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HRS union beta 16688 pair #381734425
details
property
value
status
complete
benchmark
AotoYamada_05__015.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n103.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
0.204909086227 seconds
cpu usage
0.156397159
max memory
7843840.0
stage attributes
key
value
output-size
6858
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. Alphabet: cons : [a * c] --> c false : [] --> b filter : [a -> b * c] --> c if : [b * c * c] --> c nil : [] --> c true : [] --> b Rules: if(true, x, y) => x if(false, x, y) => y filter(f, nil) => nil filter(f, cons(x, y)) => if(f x, cons(x, filter(f, y)), filter(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]): if(true, X, Y) >? X if(false, X, Y) >? Y filter(F, nil) >? nil filter(F, cons(X, Y)) >? if(F X, cons(X, filter(F, Y)), filter(F, Y)) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[nil]] = _|_ We choose Lex = {} and Mul = {@_{o -> o}, cons, false, filter, if, true}, and the following precedence: false > filter > @_{o -> o} > cons > if > true Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: if(true, X, Y) >= X if(false, X, Y) >= Y filter(F, _|_) > _|_ filter(F, cons(X, Y)) >= if(@_{o -> o}(F, X), cons(X, filter(F, Y)), filter(F, Y)) With these choices, we have: 1] if(true, X, Y) >= X because [2], by (Star) 2] if*(true, X, Y) >= X because [3], by (Select) 3] X >= X by (Meta) 4] if(false, X, Y) >= Y because [5], by (Star) 5] if*(false, X, Y) >= Y because [6], by (Select) 6] Y >= Y by (Meta) 7] filter(F, _|_) > _|_ because [8], by definition 8] filter*(F, _|_) >= _|_ by (Bot) 9] filter(F, cons(X, Y)) >= if(@_{o -> o}(F, X), cons(X, filter(F, Y)), filter(F, Y)) because [10], by (Star) 10] filter*(F, cons(X, Y)) >= if(@_{o -> o}(F, X), cons(X, filter(F, Y)), filter(F, Y)) because filter > if, [11], [18] and [19], by (Copy) 11] filter*(F, cons(X, Y)) >= @_{o -> o}(F, X) because filter > @_{o -> o}, [12] and [14], by (Copy) 12] filter*(F, cons(X, Y)) >= F because [13], by (Select) 13] F >= F by (Meta) 14] filter*(F, cons(X, Y)) >= X because [15], by (Select) 15] cons(X, Y) >= X because [16], by (Star) 16] cons*(X, Y) >= X because [17], by (Select) 17] X >= X by (Meta) 18] filter*(F, cons(X, Y)) >= cons(X, filter(F, Y)) because filter > cons, [14] and [19], by (Copy) 19] filter*(F, cons(X, Y)) >= filter(F, Y) because filter in Mul, [20] and [21], by (Stat) 20] F >= F by (Meta) 21] cons(X, Y) > Y because [22], by definition 22] cons*(X, Y) >= Y because [23], by (Select) 23] Y >= Y by (Meta) We can thus remove the following rules: filter(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]): if(true, X, Y) >? X if(false, X, Y) >? Y filter(F, cons(X, Y)) >? if(F X, cons(X, filter(F, Y)), filter(F, Y)) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {} and Mul = {@_{o -> o}, cons, false, filter, if, true}, and the following precedence: false > filter > @_{o -> o} > if > cons > true
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