details

property	value
status	complete
benchmark	AotoYamada_05__027.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n105.star.cs.uiowa.edu
space	Uncurried_Applicative_11

run statistics

property	value
solver	Wanda
configuration	HigherOrder
runtime (wallclock)	0.0468029975891 seconds
cpu usage	0.043992677
max memory	2387968.0

stage attributes

key	value
output-size	3207
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 : [] --> a cons : [a * b] --> b inc : [b] --> b map : [a -> a * b] --> b nil : [] --> b plus : [a] --> a -> a s : [a] --> a Rules: plus(0) x => x plus(s(x)) y => s(plus(x) y) inc(x) => map(plus(s(0)), x) map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(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) inc(X) >? map(plus(s(0)), 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 = 0 cons = \y0y1.3 + y1 + 2y0 inc = \y0.3 + 3y0 map = \G0y1.2 + y1 + G0(0) + 3y1G0(y1) nil = 0 plus = \y0y1.2y0 s = \y0.y0 Using this interpretation, the requirements translate to: [[plus(0) _x0]] = x0 >= x0 = [[_x0]] [[plus(s(_x0)) _x1]] = x1 + 2x0 >= x1 + 2x0 = [[s(plus(_x0) _x1)]] [[inc(_x0)]] = 3 + 3x0 > 2 + x0 = [[map(plus(s(0)), _x0)]] [[map(_F0, nil)]] = 2 + F0(0) > 0 = [[nil]] [[map(_F0, cons(_x1, _x2))]] = 5 + x2 + 2x1 + F0(0) + 3x2F0(3 + x2 + 2x1) + 6x1F0(3 + x2 + 2x1) + 9F0(3 + x2 + 2x1) >= 5 + x2 + 2x1 + F0(0) + 2F0(x1) + 3x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] We can thus remove the following rules: inc(X) => map(plus(s(0)), X) map(F, nil) => nil 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, 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 = \G0y1.3y1 + 2G0(0) + 2G0(y1) + 3y1G0(y1) plus = \y0y1.3 + y1 + 3y0 s = \y0.3 + y0 Using this interpretation, the requirements translate to: [[plus(0, _x0)]] = 12 + x0 > x0 = [[_x0]] [[plus(s(_x0), _x1)]] = 12 + x1 + 3x0 > 6 + x1 + 3x0 = [[s(plus(_x0, _x1))]] [[map(_F0, cons(_x1, _x2))]] = 9 + 3x1 + 3x2 + 2F0(0) + 3x1F0(3 + x1 + x2) + 3x2F0(3 + x1 + x2) + 11F0(3 + x1 + x2) > 3 + x1 + 3x2 + F0(x1) + 2F0(0) + 2F0(x2) + 3x2F0(x2) = [[cons(_F0 _x1, map(_F0, _x2))]] We can thus remove the following rules: plus(0, X) => X plus(s(X), Y) => s(plus(X, Y)) map(F, cons(X, Y)) => cons(F X, map(F, 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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all output return to HRS union beta 16688