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HRS union beta 16688 pair #381734860
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
n020.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
0.537838935852 seconds
cpu usage
1.096281183
max memory
8.4094976E7
stage attributes
key
value
output-size
6766
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- 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
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