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Higher Order Rewriting Union Beta pair #487093791
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
n151.star.cs.uiowa.edu
space
Kop_13
run statistics
property
value
solver
Wanda 2.2a
configuration
default
runtime (wallclock)
0.290476 seconds
cpu usage
0.127157
user time
0.109854
system time
0.017303
max virtual memory
113188.0
max residence set size
3124.0
stage attributes
key
value
starexec-result
YES
output
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 Using this interpretation, the requirements translate to: [[append(nil, _x0)]] = x0 >= x0 = [[_x0]] [[append(cons(_x0, _x1), _x2)]] = 2 + x1 + x2 + 3x0 >= 2 + x1 + x2 + 3x0 = [[cons(_x0, append(_x1, _x2))]] [[reverse(nil)]] = 3 > 0 = [[nil]] [[shuffle(nil)]] = 3 > 0 = [[nil]]
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