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Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270635
details
property
value
status
complete
benchmark
Applicative_AG01_innermost__#4.28.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n047.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda 2.1c
configuration
default
runtime (wallclock)
0.90587 seconds
cpu usage
1.73979
user time
1.62489
system time
0.114892
max virtual memory
6751936.0
max residence set size
108404.0
stage attributes
key
value
starexec-result
YES
output
1.69/0.89 YES 1.69/0.90 We consider the system theBenchmark. 1.69/0.90 1.69/0.90 Alphabet: 1.69/0.90 1.69/0.90 0 : [] --> a 1.69/0.90 bits : [a] --> a 1.69/0.90 cons : [c * d] --> d 1.69/0.90 false : [] --> b 1.69/0.90 filter : [c -> b * d] --> d 1.69/0.90 filter2 : [b * c -> b * c * d] --> d 1.69/0.90 half : [a] --> a 1.69/0.90 map : [c -> c * d] --> d 1.69/0.90 nil : [] --> d 1.69/0.90 s : [a] --> a 1.69/0.90 true : [] --> b 1.69/0.90 1.69/0.90 Rules: 1.69/0.90 1.69/0.90 half(0) => 0 1.69/0.90 half(s(0)) => 0 1.69/0.90 half(s(s(x))) => s(half(x)) 1.69/0.90 bits(0) => 0 1.69/0.90 bits(s(x)) => s(bits(half(s(x)))) 1.69/0.90 map(f, nil) => nil 1.69/0.90 map(f, cons(x, y)) => cons(f x, map(f, y)) 1.69/0.90 filter(f, nil) => nil 1.69/0.90 filter(f, cons(x, y)) => filter2(f x, f, x, y) 1.69/0.90 filter2(true, f, x, y) => cons(x, filter(f, y)) 1.69/0.90 filter2(false, f, x, y) => filter(f, y) 1.69/0.90 1.69/0.90 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 1.69/0.90 1.69/0.90 We use rule removal, following [Kop12, Theorem 2.23]. 1.69/0.90 1.69/0.90 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 1.69/0.90 1.69/0.90 half(0) >? 0 1.69/0.90 half(s(0)) >? 0 1.69/0.90 half(s(s(X))) >? s(half(X)) 1.69/0.90 bits(0) >? 0 1.69/0.90 bits(s(X)) >? s(bits(half(s(X)))) 1.69/0.90 map(F, nil) >? nil 1.69/0.90 map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 1.69/0.90 filter(F, nil) >? nil 1.69/0.90 filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) 1.69/0.90 filter2(true, F, X, Y) >? cons(X, filter(F, Y)) 1.69/0.90 filter2(false, F, X, Y) >? filter(F, Y) 1.69/0.90 1.69/0.90 We orient these requirements with a polynomial interpretation in the natural numbers. 1.69/0.90 1.69/0.90 The following interpretation satisfies the requirements: 1.69/0.90 1.69/0.90 0 = 0 1.69/0.90 bits = \y0.y0 1.69/0.90 cons = \y0y1.3 + y1 + 2y0 1.69/0.90 false = 3 1.69/0.90 filter = \G0y1.2y1 + G0(0) + 3y1G0(y1) 1.69/0.90 filter2 = \y0G1y2y3.2y0 + 2y2 + 2y3 + 2G1(0) + 3y3G1(y3) 1.69/0.90 half = \y0.y0 1.69/0.90 map = \G0y1.2 + 3y1 + G0(y1) + 2y1G0(y1) 1.69/0.90 nil = 1 1.69/0.90 s = \y0.y0 1.69/0.90 true = 3 1.69/0.90 1.69/0.90 Using this interpretation, the requirements translate to: 1.69/0.90 1.69/0.90 [[half(0)]] = 0 >= 0 = [[0]] 1.69/0.90 [[half(s(0))]] = 0 >= 0 = [[0]] 1.69/0.90 [[half(s(s(_x0)))]] = x0 >= x0 = [[s(half(_x0))]] 1.69/0.90 [[bits(0)]] = 0 >= 0 = [[0]] 1.69/0.90 [[bits(s(_x0))]] = x0 >= x0 = [[s(bits(half(s(_x0))))]] 1.69/0.90 [[map(_F0, nil)]] = 5 + 3F0(1) > 1 = [[nil]] 1.69/0.90 [[map(_F0, cons(_x1, _x2))]] = 11 + 3x2 + 6x1 + 2x2F0(3 + x2 + 2x1) + 4x1F0(3 + x2 + 2x1) + 7F0(3 + x2 + 2x1) > 5 + 2x1 + 3x2 + F0(x2) + 2x2F0(x2) + 2F0(x1) = [[cons(_F0 _x1, map(_F0, _x2))]] 1.69/0.90 [[filter(_F0, nil)]] = 2 + F0(0) + 3F0(1) > 1 = [[nil]] 1.69/0.90 [[filter(_F0, cons(_x1, _x2))]] = 6 + 2x2 + 4x1 + F0(0) + 3x2F0(3 + x2 + 2x1) + 6x1F0(3 + x2 + 2x1) + 9F0(3 + x2 + 2x1) > 2x2 + 4x1 + 2F0(0) + 2F0(x1) + 3x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] 1.69/0.90 [[filter2(true, _F0, _x1, _x2)]] = 6 + 2x1 + 2x2 + 2F0(0) + 3x2F0(x2) > 3 + 2x1 + 2x2 + F0(0) + 3x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] 1.69/0.90 [[filter2(false, _F0, _x1, _x2)]] = 6 + 2x1 + 2x2 + 2F0(0) + 3x2F0(x2) > 2x2 + F0(0) + 3x2F0(x2) = [[filter(_F0, _x2)]] 1.69/0.90 1.69/0.90 We can thus remove the following rules: 1.69/0.90 1.69/0.90 map(F, nil) => nil 1.69/0.90 map(F, cons(X, Y)) => cons(F X, map(F, Y)) 1.69/0.90 filter(F, nil) => nil 1.69/0.90 filter(F, cons(X, Y)) => filter2(F X, F, X, Y) 1.69/0.90 filter2(true, F, X, Y) => cons(X, filter(F, Y)) 1.69/0.90 filter2(false, F, X, Y) => filter(F, Y) 1.69/0.90 1.69/0.90 We use rule removal, following [Kop12, Theorem 2.23]. 1.69/0.90 1.69/0.90 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 1.69/0.90 1.69/0.90 half(0) >? 0 1.69/0.90 half(s(0)) >? 0 1.69/0.90 half(s(s(X))) >? s(half(X)) 1.69/0.90 bits(0) >? 0 1.69/0.90 bits(s(X)) >? s(bits(half(s(X)))) 1.69/0.90 1.69/0.90 We orient these requirements with a polynomial interpretation in the natural numbers. 1.69/0.90
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