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HRS union beta 16688 pair #381734714
details
property
value
status
complete
benchmark
Applicative_05__TakeDropWhile.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n045.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
0.052521944046 seconds
cpu usage
0.042058522
max memory
2519040.0
stage attributes
key
value
output-size
3324
starexec-result
YES
output
/export/starexec/sandbox2/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. Alphabet: cons : [a * c] --> c dropWhile : [a -> b * c] --> c if : [b * c * c] --> c nil : [] --> c takeWhile : [a -> b * c] --> c true : [] --> b Rules: if(true, x, y) => x if(true, x, y) => y takeWhile(f, nil) => nil takeWhile(f, cons(x, y)) => if(f x, cons(x, takeWhile(f, y)), nil) dropWhile(f, nil) => nil dropWhile(f, cons(x, y)) => if(f x, dropWhile(f, y), cons(x, 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]): if(true, X, Y) >? X if(true, X, Y) >? Y takeWhile(F, nil) >? nil takeWhile(F, cons(X, Y)) >? if(F X, cons(X, takeWhile(F, Y)), nil) dropWhile(F, nil) >? nil dropWhile(F, cons(X, Y)) >? if(F X, dropWhile(F, Y), cons(X, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: cons = \y0y1.2 + y0 + 2y1 dropWhile = \G0y1.3 + 2y1 + 2y1G0(y1) + 2G0(0) if = \y0y1y2.y0 + y1 + y2 nil = 0 takeWhile = \G0y1.3y1 + G0(0) + 2y1G0(y1) true = 3 Using this interpretation, the requirements translate to: [[if(true, _x0, _x1)]] = 3 + x0 + x1 > x0 = [[_x0]] [[if(true, _x0, _x1)]] = 3 + x0 + x1 > x1 = [[_x1]] [[takeWhile(_F0, nil)]] = F0(0) >= 0 = [[nil]] [[takeWhile(_F0, cons(_x1, _x2))]] = 6 + 3x1 + 6x2 + F0(0) + 2x1F0(2 + x1 + 2x2) + 4x2F0(2 + x1 + 2x2) + 4F0(2 + x1 + 2x2) > 2 + 2x1 + 6x2 + F0(x1) + 2F0(0) + 4x2F0(x2) = [[if(_F0 _x1, cons(_x1, takeWhile(_F0, _x2)), nil)]] [[dropWhile(_F0, nil)]] = 3 + 2F0(0) > 0 = [[nil]] [[dropWhile(_F0, cons(_x1, _x2))]] = 7 + 2x1 + 4x2 + 2x1F0(2 + x1 + 2x2) + 2F0(0) + 4x2F0(2 + x1 + 2x2) + 4F0(2 + x1 + 2x2) > 5 + 2x1 + 4x2 + F0(x1) + 2x2F0(x2) + 2F0(0) = [[if(_F0 _x1, dropWhile(_F0, _x2), cons(_x1, _x2))]] We can thus remove the following rules: if(true, X, Y) => X if(true, X, Y) => Y takeWhile(F, cons(X, Y)) => if(F X, cons(X, takeWhile(F, Y)), nil) dropWhile(F, nil) => nil dropWhile(F, cons(X, Y)) => if(F X, dropWhile(F, Y), cons(X, 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]): takeWhile(F, nil) >? nil We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: nil = 0 takeWhile = \G0y1.3 + 3y1 + G0(0) Using this interpretation, the requirements translate to: [[takeWhile(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] We can thus remove the following rules: takeWhile(F, nil) => nil All rules were succesfully removed. Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold. +++ Citations +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012.
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