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	n053.star.cs.uiowa.edu
space	Uncurried_Applicative_11

run statistics

property	value
solver	Wanda
configuration	HigherOrder
runtime (wallclock)	0.0346479415894 seconds
cpu usage	0.027818201
max memory	1732608.0

stage attributes

key	value
output-size	2680
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: 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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job log

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actions

all output return to HRS union beta 16688