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HRS union beta 16688 pair #381734905
details
property
value
status
complete
benchmark
Applicative_first_order_05__11.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n049.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
1.03104805946 seconds
cpu usage
1.027623979
max memory
3.6847616E7
stage attributes
key
value
output-size
30976
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: !facminus : [a * a] --> a !facplus : [a * a] --> a !factimes : [a * a] --> a 0 : [] --> a 1 : [] --> a 2 : [] --> a D : [a] --> a cons : [c * d] --> d constant : [] --> a div : [a * a] --> a false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d ln : [a] --> a map : [c -> c * d] --> d minus : [a] --> a nil : [] --> d pow : [a * a] --> a t : [] --> a true : [] --> b Rules: D(t) => 1 D(constant) => 0 D(!facplus(x, y)) => !facplus(D(x), D(y)) D(!factimes(x, y)) => !facplus(!factimes(y, D(x)), !factimes(x, D(y))) D(!facminus(x, y)) => !facminus(D(x), D(y)) D(minus(x)) => minus(D(x)) D(div(x, y)) => !facminus(div(D(x), y), div(!factimes(x, D(y)), pow(y, 2))) D(ln(x)) => div(D(x), x) D(pow(x, y)) => !facplus(!factimes(!factimes(y, pow(x, !facminus(y, 1))), D(x)), !factimes(!factimes(pow(x, y), ln(x)), D(y))) map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, y)) filter(f, nil) => nil filter(f, cons(x, y)) => filter2(f x, f, x, y) filter2(true, f, x, y) => cons(x, filter(f, y)) filter2(false, f, x, 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]): D(t) >? 1 D(constant) >? 0 D(!facplus(X, Y)) >? !facplus(D(X), D(Y)) D(!factimes(X, Y)) >? !facplus(!factimes(Y, D(X)), !factimes(X, D(Y))) D(!facminus(X, Y)) >? !facminus(D(X), D(Y)) D(minus(X)) >? minus(D(X)) D(div(X, Y)) >? !facminus(div(D(X), Y), div(!factimes(X, D(Y)), pow(Y, 2))) D(ln(X)) >? div(D(X), X) D(pow(X, Y)) >? !facplus(!factimes(!factimes(Y, pow(X, !facminus(Y, 1))), D(X)), !factimes(!factimes(pow(X, Y), ln(X)), D(Y))) map(F, nil) >? nil map(F, cons(X, Y)) >? cons(F X, map(F, Y)) filter(F, nil) >? nil filter(F, cons(X, Y)) >? filter2(F X, F, X, Y) filter2(true, F, X, Y) >? cons(X, filter(F, Y)) filter2(false, F, X, Y) >? filter(F, Y) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ [[1]] = _|_ [[2]] = _|_ [[@_{o -> o}(x_1, x_2)]] = @_{o -> o}(x_2, x_1) [[filter(x_1, x_2)]] = filter(x_2, x_1) [[filter2(x_1, x_2, x_3, x_4)]] = filter2(x_4, x_2, x_1, x_3) [[nil]] = _|_ We choose Lex = {@_{o -> o}, filter, filter2} and Mul = {!facminus, !facplus, !factimes, D, cons, constant, div, false, ln, map, minus, pow, t, true}, and the following precedence: false > D > !facplus > !factimes > div > pow > ln > map > !facminus > t > constant > @_{o -> o} = filter = filter2 > cons > minus > true Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: D(t) >= _|_ D(constant) >= _|_ D(!facplus(X, Y)) > !facplus(D(X), D(Y)) D(!factimes(X, Y)) >= !facplus(!factimes(Y, D(X)), !factimes(X, D(Y))) D(!facminus(X, Y)) > !facminus(D(X), D(Y)) D(minus(X)) >= minus(D(X)) D(div(X, Y)) >= !facminus(div(D(X), Y), div(!factimes(X, D(Y)), pow(Y, _|_))) D(ln(X)) >= div(D(X), X) D(pow(X, Y)) > !facplus(!factimes(!factimes(Y, pow(X, !facminus(Y, _|_))), D(X)), !factimes(!factimes(pow(X, Y), ln(X)), D(Y))) map(F, _|_) >= _|_
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