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Higher Order Rewriting Union Beta pair #487094101
details
property
value
status
complete
benchmark
AotoYamada_05__013.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n144.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda 2.2a
configuration
default
runtime (wallclock)
0.121612 seconds
cpu usage
0.116939
user time
0.097605
system time
0.019334
max virtual memory
113188.0
max residence set size
3272.0
stage attributes
key
value
starexec-result
YES
output
YES We consider the system theBenchmark. Alphabet: append : [c * c] --> c cons : [b * c] --> c flatwith : [a -> b * b] --> c flatwithsub : [a -> b * c] --> c leaf : [a] --> b nil : [] --> c node : [c] --> b Rules: append(nil, x) => x append(cons(x, y), z) => cons(x, append(y, z)) flatwith(f, leaf(x)) => cons(f x, nil) flatwith(f, node(x)) => flatwithsub(f, x) flatwithsub(f, nil) => nil flatwithsub(f, cons(x, y)) => append(flatwith(f, x), flatwithsub(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)) flatwith(F, leaf(X)) >? cons(F X, nil) flatwith(F, node(X)) >? flatwithsub(F, X) flatwithsub(F, nil) >? nil flatwithsub(F, cons(X, Y)) >? append(flatwith(F, X), flatwithsub(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 flatwith = \G0y1.y1 + G0(y1) + 2y1G0(y1) flatwithsub = \G0y1.y1 + G0(0) + G0(y1) + 2y1G0(y1) leaf = \y0.3 + 3y0 nil = 2 node = \y0.3 + 3y0 Using this interpretation, the requirements translate to: [[append(nil, _x0)]] = 2 + x0 > x0 = [[_x0]] [[append(cons(_x0, _x1), _x2)]] = 1 + x0 + x1 + x2 >= 1 + x0 + x1 + x2 = [[cons(_x0, append(_x1, _x2))]] [[flatwith(_F0, leaf(_x1))]] = 3 + 3x1 + 6x1F0(3 + 3x1) + 7F0(3 + 3x1) >= 3 + x1 + F0(x1) = [[cons(_F0 _x1, nil)]] [[flatwith(_F0, node(_x1))]] = 3 + 3x1 + 6x1F0(3 + 3x1) + 7F0(3 + 3x1) > x1 + F0(0) + F0(x1) + 2x1F0(x1) = [[flatwithsub(_F0, _x1)]] [[flatwithsub(_F0, nil)]] = 2 + F0(0) + 5F0(2) >= 2 = [[nil]] [[flatwithsub(_F0, cons(_x1, _x2))]] = 1 + x1 + x2 + F0(0) + 2x1F0(1 + x1 + x2) + 2x2F0(1 + x1 + x2) + 3F0(1 + x1 + x2) > x1 + x2 + F0(0) + F0(x1) + F0(x2) + 2x1F0(x1) + 2x2F0(x2) = [[append(flatwith(_F0, _x1), flatwithsub(_F0, _x2))]] We can thus remove the following rules: append(nil, X) => X flatwith(F, node(X)) => flatwithsub(F, X) flatwithsub(F, cons(X, Y)) => append(flatwith(F, X), flatwithsub(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(cons(X, Y), Z) >? cons(X, append(Y, Z)) flatwith(F, leaf(X)) >? cons(F X, nil) flatwithsub(F, nil) >? nil We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: append = \y0y1.y1 + 3y0 cons = \y0y1.y0 + y1 flatwith = \G0y1.3 + 3y1 + G0(0) + y1G0(y1) flatwithsub = \G0y1.3 + 3y1 + G0(0) leaf = \y0.3 + 3y0 nil = 0 Using this interpretation, the requirements translate to: [[append(cons(_x0, _x1), _x2)]] = x2 + 3x0 + 3x1 >= x0 + x2 + 3x1 = [[cons(_x0, append(_x1, _x2))]] [[flatwith(_F0, leaf(_x1))]] = 12 + 9x1 + F0(0) + 3x1F0(3 + 3x1) + 3F0(3 + 3x1) > x1 + F0(x1) = [[cons(_F0 _x1, nil)]] [[flatwithsub(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] We can thus remove the following rules: flatwith(F, leaf(X)) => cons(F X, nil) flatwithsub(F, nil) => nil 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(cons(X, Y), Z) >? cons(X, append(Y, Z)) We orient these requirements with a polynomial interpretation in the natural numbers.
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