details

property	value
status	complete
benchmark	AotoYamada_05__009.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n149.star.cs.uiowa.edu
space	Uncurried_Applicative_11

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	0.0904389 seconds
cpu usage	0.090526
user time	0.080675
system time	0.009851
max virtual memory	113188.0
max residence set size	3256.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: and : [c * c] --> c cons : [a * b] --> b false : [] --> c forall : [a -> c * b] --> c forsome : [a -> c * b] --> c nil : [] --> b or : [c * c] --> c true : [] --> c Rules: and(true, true) => true and(true, false) => false and(false, true) => false and(false, false) => false or(true, true) => true or(true, false) => true or(false, true) => true or(false, false) => false forall(f, nil) => true forall(f, cons(x, y)) => and(f x, forall(f, y)) forsome(f, nil) => false forsome(f, cons(x, y)) => or(f x, forsome(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]): and(true, true) >? true and(true, false) >? false and(false, true) >? false and(false, false) >? false or(true, true) >? true or(true, false) >? true or(false, true) >? true or(false, false) >? false forall(F, nil) >? true forall(F, cons(X, Y)) >? and(F X, forall(F, Y)) forsome(F, nil) >? false forsome(F, cons(X, Y)) >? or(F X, forsome(F, Y)) We orient these requirements with a polynomial interpretation in the natural numbers. The following interpretation satisfies the requirements: and = \y0y1.2 + y0 + y1 cons = \y0y1.3 + 3y0 + 3y1 false = 0 forall = \G0y1.3y1 + G0(0) + G0(y1) + 3y1G0(y1) forsome = \G0y1.3y1 + 3y1G0(y1) + 3G0(y1) nil = 3 or = \y0y1.2 + y0 + 2y1 true = 0 Using this interpretation, the requirements translate to: [[and(true, true)]] = 2 > 0 = [[true]] [[and(true, false)]] = 2 > 0 = [[false]] [[and(false, true)]] = 2 > 0 = [[false]] [[and(false, false)]] = 2 > 0 = [[false]] [[or(true, true)]] = 2 > 0 = [[true]] [[or(true, false)]] = 2 > 0 = [[true]] [[or(false, true)]] = 2 > 0 = [[true]] [[or(false, false)]] = 2 > 0 = [[false]] [[forall(_F0, nil)]] = 9 + F0(0) + 10F0(3) > 0 = [[true]] [[forall(_F0, cons(_x1, _x2))]] = 9 + 9x1 + 9x2 + F0(0) + 9x1F0(3 + 3x1 + 3x2) + 9x2F0(3 + 3x1 + 3x2) + 10F0(3 + 3x1 + 3x2) > 2 + x1 + 3x2 + F0(0) + F0(x1) + F0(x2) + 3x2F0(x2) = [[and(_F0 _x1, forall(_F0, _x2))]] [[forsome(_F0, nil)]] = 9 + 12F0(3) > 0 = [[false]] [[forsome(_F0, cons(_x1, _x2))]] = 9 + 9x1 + 9x2 + 9x1F0(3 + 3x1 + 3x2) + 9x2F0(3 + 3x1 + 3x2) + 12F0(3 + 3x1 + 3x2) > 2 + x1 + 6x2 + F0(x1) + 6x2F0(x2) + 6F0(x2) = [[or(_F0 _x1, forsome(_F0, _x2))]] We can thus remove the following rules: and(true, true) => true and(true, false) => false and(false, true) => false and(false, false) => false or(true, true) => true or(true, false) => true or(false, true) => true or(false, false) => false forall(F, nil) => true forall(F, cons(X, Y)) => and(F X, forall(F, Y)) forsome(F, nil) => false forsome(F, cons(X, Y)) => or(F X, forsome(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

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to Higher Order Rewriting Union Beta