Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270481
details
property
value
status
complete
benchmark
h59.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n160.star.cs.uiowa.edu
space
Hamana_Kikuchi_18
run statistics
property
value
solver
Wanda 2.1c
configuration
default
runtime (wallclock)
0.0973579 seconds
cpu usage
0.097629
user time
0.076872
system time
0.020757
max virtual memory
113176.0
max residence set size
4816.0
stage attributes
key
value
starexec-result
YES
output
0.00/0.09 YES 0.00/0.09 We consider the system theBenchmark. 0.00/0.09 0.00/0.09 Alphabet: 0.00/0.09 0.00/0.09 0 : [] --> nat 0.00/0.09 rec : [] --> nat -> a -> (nat -> a -> a) -> a 0.00/0.09 s : [] --> nat -> nat 0.00/0.09 0.00/0.09 Rules: 0.00/0.09 0.00/0.09 rec 0 x (/\y./\z.f y z) => x 0.00/0.09 rec (s x) y (/\z./\u.f z u) => f x (rec x y (/\v./\w.f v w)) 0.00/0.09 0.00/0.09 Using the transformations described in [Kop11], this system can be brought in a form without leading free variables in the left-hand side, and where the left-hand side of a variable is always a functional term or application headed by a functional term. 0.00/0.09 0.00/0.09 We now transform the resulting AFS into an AFSM by replacing all free variables by meta-variables (with arity 0). This leads to the following AFSM: 0.00/0.09 0.00/0.09 Alphabet: 0.00/0.09 0.00/0.09 0 : [] --> nat 0.00/0.09 rec : [nat * a * nat -> a -> a] --> a 0.00/0.09 s : [nat] --> nat 0.00/0.09 ~AP1 : [nat -> a -> a * nat] --> a -> a 0.00/0.09 0.00/0.09 Rules: 0.00/0.09 0.00/0.09 rec(0, X, /\x./\y.~AP1(F, x) y) => X 0.00/0.09 rec(s(X), Y, /\x./\y.~AP1(F, x) y) => ~AP1(F, X) rec(X, Y, /\z./\u.~AP1(F, z) u) 0.00/0.09 ~AP1(F, X) => F X 0.00/0.09 0.00/0.09 Symbol ~AP1 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: 0.00/0.09 0.00/0.09 Alphabet: 0.00/0.09 0.00/0.09 0 : [] --> nat 0.00/0.09 rec : [nat * a * nat -> a -> a] --> a 0.00/0.09 s : [nat] --> nat 0.00/0.09 0.00/0.09 Rules: 0.00/0.09 0.00/0.09 rec(0, X, /\x./\y.Y(x, y)) => X 0.00/0.09 rec(s(X), Y, /\x./\y.Z(x, y)) => Z(X, rec(X, Y, /\z./\u.Z(z, u))) 0.00/0.09 0.00/0.09 We use rule removal, following [Kop12, Theorem 2.23]. 0.00/0.09 0.00/0.09 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.00/0.09 0.00/0.09 rec(0, X, /\x./\y.Y(x, y)) >? X 0.00/0.09 rec(s(X), Y, /\x./\y.Z(x, y)) >? Z(X, rec(X, Y, /\z./\u.Z(z, u))) 0.00/0.09 0.00/0.09 We use a recursive path ordering as defined in [Kop12, Chapter 5]. 0.00/0.09 0.00/0.09 We choose Lex = {} and Mul = {0, rec, s}, and the following precedence: 0 > rec > s 0.00/0.09 0.00/0.09 With these choices, we have: 0.00/0.09 0.00/0.09 1] rec(0, X, /\x./\y.Y(x, y)) >= X because [2], by (Star) 0.00/0.09 2] rec*(0, X, /\x./\y.Y(x, y)) >= X because [3], by (Select) 0.00/0.09 3] X >= X by (Meta) 0.00/0.09 0.00/0.09 4] rec(s(X), Y, /\x./\y.Z(x, y)) > Z(X, rec(X, Y, /\x./\y.Z(x, y))) because [5], by definition 0.00/0.09 5] rec*(s(X), Y, /\x./\y.Z(x, y)) >= Z(X, rec(X, Y, /\x./\y.Z(x, y))) because [6], by (Select) 0.00/0.09 6] Z(rec*(s(X), Y, /\x./\y.Z(x, y)), rec*(s(X), Y, /\z./\u.Z(z, u))) >= Z(X, rec(X, Y, /\x./\y.Z(x, y))) because [7] and [11], by (Meta) 0.00/0.09 7] rec*(s(X), Y, /\x./\y.Z(x, y)) >= X because [8], by (Select) 0.00/0.09 8] s(X) >= X because [9], by (Star) 0.00/0.09 9] s*(X) >= X because [10], by (Select) 0.00/0.09 10] X >= X by (Meta) 0.00/0.09 11] rec*(s(X), Y, /\x./\y.Z(x, y)) >= rec(X, Y, /\x./\y.Z(x, y)) because rec in Mul, [12], [14] and [15], by (Stat) 0.00/0.09 12] s(X) > X because [13], by definition 0.00/0.09 13] s*(X) >= X because [10], by (Select) 0.00/0.09 14] Y >= Y by (Meta) 0.00/0.09 15] /\x./\z.Z(x, z) >= /\x./\z.Z(x, z) because [16], by (Abs) 0.00/0.09 16] /\z.Z(y, z) >= /\z.Z(y, z) because [17], by (Abs) 0.00/0.09 17] Z(y, x) >= Z(y, x) because [18] and [19], by (Meta) 0.00/0.09 18] y >= y by (Var) 0.00/0.09 19] x >= x by (Var) 0.00/0.09 0.00/0.09 We can thus remove the following rules: 0.00/0.09 0.00/0.09 rec(s(X), Y, /\x./\y.Z(x, y)) => Z(X, rec(X, Y, /\z./\u.Z(z, u))) 0.00/0.09 0.00/0.09 We use rule removal, following [Kop12, Theorem 2.23]. 0.00/0.09 0.00/0.09 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.00/0.09 0.00/0.09 rec(0, X, /\x./\y.Y(x, y)) >? X 0.00/0.09 0.00/0.09 We orient these requirements with a polynomial interpretation in the natural numbers. 0.00/0.09 0.00/0.09 The following interpretation satisfies the requirements: 0.00/0.09 0.00/0.09 0 = 3 0.00/0.09 rec = \y0y1G2.3 + y0 + y1 + G2(0,0) 0.00/0.09 0.00/0.09 Using this interpretation, the requirements translate to: 0.00/0.09 0.00/0.09 [[rec(0, _x0, /\x./\y._x1(x, y))]] = 6 + x0 + F1(0,0) > x0 = [[_x0]] 0.00/0.09 0.00/0.09 We can thus remove the following rules:
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