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Higher Order Rewriting Union Beta pair #487093695
details
property
value
status
complete
benchmark
h49.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n141.star.cs.uiowa.edu
space
Hamana_Kikuchi_18
run statistics
property
value
solver
Wanda 2.2a
configuration
default
runtime (wallclock)
20.2507 seconds
cpu usage
20.2521
user time
19.8468
system time
0.405328
max virtual memory
592772.0
max residence set size
246996.0
stage attributes
key
value
starexec-result
YES
output
YES We consider the system theBenchmark. Alphabet: 0 : [] --> nat mult : [nat * nat] --> nat plus : [nat * nat] --> nat plus3 : [nat] --> nat -> nat -> nat rec : [nat * nat * nat -> nat -> nat] --> nat s : [nat] --> nat succ2 : [] --> nat -> nat -> nat xap : [nat -> nat -> nat * nat] --> nat -> nat yap : [nat -> nat * nat] --> nat Rules: rec(0, x, /\y./\z.yap(xap(f, y), z)) => x rec(s(x), y, /\z./\u.yap(xap(f, z), u)) => yap(xap(f, x), rec(x, y, /\v./\w.yap(xap(f, v), w))) succ2 x y => s(y) plus(x, y) => rec(x, y, succ2) plus3(x) y z => plus(x, plus(y, z)) mult(x, y) => rec(x, 0, plus3(y)) xap(f, x) => f x yap(f, x) => f x This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 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: Alphabet: 0 : [] --> nat mult : [nat * nat] --> nat plus : [nat * nat] --> nat plus3 : [nat] --> nat -> nat -> nat rec : [nat * nat * nat -> nat -> nat] --> nat s : [nat] --> nat succ2 : [] --> nat -> nat -> nat yap : [nat -> nat * nat] --> nat Rules: rec(0, X, /\x./\y.yap(F(x), y)) => X rec(s(X), Y, /\x./\y.yap(F(x), y)) => yap(F(X), rec(X, Y, /\z./\u.yap(F(z), u))) succ2 X Y => s(Y) plus(X, Y) => rec(X, Y, succ2) plus3(X) Y Z => plus(X, plus(Y, Z)) mult(X, Y) => rec(X, 0, plus3(Y)) yap(F, X) => F X 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.yap(F(x), y)) >? X rec(s(X), Y, /\x./\y.yap(F(x), y)) >? yap(F(X), rec(X, Y, /\z./\u.yap(F(z), u))) succ2 X Y >? s(Y) plus(X, Y) >? rec(X, Y, succ2) plus3(X) Y Z >? plus(X, plus(Y, Z)) mult(X, Y) >? rec(X, 0, plus3(Y)) yap(F, X) >? F X We use a recursive path ordering as defined in [Kop12, Chapter 5]. Argument functions: [[0]] = _|_ We choose Lex = {} and Mul = {@_{o -> o -> o}, @_{o -> o}, mult, plus, plus3, rec, s, succ2, yap}, and the following precedence: @_{o -> o -> o} > yap > s > mult = plus3 > plus > @_{o -> o} > succ2 > rec Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: rec(_|_, X, /\x./\y.yap(F(x), y)) > X rec(s(X), Y, /\x./\y.yap(F(x), y)) >= yap(F(X), rec(X, Y, /\x./\y.yap(F(x), y))) @_{o -> o}(@_{o -> o -> o}(succ2, X), Y) >= s(Y) plus(X, Y) >= rec(X, Y, succ2) @_{o -> o}(@_{o -> o -> o}(plus3(X), Y), Z) > plus(X, plus(Y, Z)) mult(X, Y) >= rec(X, _|_, plus3(Y)) yap(F, X) >= @_{o -> o}(F, X) With these choices, we have: 1] rec(_|_, X, /\x./\y.yap(F(x), y)) > X because [2], by definition 2] rec*(_|_, X, /\x./\y.yap(F(x), y)) >= X because [3], by (Select) 3] X >= X by (Meta) 4] rec(s(X), Y, /\x./\y.yap(F(x), y)) >= yap(F(X), rec(X, Y, /\x./\y.yap(F(x), y))) because [5], by (Star) 5] rec*(s(X), Y, /\x./\y.yap(F(x), y)) >= yap(F(X), rec(X, Y, /\x./\y.yap(F(x), y))) because [6], by (Select) 6] yap(F(rec*(s(X), Y, /\x./\y.yap(F(x), y))), rec*(s(X), Y, /\z./\u.yap(F(z), u))) >= yap(F(X), rec(X, Y, /\x./\y.yap(F(x), y))) because yap in Mul, [7] and [12], by (Fun) 7] F(rec*(s(X), Y, /\x./\y.yap(F(x), y))) >= F(X) because [8], by (Meta) 8] rec*(s(X), Y, /\x./\y.yap(F(x), y)) >= X because [9], by (Select) 9] s(X) >= X because [10], by (Star) 10] s*(X) >= X because [11], by (Select) 11] X >= X by (Meta) 12] rec*(s(X), Y, /\x./\y.yap(F(x), y)) >= rec(X, Y, /\x./\y.yap(F(x), y)) because rec in Mul, [13], [15] and [16], by (Stat) 13] s(X) > X because [14], by definition 14] s*(X) >= X because [11], by (Select) 15] Y >= Y by (Meta) 16] /\x./\z.yap(F(x), z) >= /\x./\z.yap(F(x), z) because [17], by (Abs)
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