details

property	value
status	complete
benchmark	rec.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n147.star.cs.uiowa.edu
space	Mixed_HO_10

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	0.163182 seconds
cpu usage	0.163366
user time	0.144238
system time	0.019128
max virtual memory	113188.0
max residence set size	6116.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: 0 : [] --> nat rec : [nat * a * nat -> a -> a] --> a s : [nat] --> nat Rules: rec(0, x, f) => x rec(s(x), y, f) => f x rec(x, y, f) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 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, F) >? X rec(s(X), Y, F) >? F X rec(X, Y, F) We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {} and Mul = {0, @_{o -> o -> o}, @_{o -> o}, rec, s}, and the following precedence: 0 > @_{o -> o -> o} = rec > @_{o -> o} > s With these choices, we have: 1] rec(0, X, F) > X because [2], by definition 2] rec*(0, X, F) >= X because [3], by (Select) 3] X >= X by (Meta) 4] rec(s(X), Y, F) >= @_{o -> o}(@_{o -> o -> o}(F, X), rec(X, Y, F)) because [5], by (Star) 5] rec*(s(X), Y, F) >= @_{o -> o}(@_{o -> o -> o}(F, X), rec(X, Y, F)) because rec > @_{o -> o}, [6] and [11], by (Copy) 6] rec*(s(X), Y, F) >= @_{o -> o -> o}(F, X) because rec = @_{o -> o -> o}, rec in Mul, [7] and [10], by (Stat) 7] s(X) > X because [8], by definition 8] s*(X) >= X because [9], by (Select) 9] X >= X by (Meta) 10] F >= F by (Meta) 11] rec*(s(X), Y, F) >= rec(X, Y, F) because rec in Mul, [7], [12] and [10], by (Stat) 12] Y >= Y by (Meta) We can thus remove the following rules: rec(0, 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(s(X), Y, F) >? F X rec(X, Y, F) We use a recursive path ordering as defined in [Kop12, Chapter 5]. We choose Lex = {} and Mul = {@_{o -> o -> o}, @_{o -> o}, rec, s}, and the following precedence: rec > @_{o -> o -> o} > @_{o -> o} > s With these choices, we have: 1] rec(s(X), Y, F) > @_{o -> o}(@_{o -> o -> o}(F, X), rec(X, Y, F)) because [2], by definition 2] rec*(s(X), Y, F) >= @_{o -> o}(@_{o -> o -> o}(F, X), rec(X, Y, F)) because rec > @_{o -> o}, [3] and [10], by (Copy) 3] rec*(s(X), Y, F) >= @_{o -> o -> o}(F, X) because rec > @_{o -> o -> o}, [4] and [6], by (Copy) 4] rec*(s(X), Y, F) >= F because [5], by (Select) 5] F >= F by (Meta) 6] rec*(s(X), Y, F) >= X because [7], by (Select) 7] s(X) >= X because [8], by (Star) 8] s*(X) >= X because [9], by (Select) 9] X >= X by (Meta) 10] rec*(s(X), Y, F) >= rec(X, Y, F) because rec in Mul, [11], [13] and [14], by (Stat) 11] s(X) > X because [12], by definition 12] s*(X) >= X because [9], by (Select) 13] Y >= Y by (Meta) 14] F >= F by (Meta) We can thus remove the following rules: rec(s(X), Y, F) => F X rec(X, Y, F) All rules were succesfully removed. Thus, termination of the original system has been reduced to termination of the beta-rule, which is well-known to hold. +++ Citations +++ [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. popout

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actions

all output return to Higher Order Rewriting Union Beta