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HRS union beta 16688 pair #381734167
details
property
value
status
complete
benchmark
fuhkop11frocos.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n003.star.cs.uiowa.edu
space
Kop_13
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
0.084352016449 seconds
cpu usage
0.064906673
max memory
2646016.0
stage attributes
key
value
output-size
5103
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: append : [list * list] --> list cons : [nat * list] --> list map : [nat -> nat * list] --> list mirror : [list] --> list nil : [] --> list reverse : [list] --> list shuffle : [list] --> list Rules: append(nil, x) => x append(cons(x, y), z) => cons(x, append(y, z)) reverse(nil) => nil shuffle(nil) => nil shuffle(cons(x, y)) => cons(x, shuffle(reverse(y))) mirror(nil) => nil mirror(cons(x, y)) => append(cons(x, mirror(y)), cons(x, nil)) map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(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]): append(nil, X) >? X append(cons(X, Y), Z) >? cons(X, append(Y, Z)) reverse(nil) >? nil shuffle(nil) >? nil shuffle(cons(X, Y)) >? cons(X, shuffle(reverse(Y))) mirror(nil) >? nil mirror(cons(X, Y)) >? append(cons(X, mirror(Y)), cons(X, nil)) map(F, nil) >? nil map(F, cons(X, Y)) >? cons(F X, map(F, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: append = \y0y1.y0 + y1 cons = \y0y1.1 + y0 + y1 map = \G0y1.3y1 + 2G0(0) + y1G0(y1) mirror = \y0.2y0 nil = 0 reverse = \y0.y0 shuffle = \y0.2y0 Using this interpretation, the requirements translate to: [[append(nil, _x0)]] = x0 >= x0 = [[_x0]] [[append(cons(_x0, _x1), _x2)]] = 1 + x0 + x1 + x2 >= 1 + x0 + x1 + x2 = [[cons(_x0, append(_x1, _x2))]] [[reverse(nil)]] = 0 >= 0 = [[nil]] [[shuffle(nil)]] = 0 >= 0 = [[nil]] [[shuffle(cons(_x0, _x1))]] = 2 + 2x0 + 2x1 > 1 + x0 + 2x1 = [[cons(_x0, shuffle(reverse(_x1)))]] [[mirror(nil)]] = 0 >= 0 = [[nil]] [[mirror(cons(_x0, _x1))]] = 2 + 2x0 + 2x1 >= 2 + 2x0 + 2x1 = [[append(cons(_x0, mirror(_x1)), cons(_x0, nil))]] [[map(_F0, nil)]] = 2F0(0) >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 3 + 3x1 + 3x2 + F0(1 + x1 + x2) + 2F0(0) + x1F0(1 + x1 + x2) + x2F0(1 + x1 + x2) > 1 + x1 + 3x2 + F0(x1) + 2F0(0) + x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] We can thus remove the following rules: shuffle(cons(X, Y)) => cons(X, shuffle(reverse(Y))) map(F, cons(X, Y)) => cons(F X, map(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]): append(nil, X) >? X append(cons(X, Y), Z) >? cons(X, append(Y, Z)) reverse(nil) >? nil shuffle(nil) >? nil mirror(nil) >? nil mirror(cons(X, Y)) >? append(cons(X, mirror(Y)), cons(X, nil)) map(F, nil) >? nil We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: append = \y0y1.y0 + y1 cons = \y0y1.2 + y1 + 3y0 map = \G0y1.3 + 3y1 + G0(0) mirror = \y0.2 + 3y0 nil = 0 reverse = \y0.3 + 3y0 shuffle = \y0.3 + 3y0
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