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HRS union beta 16688 pair #381734902
details
property
value
status
complete
benchmark
Applicative_first_order_05__08.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n012.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
0.275604963303 seconds
cpu usage
0.272903044
max memory
1.275904E7
stage attributes
key
value
output-size
10942
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 D : [a] --> a cons : [c * d] --> d constant : [] --> a false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d map : [c -> c * d] --> d nil : [] --> d 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)) 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)) 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]] = _|_ [[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 = {filter, filter2} and Mul = {!facminus, !facplus, !factimes, @_{o -> o}, D, cons, constant, false, map, t, true}, and the following precedence: !factimes = D > !facminus > !facplus > constant > false > map > filter = filter2 > @_{o -> o} > cons > t > 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)) map(F, _|_) >= _|_ map(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), map(F, Y)) filter(F, _|_) >= _|_ filter(F, cons(X, Y)) >= filter2(@_{o -> o}(F, X), F, X, Y) filter2(true, F, X, Y) >= cons(X, filter(F, Y)) filter2(false, F, X, Y) >= filter(F, Y) With these choices, we have: 1] D(t) >= _|_ by (Bot) 2] D(constant) >= _|_ by (Bot) 3] D(!facplus(X, Y)) > !facplus(D(X), D(Y)) because [4], by definition 4] D*(!facplus(X, Y)) >= !facplus(D(X), D(Y)) because D > !facplus, [5] and [9], by (Copy) 5] D*(!facplus(X, Y)) >= D(X) because D in Mul and [6], by (Stat) 6] !facplus(X, Y) > X because [7], by definition 7] !facplus*(X, Y) >= X because [8], by (Select) 8] X >= X by (Meta) 9] D*(!facplus(X, Y)) >= D(Y) because D in Mul and [10], by (Stat)
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