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Higher Order Rewriting Union Beta pair #487093965
details
property
value
status
complete
benchmark
AotoYamada_05__011.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n150.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda 2.2a
configuration
default
runtime (wallclock)
0.150851 seconds
cpu usage
0.149362
user time
0.126572
system time
0.02279
max virtual memory
113188.0
max residence set size
2904.0
stage attributes
key
value
starexec-result
YES
output
YES We consider the system theBenchmark. Alphabet: 0 : [] --> b cons : [b * a] --> a curry : [b -> b -> b * b] --> b -> b inc : [] --> a -> a map : [b -> b] --> a -> a nil : [] --> a plus : [] --> b -> b -> b s : [b] --> b Rules: plus 0 x => x plus s(x) y => s(plus x y) map(f) nil => nil map(f) cons(x, y) => cons(f x, map(f) y) curry(f, x) y => f x y inc => map(curry(plus, s(0))) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). Symbol curry is an encoding for application that is only used in innocuous ways. We can simplify the program (without losing non-termination) by removing it. This gives: Alphabet: 0 : [] --> b cons : [b * a] --> a inc : [] --> a -> a map : [b -> b] --> a -> a nil : [] --> a plus : [b] --> b -> b s : [b] --> b Rules: plus(0) X => X plus(s(X)) Y => s(plus(X) Y) map(F) nil => nil map(F) cons(X, Y) => cons(F X, map(F) Y) inc => map(plus(s(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]): plus(0) X >? X plus(s(X)) Y >? s(plus(X) Y) map(F) nil >? nil map(F) cons(X, Y) >? cons(F X, map(F) Y) inc >? map(plus(s(0))) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: 0 = 0 cons = \y0y1.3 + y0 + y1 inc = \y0.3 + 3y0 map = \G0y1.G0(0) + y1G0(y1) nil = 3 plus = \y0y1.2y0 s = \y0.1 + y0 Using this interpretation, the requirements translate to: [[plus(0) _x0]] = x0 >= x0 = [[_x0]] [[plus(s(_x0)) _x1]] = 2 + x1 + 2x0 > 1 + x1 + 2x0 = [[s(plus(_x0) _x1)]] [[map(_F0) nil]] = 3 + F0(0) + 3F0(3) >= 3 = [[nil]] [[map(_F0) cons(_x1, _x2)]] = 3 + x1 + x2 + F0(0) + 3F0(3 + x1 + x2) + x1F0(3 + x1 + x2) + x2F0(3 + x1 + x2) >= 3 + x1 + x2 + F0(0) + F0(x1) + x2F0(x2) = [[cons(_F0 _x1, map(_F0) _x2)]] [[inc]] = \y0.3 + 3y0 > \y0.2 + 2y0 = [[map(plus(s(0)))]] We can thus remove the following rules: plus(s(X)) Y => s(plus(X) Y) inc => map(plus(s(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]): plus(0, X) >? X 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: 0 = 3 cons = \y0y1.3 + y0 + y1 map = \G0y1.3 + 3y1 + 2G0(0) + 2G0(y1) + 3y1G0(y1) nil = 0 plus = \y0y1.3 + y0 + y1 Using this interpretation, the requirements translate to:
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