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Higher Order Rewriting Union Beta pair #487094027
details
property
value
status
complete
benchmark
AotoYamada_05__012.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n140.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda 2.2a
configuration
default
runtime (wallclock)
0.094414 seconds
cpu usage
0.094539
user time
0.083958
system time
0.010581
max virtual memory
113188.0
max residence set size
3164.0
stage attributes
key
value
starexec-result
YES
output
YES We consider the system theBenchmark. Alphabet: and : [c * c] --> c cons : [a * b] --> b false : [] --> c forall : [a -> c * b] --> c forsome : [a -> c * b] --> c nil : [] --> b or : [c * c] --> c true : [] --> c Rules: and(true, true) => true and(x, false) => false and(false, x) => false or(true, x) => true or(x, true) => true or(false, false) => false forall(f, nil) => true forall(f, cons(x, y)) => and(f x, forall(f, y)) forsome(f, nil) => false forsome(f, cons(x, y)) => or(f x, forsome(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]): and(true, true) >? true and(X, false) >? false and(false, X) >? false or(true, X) >? true or(X, true) >? true or(false, false) >? false forall(F, nil) >? true forall(F, cons(X, Y)) >? and(F X, forall(F, Y)) forsome(F, nil) >? false forsome(F, cons(X, Y)) >? or(F X, forsome(F, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: and = \y0y1.1 + y0 + y1 cons = \y0y1.3 + 3y0 + 3y1 false = 0 forall = \G0y1.3y1 + 2G0(y1) + 3y1G0(y1) forsome = \G0y1.3y1 + G0(0) + 3y1G0(y1) + 3G0(y1) nil = 3 or = \y0y1.y1 + 2y0 true = 0 Using this interpretation, the requirements translate to: [[and(true, true)]] = 1 > 0 = [[true]] [[and(_x0, false)]] = 1 + x0 > 0 = [[false]] [[and(false, _x0)]] = 1 + x0 > 0 = [[false]] [[or(true, _x0)]] = x0 >= 0 = [[true]] [[or(_x0, true)]] = 2x0 >= 0 = [[true]] [[or(false, false)]] = 0 >= 0 = [[false]] [[forall(_F0, nil)]] = 9 + 11F0(3) > 0 = [[true]] [[forall(_F0, cons(_x1, _x2))]] = 9 + 9x1 + 9x2 + 9x1F0(3 + 3x1 + 3x2) + 9x2F0(3 + 3x1 + 3x2) + 11F0(3 + 3x1 + 3x2) > 1 + x1 + 3x2 + F0(x1) + 2F0(x2) + 3x2F0(x2) = [[and(_F0 _x1, forall(_F0, _x2))]] [[forsome(_F0, nil)]] = 9 + F0(0) + 12F0(3) > 0 = [[false]] [[forsome(_F0, cons(_x1, _x2))]] = 9 + 9x1 + 9x2 + F0(0) + 9x1F0(3 + 3x1 + 3x2) + 9x2F0(3 + 3x1 + 3x2) + 12F0(3 + 3x1 + 3x2) > 2x1 + 3x2 + F0(0) + 2F0(x1) + 3x2F0(x2) + 3F0(x2) = [[or(_F0 _x1, forsome(_F0, _x2))]] We can thus remove the following rules: and(true, true) => true and(X, false) => false and(false, X) => false forall(F, nil) => true forall(F, cons(X, Y)) => and(F X, forall(F, Y)) forsome(F, nil) => false forsome(F, cons(X, Y)) => or(F X, forsome(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]): or(true, X) >? true or(X, true) >? true or(false, false) >? false We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: false = 0 or = \y0y1.3 + 3y0 + 3y1 true = 0 Using this interpretation, the requirements translate to: [[or(true, _x0)]] = 3 + 3x0 > 0 = [[true]] [[or(_x0, true)]] = 3 + 3x0 > 0 = [[true]]
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