = [[_x1]] 0.00/0.05 0.00/0.05 We can thus remove the following rules: 0.00/0.05 0.00/0.05 rand(X) => X 0.00/0.05 rand(s(X)) => rand(X) 0.00/0.05 bool(0) => false 0.00/0.05 bool(s(0)) => true 0.00/0.05 filter(F, nil) => nil 0.00/0.05 filter(F, cons(X, Y)) => consif(F X, X, filter(F, Y)) 0.00/0.05 consif(true, X, Y) => cons(X, Y) 0.00/0.05 consif(false, X, Y) => Y 0.00/0.05 0.00/0.05 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. 0.00/0.05 0.00/0.05 0.00/0.05 +++ Citations +++ 0.00/0.05 0.00/0.05 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 0.00/0.05 EOF
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