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Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270521
details
property
value
status
complete
benchmark
AotoYamada_05__010.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n061.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda 2.1c
configuration
default
runtime (wallclock)
0.637541 seconds
cpu usage
0.638242
user time
0.560713
system time
0.077529
max virtual memory
131756.0
max residence set size
16308.0
stage attributes
key
value
starexec-result
YES
output
0.54/0.59 YES 0.61/0.63 We consider the system theBenchmark. 0.61/0.63 0.61/0.63 Alphabet: 0.61/0.63 0.61/0.63 0 : [] --> b 0.61/0.63 cons : [b * a] --> a 0.61/0.63 curry : [b -> b -> b * b] --> b -> b 0.61/0.63 double : [] --> a -> a 0.61/0.63 inc : [] --> a -> a 0.61/0.63 map : [b -> b] --> a -> a 0.61/0.63 nil : [] --> a 0.61/0.63 plus : [] --> b -> b -> b 0.61/0.63 s : [b] --> b 0.61/0.63 times : [] --> b -> b -> b 0.61/0.63 0.61/0.63 Rules: 0.61/0.63 0.61/0.63 plus 0 x => x 0.61/0.63 plus s(x) y => s(plus x y) 0.61/0.63 times 0 x => 0 0.61/0.63 times s(x) y => plus (times x y) y 0.61/0.63 curry(f, x) y => f x y 0.61/0.63 map(f) nil => nil 0.61/0.63 map(f) cons(x, y) => cons(f x, map(f) y) 0.61/0.63 inc => map(curry(plus, s(0))) 0.61/0.63 double => map(curry(times, s(s(0)))) 0.61/0.63 0.61/0.63 This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). 0.61/0.63 0.61/0.63 Symbol curry 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: 0.61/0.63 0.61/0.63 Alphabet: 0.61/0.63 0.61/0.63 0 : [] --> b 0.61/0.63 cons : [b * a] --> a 0.61/0.63 double : [] --> a -> a 0.61/0.63 inc : [] --> a -> a 0.61/0.63 map : [b -> b] --> a -> a 0.61/0.63 nil : [] --> a 0.61/0.63 plus : [b] --> b -> b 0.61/0.63 s : [b] --> b 0.61/0.63 times : [b] --> b -> b 0.61/0.63 0.61/0.63 Rules: 0.61/0.63 0.61/0.63 plus(0) X => X 0.61/0.63 plus(s(X)) Y => s(plus(X) Y) 0.61/0.63 times(0) X => 0 0.61/0.63 times(s(X)) Y => plus(times(X) Y) Y 0.61/0.63 map(F) nil => nil 0.61/0.63 map(F) cons(X, Y) => cons(F X, map(F) Y) 0.61/0.63 inc => map(plus(s(0))) 0.61/0.63 double => map(times(s(s(0)))) 0.61/0.63 0.61/0.63 We use rule removal, following [Kop12, Theorem 2.23]. 0.61/0.63 0.61/0.63 This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]): 0.61/0.63 0.61/0.63 plus(0) X >? X 0.61/0.63 plus(s(X)) Y >? s(plus(X) Y) 0.61/0.63 times(0) X >? 0 0.61/0.63 times(s(X)) Y >? plus(times(X) Y) Y 0.61/0.63 map(F) nil >? nil 0.61/0.63 map(F) cons(X, Y) >? cons(F X, map(F) Y) 0.61/0.63 inc >? map(plus(s(0))) 0.61/0.63 double >? map(times(s(s(0)))) 0.61/0.63 0.61/0.63 We use a recursive path ordering as defined in [Kop12, Chapter 5]. 0.61/0.63 0.61/0.63 Argument functions: 0.61/0.63 0.61/0.63 [[0]] = _|_ 0.61/0.63 [[nil]] = _|_ 0.61/0.63 0.61/0.63 We choose Lex = {} and Mul = {@_{o -> o}, cons, double, inc, map, plus, s, times}, and the following precedence: double > inc > times > plus > @_{o -> o} > cons > map > s 0.61/0.63 0.61/0.63 Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows: 0.61/0.63 0.61/0.63 @_{o -> o}(plus(_|_), X) >= X 0.61/0.63 @_{o -> o}(plus(s(X)), Y) >= s(@_{o -> o}(plus(X), Y)) 0.61/0.63 @_{o -> o}(times(_|_), X) >= _|_ 0.61/0.63 @_{o -> o}(times(s(X)), Y) >= @_{o -> o}(plus(@_{o -> o}(times(X), Y)), Y) 0.61/0.63 @_{o -> o}(map(F), _|_) >= _|_ 0.61/0.63 @_{o -> o}(map(F), cons(X, Y)) > cons(@_{o -> o}(F, X), @_{o -> o}(map(F), Y)) 0.61/0.63 inc > map(plus(s(_|_))) 0.61/0.63 double > map(times(s(s(_|_)))) 0.61/0.63 0.61/0.63 With these choices, we have: 0.61/0.63 0.61/0.63 1] @_{o -> o}(plus(_|_), X) >= X because [2], by (Star) 0.61/0.63 2] @_{o -> o}*(plus(_|_), X) >= X because [3], by (Select) 0.61/0.63 3] X >= X by (Meta) 0.61/0.63 0.61/0.63 4] @_{o -> o}(plus(s(X)), Y) >= s(@_{o -> o}(plus(X), Y)) because [5], by (Star) 0.61/0.63 5] @_{o -> o}*(plus(s(X)), Y) >= s(@_{o -> o}(plus(X), Y)) because @_{o -> o} > s and [6], by (Copy) 0.61/0.63 6] @_{o -> o}*(plus(s(X)), Y) >= @_{o -> o}(plus(X), Y) because [7], by (Select) 0.61/0.63 7] plus(s(X)) @_{o -> o}*(plus(s(X)), Y) >= @_{o -> o}(plus(X), Y) because [8] 0.61/0.63 8] plus*(s(X), @_{o -> o}*(plus(s(X)), Y)) >= @_{o -> o}(plus(X), Y) because plus > @_{o -> o}, [9] and [13], by (Copy) 0.61/0.63 9] plus*(s(X), @_{o -> o}*(plus(s(X)), Y)) >= plus(X) because plus in Mul and [10], by (Stat)
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