details

property	value
status	complete
benchmark	AotoYamada_05__004.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)	0.050166 seconds
cpu usage	0.049679
user time	0.039463
system time	0.010216
max virtual memory	113188.0
max residence set size	2260.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: 0 : [] --> b cons : [a * b] --> b nil : [] --> b plus : [b * b] --> b s : [b] --> b sumwith : [a -> b * b] --> b Rules: plus(0, x) => x plus(s(x), y) => s(plus(x, y)) sumwith(f, nil) => nil sumwith(f, cons(x, y)) => plus(f x, sumwith(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]): plus(0, X) >? X plus(s(X), Y) >? s(plus(X, Y)) sumwith(F, nil) >? nil sumwith(F, cons(X, Y)) >? plus(F X, sumwith(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 + 3y0 + 3y1 nil = 0 plus = \y0y1.1 + y0 + y1 s = \y0.y0 sumwith = \G0y1.3 + 3y1 + G0(0) + 2y1G0(y1) Using this interpretation, the requirements translate to: [[plus(0, _x0)]] = 4 + x0 > x0 = [[_x0]] [[plus(s(_x0), _x1)]] = 1 + x0 + x1 >= 1 + x0 + x1 = [[s(plus(_x0, _x1))]] [[sumwith(_F0, nil)]] = 3 + F0(0) > 0 = [[nil]] [[sumwith(_F0, cons(_x1, _x2))]] = 12 + 9x1 + 9x2 + F0(0) + 6x1F0(3 + 3x1 + 3x2) + 6x2F0(3 + 3x1 + 3x2) + 6F0(3 + 3x1 + 3x2) > 4 + x1 + 3x2 + F0(0) + F0(x1) + 2x2F0(x2) = [[plus(_F0 _x1, sumwith(_F0, _x2))]] We can thus remove the following rules: plus(0, X) => X sumwith(F, nil) => nil sumwith(F, cons(X, Y)) => plus(F X, sumwith(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]): plus(s(X), Y) >? s(plus(X, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: plus = \y0y1.y1 + 3y0 s = \y0.1 + y0 Using this interpretation, the requirements translate to: [[plus(s(_x0), _x1)]] = 3 + x1 + 3x0 > 1 + x1 + 3x0 = [[s(plus(_x0, _x1))]] We can thus remove the following rules: plus(s(X), Y) => s(plus(X, Y)) 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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actions

all output return to Higher Order Rewriting Union Beta