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Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270603
details
property
value
status
complete
benchmark
Applicative_05__TreeFlatten.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n118.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda 2.1c
configuration
default
runtime (wallclock)
0.168582 seconds
cpu usage
0.14737
user time
0.126812
system time
0.020558
max virtual memory
113176.0
max residence set size
3460.0
stage attributes
key
value
starexec-result
YES
output
0.00/0.16 YES 0.00/0.16 We consider the system theBenchmark. 0.00/0.16 0.00/0.16 Alphabet: 0.00/0.16 0.00/0.16 append : [a * a] --> a 0.00/0.16 concat : [a] --> a 0.00/0.16 cons : [a * a] --> a 0.00/0.16 flatten : [] --> a -> a 0.00/0.16 map : [a -> a * a] --> a 0.00/0.16 nil : [] --> a 0.00/0.16 node : [a * a] --> a 0.00/0.16 0.00/0.16 Rules: 0.00/0.16 0.00/0.16 map(f, nil) => nil 0.00/0.16 map(f, cons(x, y)) => cons(f x, map(f, y)) 0.00/0.16 flatten node(x, y) => cons(x, concat(map(flatten, y))) 0.00/0.16 concat(nil) => nil 0.00/0.16 concat(cons(x, y)) => append(x, concat(y)) 0.00/0.16 append(nil, x) => x 0.00/0.16 append(cons(x, y), z) => cons(x, append(y, z)) 0.00/0.16 0.00/0.16 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 0.00/0.16 0.00/0.16 We use rule removal, following [Kop12, Theorem 2.23]. 0.00/0.16 0.00/0.16 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.00/0.16 0.00/0.16 map(F, nil) >? nil 0.00/0.16 map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 0.00/0.16 flatten node(X, Y) >? cons(X, concat(map(flatten, Y))) 0.00/0.16 concat(nil) >? nil 0.00/0.16 concat(cons(X, Y)) >? append(X, concat(Y)) 0.00/0.16 append(nil, X) >? X 0.00/0.16 append(cons(X, Y), Z) >? cons(X, append(Y, Z)) 0.00/0.16 0.00/0.16 We orient these requirements with a polynomial interpretation in the natural numbers. 0.00/0.16 0.00/0.16 The following interpretation satisfies the requirements: 0.00/0.16 0.00/0.16 append = \y0y1.y0 + y1 0.00/0.16 concat = \y0.y0 0.00/0.16 cons = \y0y1.1 + y0 + y1 0.00/0.16 flatten = \y0.0 0.00/0.16 map = \G0y1.2 + y1 + G0(y1) + 2y1G0(y1) 0.00/0.16 nil = 0 0.00/0.16 node = \y0y1.3 + 3y0 + 3y1 0.00/0.16 0.00/0.16 Using this interpretation, the requirements translate to: 0.00/0.16 0.00/0.16 [[map(_F0, nil)]] = 2 + F0(0) > 0 = [[nil]] 0.00/0.16 [[map(_F0, cons(_x1, _x2))]] = 3 + x1 + x2 + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 3F0(1 + x1 + x2) >= 3 + x1 + x2 + F0(x1) + F0(x2) + 2x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 0.00/0.16 [[flatten node(_x0, _x1)]] = 3 + 3x0 + 3x1 >= 3 + x0 + x1 = [[cons(_x0, concat(map(flatten, _x1)))]] 0.00/0.16 [[concat(nil)]] = 0 >= 0 = [[nil]] 0.00/0.16 [[concat(cons(_x0, _x1))]] = 1 + x0 + x1 > x0 + x1 = [[append(_x0, concat(_x1))]] 0.00/0.16 [[append(nil, _x0)]] = x0 >= x0 = [[_x0]] 0.00/0.16 [[append(cons(_x0, _x1), _x2)]] = 1 + x0 + x1 + x2 >= 1 + x0 + x1 + x2 = [[cons(_x0, append(_x1, _x2))]] 0.00/0.16 0.00/0.16 We can thus remove the following rules: 0.00/0.16 0.00/0.16 map(F, nil) => nil 0.00/0.16 concat(cons(X, Y)) => append(X, concat(Y)) 0.00/0.16 0.00/0.16 We use rule removal, following [Kop12, Theorem 2.23]. 0.00/0.16 0.00/0.16 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.00/0.16 0.00/0.16 map(F, cons(X, Y)) >? cons(F X, map(F, Y)) 0.00/0.16 flatten node(X, Y) >? cons(X, concat(map(flatten, Y))) 0.00/0.16 concat(nil) >? nil 0.00/0.16 append(nil, X) >? X 0.00/0.16 append(cons(X, Y), Z) >? cons(X, append(Y, Z)) 0.00/0.16 0.00/0.16 We orient these requirements with a polynomial interpretation in the natural numbers. 0.00/0.16 0.00/0.16 The following interpretation satisfies the requirements: 0.00/0.16 0.00/0.16 append = \y0y1.3 + y1 + 3y0 0.00/0.16 concat = \y0.y0 0.00/0.16 cons = \y0y1.3 + y0 + y1 0.00/0.16 flatten = \y0.0 0.00/0.16 map = \G0y1.2y1 + G0(0) + 3y1G0(y1) 0.00/0.16 nil = 0 0.00/0.16 node = \y0y1.3 + 3y0 + 3y1 0.00/0.16 0.00/0.16 Using this interpretation, the requirements translate to: 0.00/0.16 0.00/0.16 [[map(_F0, cons(_x1, _x2))]] = 6 + 2x1 + 2x2 + F0(0) + 3x1F0(3 + x1 + x2) + 3x2F0(3 + x1 + x2) + 9F0(3 + x1 + x2) > 3 + x1 + 2x2 + F0(0) + F0(x1) + 3x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] 0.00/0.16 [[flatten node(_x0, _x1)]] = 3 + 3x0 + 3x1 >= 3 + x0 + 2x1 = [[cons(_x0, concat(map(flatten, _x1)))]] 0.00/0.16 [[concat(nil)]] = 0 >= 0 = [[nil]] 0.00/0.16 [[append(nil, _x0)]] = 3 + x0 > x0 = [[_x0]] 0.00/0.16 [[append(cons(_x0, _x1), _x2)]] = 12 + x2 + 3x0 + 3x1 > 6 + x0 + x2 + 3x1 = [[cons(_x0, append(_x1, _x2))]] 0.00/0.16 0.00/0.16 We can thus remove the following rules: 0.00/0.16 0.00/0.16 map(F, cons(X, Y)) => cons(F X, map(F, Y)) 0.00/0.16 append(nil, X) => X 0.00/0.16 append(cons(X, Y), Z) => cons(X, append(Y, Z)) 0.00/0.16
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