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HRS union beta 16688 pair #381734617
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
n081.star.cs.uiowa.edu
space
Mixed_HO_10
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
0.081209897995 seconds
cpu usage
0.077781284
max memory
4771840.0
stage attributes
key
value
output-size
3380
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- 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) about to try horpo 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) about to try horpo 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.
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