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Higher Order Rewriting Union Beta pair #487093733
details
property
value
status
complete
benchmark
h46.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n145.star.cs.uiowa.edu
space
Hamana_Kikuchi_18
run statistics
property
value
solver
Wanda 2.2a
configuration
default
runtime (wallclock)
0.185558 seconds
cpu usage
0.18594
user time
0.155746
system time
0.030194
max virtual memory
113188.0
max residence set size
6648.0
stage attributes
key
value
starexec-result
YES
output
YES We consider the system theBenchmark. Alphabet: 0 : [] --> nat plus : [] --> nat -> nat -> nat rec : [] --> nat -> nat -> (nat -> nat -> nat) -> nat s : [] --> nat -> nat succ : [] --> nat -> nat -> nat Rules: rec 0 x (/\y./\z.f y z) => x rec (s x) y (/\z./\u.f z u) => f x (rec x y (/\v./\w.f v w)) succ x y => s y plus x y => rec x y (/\z./\u.succ z u) 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. 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: Alphabet: 0 : [] --> nat plus : [nat * nat] --> nat rec : [nat * nat * nat -> nat -> nat] --> nat s : [nat] --> nat succ : [nat * nat] --> nat ~AP1 : [nat -> nat -> nat * nat] --> nat -> nat Rules: rec(0, X, /\x./\y.~AP1(F, x) y) => X rec(s(X), Y, /\x./\y.~AP1(F, x) y) => ~AP1(F, X) rec(X, Y, /\z./\u.~AP1(F, z) u) succ(X, Y) => s(Y) plus(X, Y) => rec(X, Y, /\x./\y.succ(x, y)) rec(0, X, /\x./\y.plus(x, y)) => X rec(0, X, /\x./\y.succ(x, y)) => X rec(s(X), Y, /\x./\y.plus(x, y)) => plus(X, rec(X, Y, /\z./\u.plus(z, u))) rec(s(X), Y, /\x./\y.succ(x, y)) => succ(X, rec(X, Y, /\z./\u.succ(z, u))) ~AP1(F, X) => F X 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. Additionally, we can remove some (now-)redundant rules. This gives: Alphabet: 0 : [] --> nat plus : [nat * nat] --> nat rec : [nat * nat * nat -> nat -> nat] --> nat s : [nat] --> nat succ : [nat * nat] --> nat Rules: rec(0, X, /\x./\y.Y(x, y)) => X rec(s(X), Y, /\x./\y.Z(x, y)) => Z(X, rec(X, Y, /\z./\u.Z(z, u))) succ(X, Y) => s(Y) plus(X, Y) => rec(X, Y, /\x./\y.succ(x, y)) We use rule removal, following [Kop12, Theorem 2.23]. This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): rec(0, X, /\x./\y.Y(x, y)) >? X rec(s(X), Y, /\x./\y.Z(x, y)) >? Z(X, rec(X, Y, /\z./\u.Z(z, u))) succ(X, Y) >? s(Y) plus(X, Y) >? rec(X, Y, /\x./\y.succ(x, y)) We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {} and Mul = {0, plus, rec, s, succ}, and the following precedence: 0 > plus > rec > succ > s With these choices, we have: 1] rec(0, X, /\x./\y.Y(x, y)) >= X because [2], by (Star) 2] rec*(0, X, /\x./\y.Y(x, y)) >= X because [3], by (Select) 3] X >= X by (Meta) 4] rec(s(X), Y, /\x./\y.Z(x, y)) > Z(X, rec(X, Y, /\x./\y.Z(x, y))) because [5], by definition 5] rec*(s(X), Y, /\x./\y.Z(x, y)) >= Z(X, rec(X, Y, /\x./\y.Z(x, y))) because [6], by (Select) 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) 7] rec*(s(X), Y, /\x./\y.Z(x, y)) >= X because [8], by (Select) 8] s(X) >= X because [9], by (Star) 9] s*(X) >= X because [10], by (Select) 10] X >= X by (Meta) 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) 12] s(X) > X because [13], by definition 13] s*(X) >= X because [10], by (Select) 14] Y >= Y by (Meta) 15] /\x./\z.Z(x, z) >= /\x./\z.Z(x, z) because [16], by (Abs) 16] /\z.Z(y, z) >= /\z.Z(y, z) because [17], by (Abs) 17] Z(y, x) >= Z(y, x) because [18] and [19], by (Meta) 18] y >= y by (Var) 19] x >= x by (Var) 20] succ(X, Y) >= s(Y) because [21], by (Star) 21] succ*(X, Y) >= s(Y) because succ > s and [22], by (Copy) 22] succ*(X, Y) >= Y because [23], by (Select) 23] Y >= Y by (Meta)
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