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Higher Order Rewriting Union Beta pair #487093961
details
property
value
status
complete
benchmark
AotoYamada_05__016.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n148.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda 2.2a
configuration
default
runtime (wallclock)
0.749303 seconds
cpu usage
0.749293
user time
0.691603
system time
0.05769
max virtual memory
137936.0
max residence set size
23148.0
stage attributes
key
value
starexec-result
YES
output
YES We consider the system theBenchmark. Alphabet: 0 : [] --> a cons : [a * c] --> c false : [] --> b filter : [a -> b] --> c -> c filtersub : [b * a -> b * c] --> c neq : [a] --> a -> b nil : [] --> c nonzero : [] --> c -> c s : [a] --> a true : [] --> b Rules: neq(0) 0 => false neq(0) s(x) => true neq(s(x)) 0 => true neq(s(x)) s(y) => neq(x) y filter(f) nil => nil filter(f) cons(x, y) => filtersub(f x, f, cons(x, y)) filtersub(true, f, cons(x, y)) => cons(x, filter(f) y) filtersub(false, f, cons(x, y)) => filter(f) y nonzero => filter(neq(0)) 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]): neq(0) 0 >? false neq(0) s(X) >? true neq(s(X)) 0 >? true neq(s(X)) s(Y) >? neq(X) Y filter(F) nil >? nil filter(F) cons(X, Y) >? filtersub(F X, F, cons(X, Y)) filtersub(true, F, cons(X, Y)) >? cons(X, filter(F) Y) filtersub(false, F, cons(X, Y)) >? filter(F) Y nonzero >? filter(neq(0)) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[@_{o -> o}(x_1, x_2)]] = @_{o -> o}(x_2, x_1) [[false]] = _|_ [[filtersub(x_1, x_2, x_3)]] = filtersub(x_3, x_2, x_1) [[nil]] = _|_ [[true]] = _|_ We choose Lex = {@_{o -> o}, filtersub} and Mul = {cons, filter, neq, nonzero, s}, and the following precedence: nonzero > neq > @_{o -> o} = filtersub > filter > cons > s Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: @_{o -> o}(neq(_|_), _|_) >= _|_ @_{o -> o}(neq(_|_), s(X)) >= _|_ @_{o -> o}(neq(s(X)), _|_) >= _|_ @_{o -> o}(neq(s(X)), s(Y)) >= @_{o -> o}(neq(X), Y) @_{o -> o}(filter(F), _|_) >= _|_ @_{o -> o}(filter(F), cons(X, Y)) >= filtersub(@_{o -> o}(F, X), F, cons(X, Y)) filtersub(_|_, F, cons(X, Y)) >= cons(X, @_{o -> o}(filter(F), Y)) filtersub(_|_, F, cons(X, Y)) > @_{o -> o}(filter(F), Y) nonzero >= filter(neq(_|_)) With these choices, we have: 1] @_{o -> o}(neq(_|_), _|_) >= _|_ by (Bot) 2] @_{o -> o}(neq(_|_), s(X)) >= _|_ by (Bot) 3] @_{o -> o}(neq(s(X)), _|_) >= _|_ by (Bot) 4] @_{o -> o}(neq(s(X)), s(Y)) >= @_{o -> o}(neq(X), Y) because [5] and [10], by (Fun) 5] neq(s(X)) >= neq(X) because [6], by (Star) 6] neq*(s(X)) >= neq(X) because neq in Mul and [7], by (Stat) 7] s(X) > X because [8], by definition 8] s*(X) >= X because [9], by (Select) 9] X >= X by (Meta) 10] s(Y) >= Y because [11], by (Star) 11] s*(Y) >= Y because [12], by (Select) 12] Y >= Y by (Meta) 13] @_{o -> o}(filter(F), _|_) >= _|_ by (Bot) 14] @_{o -> o}(filter(F), cons(X, Y)) >= filtersub(@_{o -> o}(F, X), F, cons(X, Y)) because [15], by (Star) 15] @_{o -> o}*(filter(F), cons(X, Y)) >= filtersub(@_{o -> o}(F, X), F, cons(X, Y)) because @_{o -> o} = filtersub, [16], [19], [22], [25] and [28], by (Stat) 16] filter(F) > F because [17], by definition 17] filter*(F) >= F because [18], by (Select) 18] F >= F by (Meta) 19] cons(X, Y) >= cons(X, Y) because cons in Mul, [20] and [21], by (Fun) 20] X >= X by (Meta) 21] Y >= Y by (Meta) 22] @_{o -> o}*(filter(F), cons(X, Y)) >= @_{o -> o}(F, X) because [16], [23], [25] and [27], by (Stat) 23] cons(X, Y) >= X because [24], by (Star) 24] cons*(X, Y) >= X because [20], by (Select)
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