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Higher Order Rewriting Union Beta pair #487093779
details
property
value
status
complete
benchmark
kop12thesis_ex2.11.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n146.star.cs.uiowa.edu
space
Kop_13
run statistics
property
value
solver
Wanda 2.2a
configuration
default
runtime (wallclock)
0.349735 seconds
cpu usage
0.350263
user time
0.312466
system time
0.037797
max virtual memory
113188.0
max residence set size
11656.0
stage attributes
key
value
starexec-result
YES
output
YES We consider the system theBenchmark. Alphabet: cons : [nat * list] --> list emap : [nat -> nat * list] --> list nil : [] --> list twice : [nat -> nat] --> nat -> nat Rules: emap(f, nil) => nil emap(f, cons(x, y)) => cons(f x, emap(/\z.twice(f) z, y)) twice(f) x => f (f 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]): emap(F, nil) >? nil emap(F, cons(X, Y)) >? cons(F X, emap(/\x.twice(F, x), Y)) twice(F, X) >? F (F X) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[emap(x_1, x_2)]] = emap(x_2, x_1) [[nil]] = _|_ We choose Lex = {emap} and Mul = {@_{o -> o}, cons, twice}, and the following precedence: emap > cons > twice > @_{o -> o} Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: emap(F, _|_) > _|_ emap(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), emap(/\x.twice(F, x), Y)) twice(F, X) >= @_{o -> o}(F, @_{o -> o}(F, X)) With these choices, we have: 1] emap(F, _|_) > _|_ because [2], by definition 2] emap*(F, _|_) >= _|_ by (Bot) 3] emap(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), emap(/\x.twice(F, x), Y)) because [4], by (Star) 4] emap*(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), emap(/\x.twice(F, x), Y)) because emap > cons, [5] and [12], by (Copy) 5] emap*(F, cons(X, Y)) >= @_{o -> o}(F, X) because emap > @_{o -> o}, [6] and [8], by (Copy) 6] emap*(F, cons(X, Y)) >= F because [7], by (Select) 7] F >= F by (Meta) 8] emap*(F, cons(X, Y)) >= X because [9], by (Select) 9] cons(X, Y) >= X because [10], by (Star) 10] cons*(X, Y) >= X because [11], by (Select) 11] X >= X by (Meta) 12] emap*(F, cons(X, Y)) >= emap(/\x.twice(F, x), Y) because [13], [16] and [21], by (Stat) 13] cons(X, Y) > Y because [14], by definition 14] cons*(X, Y) >= Y because [15], by (Select) 15] Y >= Y by (Meta) 16] emap*(F, cons(X, Y)) >= /\y.twice(F, y) because [17], by (F-Abs) 17] emap*(F, cons(X, Y), x) >= twice(F, x) because emap > twice, [18] and [19], by (Copy) 18] emap*(F, cons(X, Y), x) >= F because [7], by (Select) 19] emap*(F, cons(X, Y), x) >= x because [20], by (Select) 20] x >= x by (Var) 21] emap*(F, cons(X, Y)) >= Y because [22], by (Select) 22] cons(X, Y) >= Y because [14], by (Star) 23] twice(F, X) >= @_{o -> o}(F, @_{o -> o}(F, X)) because [24], by (Star) 24] twice*(F, X) >= @_{o -> o}(F, @_{o -> o}(F, X)) because twice > @_{o -> o}, [25] and [27], by (Copy) 25] twice*(F, X) >= F because [26], by (Select) 26] F >= F by (Meta) 27] twice*(F, X) >= @_{o -> o}(F, X) because twice > @_{o -> o}, [25] and [28], by (Copy) 28] twice*(F, X) >= X because [29], by (Select) 29] X >= X by (Meta) We can thus remove the following rules: emap(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]): emap(F, cons(X, Y)) >? cons(F X, emap(/\x.twice(F, x), Y)) twice(F, X) >? F (F X) We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[emap(x_1, x_2)]] = emap(x_2, x_1) We choose Lex = {emap} and Mul = {@_{o -> o}, cons, twice}, and the following precedence: emap > cons > twice > @_{o -> o} Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: emap(F, cons(X, Y)) >= cons(@_{o -> o}(F, X), emap(/\x.twice(F, x), Y)) twice(F, X) > @_{o -> o}(F, @_{o -> o}(F, X)) With these choices, we have:
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