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HRS union beta 16688 pair #381734923
details
property
value
status
complete
benchmark
Applicative_first_order_05__29.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n083.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
0.465509176254 seconds
cpu usage
0.461998363
max memory
1.8870272E7
stage attributes
key
value
output-size
19969
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: 0 : [] --> a ack : [a * a] --> a cons : [c * d] --> d false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d map : [c -> c * d] --> d nil : [] --> d succ : [a] --> a true : [] --> b Rules: ack(0, x) => succ(x) ack(succ(x), y) => ack(x, succ(0)) ack(succ(x), succ(y)) => ack(x, ack(succ(x), 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]): ack(0, X) >? succ(X) ack(succ(X), Y) >? ack(X, succ(0)) ack(succ(X), succ(Y)) >? ack(X, ack(succ(X), 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]] = _|_ [[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 = {ack, filter, filter2} and Mul = {@_{o -> o}, cons, false, map, succ, true}, and the following precedence: ack > false > filter = filter2 > map > @_{o -> o} > cons > succ > true Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: ack(_|_, X) >= succ(X) ack(succ(X), Y) >= ack(X, succ(_|_)) ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), 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] ack(_|_, X) >= succ(X) because [2], by (Star) 2] ack*(_|_, X) >= succ(X) because ack > succ and [3], by (Copy) 3] ack*(_|_, X) >= X because [4], by (Select) 4] X >= X by (Meta) 5] ack(succ(X), Y) >= ack(X, succ(_|_)) because [6], by (Star) 6] ack*(succ(X), Y) >= ack(X, succ(_|_)) because [7], [10] and [12], by (Stat) 7] succ(X) > X because [8], by definition 8] succ*(X) >= X because [9], by (Select) 9] X >= X by (Meta) 10] ack*(succ(X), Y) >= X because [11], by (Select) 11] succ(X) >= X because [8], by (Star) 12] ack*(succ(X), Y) >= succ(_|_) because ack > succ and [13], by (Copy) 13] ack*(succ(X), Y) >= _|_ by (Bot) 14] ack(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [15], by (Star) 15] ack*(succ(X), succ(Y)) >= ack(X, ack(succ(X), Y)) because [16], [19] and [21], by (Stat) 16] succ(X) > X because [17], by definition 17] succ*(X) >= X because [18], by (Select) 18] X >= X by (Meta) 19] ack*(succ(X), succ(Y)) >= X because [20], by (Select) 20] succ(X) >= X because [17], by (Star) 21] ack*(succ(X), succ(Y)) >= ack(succ(X), Y) because [22], [24], [27] and [28], by (Stat)
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