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Higher Order Rewriting Union Beta pair #487093929
details
property
value
status
complete
benchmark
Applicative_first_order_05__#3.32.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n150.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda 2.2a
configuration
default
runtime (wallclock)
0.386598 seconds
cpu usage
0.360675
user time
0.304957
system time
0.055718
max virtual memory
113188.0
max residence set size
5856.0
stage attributes
key
value
starexec-result
YES
output
YES We consider the system theBenchmark. Alphabet: a : [] --> a b : [] --> a cons : [e * f] --> f f : [a] --> b false : [] --> d filter : [e -> d * f] --> f filter2 : [d * e -> d * e * f] --> f g : [a] --> c map : [e -> e * f] --> f nil : [] --> f true : [] --> d Rules: f(a) => f(b) g(b) => g(a) map(h, nil) => nil map(h, cons(x, y)) => cons(h x, map(h, y)) filter(h, nil) => nil filter(h, cons(x, y)) => filter2(h x, h, x, y) filter2(true, h, x, y) => cons(x, filter(h, y)) filter2(false, h, x, y) => filter(h, 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]): f(a) >? f(b) g(b) >? g(a) map(F, nil) >? nil map(F, cons(X, Y)) >? cons(F X, map(F, Y)) filter(F, nil) >? nil filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) filter2(true, F, X, Y) >? cons(X, filter(F, Y)) filter2(false, F, X, Y) >? filter(F, Y) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: a = 0 b = 0 cons = \y0y1.2 + y1 + 2y0 f = \y0.y0 false = 3 filter = \G0y1.2 + 2y1 + G0(0) + y1G0(y1) filter2 = \y0G1y2y3.1 + y0 + 2y3 + 3y2 + G1(y2) + y3G1(y3) g = \y0.2y0 map = \G0y1.1 + 3y1 + G0(0) + 2y1G0(y1) nil = 0 true = 3 Using this interpretation, the requirements translate to: [[f(a)]] = 0 >= 0 = [[f(b)]] [[g(b)]] = 0 >= 0 = [[g(a)]] [[map(_F0, nil)]] = 1 + F0(0) > 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 7 + 3x2 + 6x1 + F0(0) + 2x2F0(2 + x2 + 2x1) + 4x1F0(2 + x2 + 2x1) + 4F0(2 + x2 + 2x1) > 3 + 2x1 + 3x2 + F0(0) + 2x2F0(x2) + 2F0(x1) = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 2 + F0(0) > 0 = [[nil]] [[filter(_F0, cons(_x1, _x2))]] = 6 + 2x2 + 4x1 + F0(0) + 2x1F0(2 + x2 + 2x1) + 2F0(2 + x2 + 2x1) + x2F0(2 + x2 + 2x1) > 1 + 2x2 + 4x1 + 2F0(x1) + x2F0(x2) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 4 + 2x2 + 3x1 + F0(x1) + x2F0(x2) >= 4 + 2x1 + 2x2 + F0(0) + x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 4 + 2x2 + 3x1 + F0(x1) + x2F0(x2) > 2 + 2x2 + F0(0) + x2F0(x2) = [[filter(_F0, _x2)]] We can thus remove the following rules: map(F, nil) => nil map(F, cons(X, Y)) => cons(F X, map(F, Y)) filter(F, nil) => nil filter(F, cons(X, Y)) => filter2(F X, F, X, Y) filter2(false, F, X, Y) => filter(F, 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]): f(a) >? f(b) g(b) >? g(a) filter2(true, F, X, Y) >? cons(X, filter(F, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: a = 0 b = 0 cons = \y0y1.y0 + y1 f = \y0.2y0 filter = \G0y1.y1 + G0(0) filter2 = \y0G1y2y3.3 + 3y0 + 3y2 + 3y3 + G1(y0) + 2G1(0) + 2G1(y3) g = \y0.y0 true = 3 Using this interpretation, the requirements translate to:
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