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HRS union beta 16688 pair #381734815
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
n111.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
0.588150978088 seconds
cpu usage
1.163360197
max memory
8.5434368E7
stage attributes
key
value
output-size
8348
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: 0 : [] --> a bits : [a] --> a cons : [c * d] --> d false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d half : [a] --> a map : [c -> c * d] --> d nil : [] --> d s : [a] --> a true : [] --> b Rules: half(0) => 0 half(s(0)) => 0 half(s(s(x))) => s(half(x)) bits(0) => 0 bits(s(x)) => s(bits(half(s(x)))) 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) 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]): half(0) >? 0 half(s(0)) >? 0 half(s(s(X))) >? s(half(X)) bits(0) >? 0 bits(s(X)) >? s(bits(half(s(X)))) 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: 0 = 0 bits = \y0.y0 cons = \y0y1.3 + y1 + 2y0 false = 3 filter = \G0y1.2y1 + G0(0) + 3y1G0(y1) filter2 = \y0G1y2y3.2y0 + 2y2 + 2y3 + 2G1(0) + 3y3G1(y3) half = \y0.y0 map = \G0y1.2 + 3y1 + G0(y1) + 2y1G0(y1) nil = 1 s = \y0.y0 true = 3 Using this interpretation, the requirements translate to: [[half(0)]] = 0 >= 0 = [[0]] [[half(s(0))]] = 0 >= 0 = [[0]] [[half(s(s(_x0)))]] = x0 >= x0 = [[s(half(_x0))]] [[bits(0)]] = 0 >= 0 = [[0]] [[bits(s(_x0))]] = x0 >= x0 = [[s(bits(half(s(_x0))))]] [[map(_F0, nil)]] = 5 + 3F0(1) > 1 = [[nil]] [[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))]] [[filter(_F0, nil)]] = 2 + F0(0) + 3F0(1) > 1 = [[nil]] [[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)]] [[filter2(true, _F0, _x1, _x2)]] = 6 + 2x1 + 2x2 + 2F0(0) + 3x2F0(x2) > 3 + 2x1 + 2x2 + F0(0) + 3x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 6 + 2x1 + 2x2 + 2F0(0) + 3x2F0(x2) > 2x2 + F0(0) + 3x2F0(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(true, F, X, Y) => cons(X, filter(F, 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]): half(0) >? 0 half(s(0)) >? 0
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