Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270389
details
property
value
status
complete
benchmark
h13.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n073.star.cs.uiowa.edu
space
Hamana_Kikuchi_18
run statistics
property
value
solver
Wanda 2.1c
configuration
default
runtime (wallclock)
5.79816 seconds
cpu usage
5.79787
user time
5.67991
system time
0.117959
max virtual memory
187240.0
max residence set size
70820.0
stage attributes
key
value
starexec-result
YES
output
5.72/5.78 YES 5.72/5.79 We consider the system theBenchmark. 5.72/5.79 5.72/5.79 Alphabet: 5.72/5.79 5.72/5.79 0 : [] --> a 5.72/5.79 rec : [a -> b -> b * b * a] --> b 5.72/5.79 s : [a] --> a 5.72/5.79 xap : [a -> b -> b * a] --> b -> b 5.72/5.79 yap : [b -> b * b] --> b 5.72/5.79 5.72/5.79 Rules: 5.72/5.79 5.72/5.79 rec(/\x./\y.yap(xap(f, x), y), z, 0) => z 5.72/5.79 rec(/\x./\y.yap(xap(f, x), y), z, s(u)) => yap(xap(f, s(u)), rec(/\v./\w.yap(xap(f, v), w), z, u)) 5.72/5.79 xap(f, x) => f x 5.72/5.79 yap(f, x) => f x 5.72/5.79 5.72/5.79 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 5.72/5.79 5.72/5.79 Symbol xap is an encoding for application that is only used in innocuous ways. We can simplify the program (without losing non-termination) by removing it. This gives: 5.72/5.79 5.72/5.79 Alphabet: 5.72/5.79 5.72/5.79 0 : [] --> a 5.72/5.79 rec : [a -> b -> b * b * a] --> b 5.72/5.79 s : [a] --> a 5.72/5.79 yap : [b -> b * b] --> b 5.72/5.79 5.72/5.79 Rules: 5.72/5.79 5.72/5.79 rec(/\x./\y.yap(F(x), y), X, 0) => X 5.72/5.79 rec(/\x./\y.yap(F(x), y), X, s(Y)) => yap(F(s(Y)), rec(/\z./\u.yap(F(z), u), X, Y)) 5.72/5.79 yap(F, X) => F X 5.72/5.79 5.72/5.79 We use rule removal, following [Kop12, Theorem 2.23]. 5.72/5.79 5.72/5.79 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 5.72/5.79 5.72/5.79 rec(/\x./\y.yap(F(x), y), X, 0) >? X 5.72/5.79 rec(/\x./\y.yap(F(x), y), X, s(Y)) >? yap(F(s(Y)), rec(/\z./\u.yap(F(z), u), X, Y)) 5.72/5.79 yap(F, X) >? F X 5.72/5.79 5.72/5.79 We use a recursive path ordering as defined in [Kop12, Chapter 5]. 5.72/5.79 5.72/5.79 We choose Lex = {} and Mul = {0, @_{o -> o}, rec, s, yap}, and the following precedence: 0 > rec > s > yap > @_{o -> o} 5.72/5.79 5.72/5.79 With these choices, we have: 5.72/5.79 5.72/5.79 1] rec(/\x./\y.yap(F(x), y), X, 0) > X because [2], by definition 5.72/5.79 2] rec*(/\x./\y.yap(F(x), y), X, 0) >= X because [3], by (Select) 5.72/5.79 3] X >= X by (Meta) 5.72/5.79 5.72/5.79 4] rec(/\x./\y.yap(F(x), y), X, s(Y)) >= yap(F(s(Y)), rec(/\x./\y.yap(F(x), y), X, Y)) because [5], by (Star) 5.72/5.79 5] rec*(/\x./\y.yap(F(x), y), X, s(Y)) >= yap(F(s(Y)), rec(/\x./\y.yap(F(x), y), X, Y)) because rec > yap, [6] and [14], by (Copy) 5.72/5.79 6] rec*(/\x./\y.yap(F(x), y), X, s(Y)) >= F(s(Y)) because [7], by (Select) 5.72/5.79 7] /\x.yap(F(rec*(/\y./\z.yap(F(y), z), X, s(Y))), x) >= F(s(Y)) because [8], by (Eta)[Kop13:2] 5.72/5.79 8] F(rec*(/\x./\y.yap(F(x), y), X, s(Y))) >= F(s(Y)) because [9], by (Meta) 5.72/5.79 9] rec*(/\x./\y.yap(F(x), y), X, s(Y)) >= s(Y) because rec > s and [10], by (Copy) 5.72/5.79 10] rec*(/\x./\y.yap(F(x), y), X, s(Y)) >= Y because [11], by (Select) 5.72/5.79 11] s(Y) >= Y because [12], by (Star) 5.72/5.79 12] s*(Y) >= Y because [13], by (Select) 5.72/5.79 13] Y >= Y by (Meta) 5.72/5.79 14] rec*(/\x./\y.yap(F(x), y), X, s(Y)) >= rec(/\x./\y.yap(F(x), y), X, Y) because rec in Mul, [15], [21] and [22], by (Stat) 5.72/5.79 15] /\x./\z.yap(F(x), z) >= /\x./\z.yap(F(x), z) because [16], by (Abs) 5.72/5.79 16] /\z.yap(F(y), z) >= /\z.yap(F(y), z) because [17], by (Abs) 5.72/5.79 17] yap(F(y), x) >= yap(F(y), x) because yap in Mul, [18] and [20], by (Fun) 5.72/5.79 18] F(y) >= F(y) because [19], by (Meta) 5.72/5.79 19] y >= y by (Var) 5.72/5.79 20] x >= x by (Var) 5.72/5.79 21] X >= X by (Meta) 5.72/5.79 22] s(Y) > Y because [23], by definition 5.72/5.79 23] s*(Y) >= Y because [13], by (Select) 5.72/5.79 5.72/5.79 24] yap(F, X) > @_{o -> o}(F, X) because [25], by definition 5.72/5.79 25] yap*(F, X) >= @_{o -> o}(F, X) because yap > @_{o -> o}, [26] and [28], by (Copy) 5.72/5.79 26] yap*(F, X) >= F because [27], by (Select) 5.72/5.79 27] F >= F by (Meta) 5.72/5.79 28] yap*(F, X) >= X because [29], by (Select) 5.72/5.79 29] X >= X by (Meta) 5.72/5.79 5.72/5.79 We can thus remove the following rules: 5.72/5.79 5.72/5.79 rec(/\x./\y.yap(F(x), y), X, 0) => X 5.72/5.79 yap(F, X) => F X 5.72/5.79 5.72/5.79 We use rule removal, following [Kop12, Theorem 2.23]. 5.72/5.79 5.72/5.79 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 5.72/5.79 5.72/5.79 rec(/\x./\y.yap(F(x), y), X, s(Y)) >? yap(F(s(Y)), rec(/\z./\u.yap(F(z), u), X, Y)) 5.72/5.79 5.72/5.79 We orient these requirements with a polynomial interpretation in the natural numbers. 5.72/5.79 5.72/5.79 The following interpretation satisfies the requirements: 5.72/5.79 5.72/5.79 rec = \G0y1y2.y1 + y2 + G0(0,0) + 2y2y2G0(y2,y2) + 2G0(y2,y1) 5.72/5.79 s = \y0.3 + 3y0 5.72/5.79 yap = \G0y1.y1 + 2G0(0) 5.72/5.79
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10