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Higher Order Rewriting Union Beta pair #487094069
details
property
value
status
complete
benchmark
Applicative_first_order_05__17.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n141.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda 2.2a
configuration
default
runtime (wallclock)
0.235081 seconds
cpu usage
0.235385
user time
0.20784
system time
0.027545
max virtual memory
113188.0
max residence set size
4920.0
stage attributes
key
value
starexec-result
YES
output
YES We consider the system theBenchmark. Alphabet: !facdot : [a * a] --> a 1 : [] --> a cons : [c * d] --> d false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d i : [a] --> a map : [c -> c * d] --> d nil : [] --> d true : [] --> b Rules: !facdot(1, x) => x !facdot(x, 1) => x !facdot(i(x), x) => 1 !facdot(x, i(x)) => 1 !facdot(i(x), !facdot(x, y)) => y !facdot(x, !facdot(i(x), y)) => y !facdot(!facdot(x, y), z) => !facdot(x, !facdot(y, z)) i(1) => 1 i(i(x)) => x i(!facdot(x, y)) => !facdot(i(y), i(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]): !facdot(1, X) >? X !facdot(X, 1) >? X !facdot(i(X), X) >? 1 !facdot(X, i(X)) >? 1 !facdot(i(X), !facdot(X, Y)) >? Y !facdot(X, !facdot(i(X), Y)) >? Y !facdot(!facdot(X, Y), Z) >? !facdot(X, !facdot(Y, Z)) i(1) >? 1 i(i(X)) >? X i(!facdot(X, Y)) >? !facdot(i(Y), i(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: !facdot = \y0y1.y0 + y1 1 = 0 cons = \y0y1.3 + y1 + 2y0 false = 3 filter = \G0y1.1 + 3y1 + G0(0) + 2y1G0(y1) filter2 = \y0G1y2y3.1 + 2y0 + 2y2 + 3y3 + G1(0) + 2y3G1(y3) i = \y0.y0 map = \G0y1.3y1 + G0(0) + 3y1G0(y1) nil = 0 true = 3 Using this interpretation, the requirements translate to: [[!facdot(1, _x0)]] = x0 >= x0 = [[_x0]] [[!facdot(_x0, 1)]] = x0 >= x0 = [[_x0]] [[!facdot(i(_x0), _x0)]] = 2x0 >= 0 = [[1]] [[!facdot(_x0, i(_x0))]] = 2x0 >= 0 = [[1]] [[!facdot(i(_x0), !facdot(_x0, _x1))]] = x1 + 2x0 >= x1 = [[_x1]] [[!facdot(_x0, !facdot(i(_x0), _x1))]] = x1 + 2x0 >= x1 = [[_x1]] [[!facdot(!facdot(_x0, _x1), _x2)]] = x0 + x1 + x2 >= x0 + x1 + x2 = [[!facdot(_x0, !facdot(_x1, _x2))]] [[i(1)]] = 0 >= 0 = [[1]] [[i(i(_x0))]] = x0 >= x0 = [[_x0]] [[i(!facdot(_x0, _x1))]] = x0 + x1 >= x0 + x1 = [[!facdot(i(_x1), i(_x0))]] [[map(_F0, nil)]] = F0(0) >= 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 9 + 3x2 + 6x1 + F0(0) + 3x2F0(3 + x2 + 2x1) + 6x1F0(3 + x2 + 2x1) + 9F0(3 + x2 + 2x1) > 3 + 2x1 + 3x2 + F0(0) + 2F0(x1) + 3x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] [[filter(_F0, nil)]] = 1 + F0(0) > 0 = [[nil]] [[filter(_F0, cons(_x1, _x2))]] = 10 + 3x2 + 6x1 + F0(0) + 2x2F0(3 + x2 + 2x1) + 4x1F0(3 + x2 + 2x1) + 6F0(3 + x2 + 2x1) > 1 + 3x2 + 4x1 + F0(0) + 2x2F0(x2) + 2F0(x1) = [[filter2(_F0 _x1, _F0, _x1, _x2)]] [[filter2(true, _F0, _x1, _x2)]] = 7 + 2x1 + 3x2 + F0(0) + 2x2F0(x2) > 4 + 2x1 + 3x2 + F0(0) + 2x2F0(x2) = [[cons(_x1, filter(_F0, _x2))]] [[filter2(false, _F0, _x1, _x2)]] = 7 + 2x1 + 3x2 + F0(0) + 2x2F0(x2) > 1 + 3x2 + F0(0) + 2x2F0(x2) = [[filter(_F0, _x2)]] We can thus remove the following rules: 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)
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