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	n142.star.cs.uiowa.edu
space	Uncurried_Applicative_11

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	0.0871709 seconds
cpu usage	0.087263
user time	0.076093
system time	0.01117
max virtual memory	113188.0
max residence set size	2876.0

stage attributes

key	value
starexec-result	YES

output

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. popout

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all output return to Higher Order Rewriting Union Beta