details

property	value
status	complete
benchmark	01GoedelT.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n137.star.cs.uiowa.edu
space	Blanqui_15

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	0.183529 seconds
cpu usage	0.18376
user time	0.157506
system time	0.026254
max virtual memory	113188.0
max residence set size	7248.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: rec : [] --> N -> a -> (N -> a -> a) -> a s : [] --> N -> N z : [] --> N Rules: rec z x (/\y.f y) => x rec (s x) y (/\u.f u) => f x (rec x y (/\v.f v)) 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: rec : [N * a * N -> a -> a] --> a s : [N] --> N z : [] --> N ~AP1 : [N -> a -> a * N] --> a -> a Rules: rec(z, X, /\x.~AP1(F, x)) => X rec(s(X), Y, /\x.~AP1(F, x)) => ~AP1(F, X) rec(X, Y, /\y.~AP1(F, y)) ~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. This gives: Alphabet: rec : [N * a * N -> a -> a] --> a s : [N] --> N z : [] --> N Rules: rec(z, X, /\x.F(x)) => X rec(s(X), Y, /\x.F(x)) => F(X) rec(X, Y, /\y.F(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(z, X, /\x.F(x)) >? X rec(s(X), Y, /\x.F(x)) >? F(X) rec(X, Y, /\y.F(y)) We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {} and Mul = {@_{o -> o}, rec, s, z}, and the following precedence: rec > @_{o -> o} > s > z With these choices, we have: 1] rec(z, X, /\x.F(x)) > X because [2], by definition 2] rec*(z, X, /\x.F(x)) >= X because [3], by (Select) 3] X >= X by (Meta) 4] rec(s(X), Y, /\x.F(x)) >= @_{o -> o}(F(X), rec(X, Y, /\x.F(x))) because [5], by (Star) 5] rec*(s(X), Y, /\x.F(x)) >= @_{o -> o}(F(X), rec(X, Y, /\x.F(x))) because rec > @_{o -> o}, [6] and [12], by (Copy) 6] rec*(s(X), Y, /\x.F(x)) >= F(X) because [7], by (Select) 7] F(rec*(s(X), Y, /\x.F(x))) >= F(X) because [8], by (Meta) 8] rec*(s(X), Y, /\x.F(x)) >= 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.F(x)) >= rec(X, Y, /\x.F(x)) 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] /\y.F(y) >= /\y.F(y) because [17], by (Abs) 17] F(x) >= F(x) because [18], by (Meta) 18] x >= x by (Var) We can thus remove the following rules: rec(z, X, /\x.F(x)) => 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(s(X), Y, /\x.F(x)) >? F(X) rec(X, Y, /\y.F(y)) We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {rec} and Mul = {@_{o -> o}, s}, and the following precedence: rec > @_{o -> o} > s With these choices, we have: 1] rec(s(X), Y, /\x.F(x)) > @_{o -> o}(F(X), rec(X, Y, /\x.F(x))) because [2], by definition 2] rec*(s(X), Y, /\x.F(x)) >= @_{o -> o}(F(X), rec(X, Y, /\x.F(x))) because rec > @_{o -> o}, [3] and [9], by (Copy) 3] rec*(s(X), Y, /\x.F(x)) >= F(X) because [4], by (Select) 4] F(rec*(s(X), Y, /\x.F(x))) >= F(X) because [5], by (Meta) 5] rec*(s(X), Y, /\x.F(x)) >= X because [6], by (Select) 6] s(X) >= X because [7], by (Star) 7] s*(X) >= X because [8], by (Select) popout

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all output return to Higher Order Rewriting Union Beta