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HRS union beta 16688 pair #381734626
details
property
value
status
complete
benchmark
reverse.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n191.star.cs.uiowa.edu
space
Mixed_HO_10
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
0.369218111038 seconds
cpu usage
0.318638476
max memory
1.4675968E7
stage attributes
key
value
output-size
11463
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: app : [list * list] --> list cons : [nat * list] --> list foldl : [list -> nat -> list * list * list] --> list iconsc : [] --> list -> nat -> list nil : [] --> list reverse : [list] --> list reverse1 : [list] --> list Rules: app(nil, x) => x app(cons(x, y), z) => cons(x, app(y, z)) foldl(f, x, nil) => x foldl(f, x, cons(y, z)) => foldl(f, f x y, z) iconsc => /\x./\y.cons(y, x) reverse(x) => foldl(iconsc, nil, x) reverse1(x) => foldl(/\y./\z.app(cons(z, nil), y), nil, x) 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]): app(nil, X) >? X app(cons(X, Y), Z) >? cons(X, app(Y, Z)) foldl(F, X, nil) >? X foldl(F, X, cons(Y, Z)) >? foldl(F, F X Y, Z) iconsc >? /\x./\y.cons(y, x) reverse(X) >? foldl(iconsc, nil, X) reverse1(X) >? foldl(/\x./\y.app(cons(y, nil), x), nil, X) about to try horpo We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[foldl(x_1, x_2, x_3)]] = foldl(x_3, x_1, x_2) [[nil]] = _|_ We choose Lex = {foldl} and Mul = {@_{o -> o -> o}, @_{o -> o}, app, cons, iconsc, reverse, reverse1}, and the following precedence: reverse > iconsc > reverse1 > app > cons > foldl > @_{o -> o} > @_{o -> o -> o} Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: app(_|_, X) >= X app(cons(X, Y), Z) > cons(X, app(Y, Z)) foldl(F, X, _|_) >= X foldl(F, X, cons(Y, Z)) >= foldl(F, @_{o -> o}(@_{o -> o -> o}(F, X), Y), Z) iconsc > /\x./\y.cons(y, x) reverse(X) >= foldl(iconsc, _|_, X) reverse1(X) >= foldl(/\x./\y.app(cons(y, _|_), x), _|_, X) With these choices, we have: 1] app(_|_, X) >= X because [2], by (Star) 2] app*(_|_, X) >= X because [3], by (Select) 3] X >= X by (Meta) 4] app(cons(X, Y), Z) > cons(X, app(Y, Z)) because [5], by definition 5] app*(cons(X, Y), Z) >= cons(X, app(Y, Z)) because app > cons, [6] and [10], by (Copy) 6] app*(cons(X, Y), Z) >= X because [7], by (Select) 7] cons(X, Y) >= X because [8], by (Star) 8] cons*(X, Y) >= X because [9], by (Select) 9] X >= X by (Meta) 10] app*(cons(X, Y), Z) >= app(Y, Z) because app in Mul, [11] and [14], by (Stat) 11] cons(X, Y) > Y because [12], by definition 12] cons*(X, Y) >= Y because [13], by (Select) 13] Y >= Y by (Meta) 14] Z >= Z by (Meta) 15] foldl(F, X, _|_) >= X because [16], by (Star) 16] foldl*(F, X, _|_) >= X because [17], by (Select) 17] X >= X by (Meta) 18] foldl(F, X, cons(Y, Z)) >= foldl(F, @_{o -> o}(@_{o -> o -> o}(F, X), Y), Z) because [19], by (Star) 19] foldl*(F, X, cons(Y, Z)) >= foldl(F, @_{o -> o}(@_{o -> o -> o}(F, X), Y), Z) because [20], [23], [25] and [33], by (Stat) 20] cons(Y, Z) > Z because [21], by definition 21] cons*(Y, Z) >= Z because [22], by (Select) 22] Z >= Z by (Meta) 23] foldl*(F, X, cons(Y, Z)) >= F because [24], by (Select) 24] F >= F by (Meta) 25] foldl*(F, X, cons(Y, Z)) >= @_{o -> o}(@_{o -> o -> o}(F, X), Y) because foldl > @_{o -> o}, [26] and [29], by (Copy) 26] foldl*(F, X, cons(Y, Z)) >= @_{o -> o -> o}(F, X) because foldl > @_{o -> o -> o}, [23] and [27], by (Copy) 27] foldl*(F, X, cons(Y, Z)) >= X because [28], by (Select) 28] X >= X by (Meta) 29] foldl*(F, X, cons(Y, Z)) >= Y because [30], by (Select) 30] cons(Y, Z) >= Y because [31], by (Star) 31] cons*(Y, Z) >= Y because [32], by (Select)
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