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Higher Order Rewriting Union Beta pair #487093989
details
property
value
status
complete
benchmark
Applicative_05__TreeLevels.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n145.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda 2.2a
configuration
default
runtime (wallclock)
0.115729 seconds
cpu usage
0.115861
user time
0.099702
system time
0.016159
max virtual memory
113188.0
max residence set size
3548.0
stage attributes
key
value
starexec-result
YES
output
YES We consider the system theBenchmark. Alphabet: append : [a * a] --> a combine : [a * a] --> a cons : [a * a] --> a levels : [] --> a -> a map : [a -> a * a] --> a nil : [] --> a node : [a * a] --> a zip : [a * a] --> a Rules: map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, y)) append(x, nil) => x append(nil, x) => x append(cons(x, y), z) => cons(x, append(y, z)) zip(nil, x) => x zip(x, nil) => x zip(cons(x, y), cons(z, u)) => cons(append(x, z), zip(y, u)) combine(x, nil) => x combine(x, cons(y, z)) => combine(zip(x, y), z) levels node(x, y) => cons(cons(x, nil), combine(nil, map(levels, 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]): map(F, nil) >? nil map(F, cons(X, Y)) >? cons(F X, map(F, Y)) append(X, nil) >? X append(nil, X) >? X append(cons(X, Y), Z) >? cons(X, append(Y, Z)) zip(nil, X) >? X zip(X, nil) >? X zip(cons(X, Y), cons(Z, U)) >? cons(append(X, Z), zip(Y, U)) combine(X, nil) >? X combine(X, cons(Y, Z)) >? combine(zip(X, Y), Z) levels node(X, Y) >? cons(cons(X, nil), combine(nil, map(levels, Y))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: append = \y0y1.y0 + y1 combine = \y0y1.y0 + y1 cons = \y0y1.1 + y0 + y1 levels = \y0.0 map = \G0y1.2y1 + 2y1G0(y1) + 2G0(0) nil = 0 node = \y0y1.3 + 3y0 + 3y1 zip = \y0y1.y0 + y1 Using this interpretation, the requirements translate to: [[map(_F0, nil)]] = 2F0(0) >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 2 + 2x1 + 2x2 + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 2F0(0) + 2F0(1 + x1 + x2) > 1 + x1 + 2x2 + F0(x1) + 2x2F0(x2) + 2F0(0) = [[cons(_F0 _x1, map(_F0, _x2))]] [[append(_x0, nil)]] = x0 >= x0 = [[_x0]] [[append(nil, _x0)]] = x0 >= x0 = [[_x0]] [[append(cons(_x0, _x1), _x2)]] = 1 + x0 + x1 + x2 >= 1 + x0 + x1 + x2 = [[cons(_x0, append(_x1, _x2))]] [[zip(nil, _x0)]] = x0 >= x0 = [[_x0]] [[zip(_x0, nil)]] = x0 >= x0 = [[_x0]] [[zip(cons(_x0, _x1), cons(_x2, _x3))]] = 2 + x0 + x1 + x2 + x3 > 1 + x0 + x1 + x2 + x3 = [[cons(append(_x0, _x2), zip(_x1, _x3))]] [[combine(_x0, nil)]] = x0 >= x0 = [[_x0]] [[combine(_x0, cons(_x1, _x2))]] = 1 + x0 + x1 + x2 > x0 + x1 + x2 = [[combine(zip(_x0, _x1), _x2)]] [[levels node(_x0, _x1)]] = 3 + 3x0 + 3x1 > 2 + x0 + 2x1 = [[cons(cons(_x0, nil), combine(nil, map(levels, _x1)))]] We can thus remove the following rules: map(F, cons(X, Y)) => cons(F X, map(F, Y)) zip(cons(X, Y), cons(Z, U)) => cons(append(X, Z), zip(Y, U)) combine(X, cons(Y, Z)) => combine(zip(X, Y), Z) levels node(X, Y) => cons(cons(X, nil), combine(nil, map(levels, Y))) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): map(F, nil) >? nil append(X, nil) >? X append(nil, X) >? X append(cons(X, Y), Z) >? cons(X, append(Y, Z)) zip(nil, X) >? X zip(X, nil) >? X combine(X, nil) >? X We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: append = \y0y1.3 + y1 + 3y0 combine = \y0y1.3 + y0 + y1 cons = \y0y1.3 + y0 + y1 map = \G0y1.3 + 3y1 + G0(0)
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