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Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270579
details
property
value
status
complete
benchmark
Applicative_05__Ex6Recursor.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n102.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda 2.1c
configuration
default
runtime (wallclock)
0.115374 seconds
cpu usage
0.115516
user time
0.096441
system time
0.019075
max virtual memory
113176.0
max residence set size
4680.0
stage attributes
key
value
starexec-result
YES
output
0.00/0.11 YES 0.00/0.11 We consider the system theBenchmark. 0.00/0.11 0.00/0.11 Alphabet: 0.00/0.11 0.00/0.11 0 : [] --> a 0.00/0.11 rec : [a -> b -> b * b * a] --> b 0.00/0.11 s : [a] --> a 0.00/0.11 0.00/0.11 Rules: 0.00/0.11 0.00/0.11 rec(f, x, 0) => x 0.00/0.11 rec(f, x, s(y)) => f s(y) rec(f, x, y) 0.00/0.11 0.00/0.11 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 0.00/0.11 0.00/0.11 We use rule removal, following [Kop12, Theorem 2.23]. 0.00/0.11 0.00/0.11 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.00/0.11 0.00/0.11 rec(F, X, 0) >? X 0.00/0.11 rec(F, X, s(Y)) >? F s(Y) rec(F, X, Y) 0.00/0.11 0.00/0.11 We use a recursive path ordering as defined in [Kop12, Chapter 5]. 0.00/0.11 0.00/0.11 We choose Lex = {} and Mul = {0, @_{o -> o -> o}, @_{o -> o}, rec, s}, and the following precedence: 0 > rec > @_{o -> o -> o} > @_{o -> o} > s 0.00/0.11 0.00/0.11 With these choices, we have: 0.00/0.11 0.00/0.11 1] rec(F, X, 0) > X because [2], by definition 0.00/0.11 2] rec*(F, X, 0) >= X because [3], by (Select) 0.00/0.11 3] X >= X by (Meta) 0.00/0.11 0.00/0.11 4] rec(F, X, s(Y)) > @_{o -> o}(@_{o -> o -> o}(F, s(Y)), rec(F, X, Y)) because [5], by definition 0.00/0.11 5] rec*(F, X, s(Y)) >= @_{o -> o}(@_{o -> o -> o}(F, s(Y)), rec(F, X, Y)) because rec > @_{o -> o}, [6] and [12], by (Copy) 0.00/0.11 6] rec*(F, X, s(Y)) >= @_{o -> o -> o}(F, s(Y)) because rec > @_{o -> o -> o}, [7] and [9], by (Copy) 0.00/0.11 7] rec*(F, X, s(Y)) >= F because [8], by (Select) 0.00/0.11 8] F >= F by (Meta) 0.00/0.11 9] rec*(F, X, s(Y)) >= s(Y) because [10], by (Select) 0.00/0.11 10] s(Y) >= s(Y) because s in Mul and [11], by (Fun) 0.00/0.11 11] Y >= Y by (Meta) 0.00/0.11 12] rec*(F, X, s(Y)) >= rec(F, X, Y) because rec in Mul, [13], [14] and [15], by (Stat) 0.00/0.11 13] F >= F by (Meta) 0.00/0.11 14] X >= X by (Meta) 0.00/0.11 15] s(Y) > Y because [16], by definition 0.00/0.11 16] s*(Y) >= Y because [11], by (Select) 0.00/0.11 0.00/0.11 We can thus remove the following rules: 0.00/0.11 0.00/0.11 rec(F, X, 0) => X 0.00/0.11 rec(F, X, s(Y)) => F s(Y) rec(F, X, Y) 0.00/0.11 0.00/0.11 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. 0.00/0.11 0.00/0.11 0.00/0.11 +++ Citations +++ 0.00/0.11 0.00/0.11 [Kop12] C. Kop. Higher Order Termination. PhD Thesis, 2012. 0.00/0.11 EOF
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