Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270367

details

property	value
status	complete
benchmark	h02.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n168.star.cs.uiowa.edu
space	Hamana_Kikuchi_18

run statistics

property	value
solver	Wanda 2.1c
configuration	default
runtime (wallclock)	0.30069 seconds
cpu usage	0.300945
user time	0.269933
system time	0.031012
max virtual memory	113176.0
max residence set size	10084.0

stage attributes

key	value
starexec-result	YES

output

0.00/0.28	YES
0.00/0.29	We consider the system theBenchmark.
0.00/0.29	
0.00/0.29	  Alphabet:
0.00/0.29	
0.00/0.29	    0 : [] --> c 
0.00/0.29	    1 : [] --> c 
0.00/0.29	    add : [] --> c -> a -> c 
0.00/0.29	    cons : [] --> a -> b -> b 
0.00/0.29	    fold : [] --> (c -> a -> c) -> b -> c -> c 
0.00/0.29	    mul : [] --> c -> a -> c 
0.00/0.29	    nil : [] --> b 
0.00/0.29	    prod : [] --> b -> c 
0.00/0.29	    sum : [] --> b -> c 
0.00/0.29	
0.00/0.29	  Rules:
0.00/0.29	
0.00/0.29	    fold (/\x./\y.f x y) nil z => z 
0.00/0.29	    fold (/\x./\y.f x y) (cons z u) v => fold (/\w./\x'.f w x') u (f v z) 
0.00/0.29	    sum x => fold (/\y./\z.add y z) x 0 
0.00/0.29	    fold (/\x./\y.mul x y) z 1 => prod z 
0.00/0.29	
0.00/0.29	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.
0.00/0.29	
0.00/0.29	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:
0.00/0.29	
0.00/0.29	  Alphabet:
0.00/0.29	
0.00/0.29	    0 : [] --> c 
0.00/0.29	    1 : [] --> c 
0.00/0.29	    add : [] --> c -> a -> c 
0.00/0.29	    cons : [a * b] --> b 
0.00/0.29	    fold : [c -> a -> c * b * c] --> c 
0.00/0.29	    mul : [c * a] --> c 
0.00/0.29	    nil : [] --> b 
0.00/0.29	    prod : [b] --> c 
0.00/0.29	    sum : [b] --> c 
0.00/0.29	    ~AP1 : [c -> a -> c * c] --> a -> c 
0.00/0.29	
0.00/0.29	  Rules:
0.00/0.29	
0.00/0.29	    fold(/\x./\y.~AP1(F, x) y, nil, X) => X 
0.00/0.29	    fold(/\x./\y.~AP1(F, x) y, cons(X, Y), Z) => fold(/\z./\u.~AP1(F, z) u, Y, ~AP1(F, Z) X) 
0.00/0.29	    sum(X) => fold(/\x./\y.~AP1(add, x) y, X, 0) 
0.00/0.29	    fold(/\x./\y.mul(x, y), X, 1) => prod(X) 
0.00/0.29	    fold(/\x./\y.mul(x, y), nil, X) => X 
0.00/0.29	    fold(/\x./\y.mul(x, y), cons(X, Y), Z) => fold(/\z./\u.mul(z, u), Y, mul(Z, X)) 
0.00/0.29	    ~AP1(F, X) => F X 
0.00/0.29	
0.00/0.29	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.  Additionally, we can remove some (now-)redundant rules.  This gives:
0.00/0.29	
0.00/0.29	  Alphabet:
0.00/0.29	
0.00/0.29	    0 : [] --> c 
0.00/0.29	    1 : [] --> c 
0.00/0.29	    add : [c * a] --> c 
0.00/0.29	    cons : [a * b] --> b 
0.00/0.29	    fold : [c -> a -> c * b * c] --> c 
0.00/0.29	    mul : [c * a] --> c 
0.00/0.29	    nil : [] --> b 
0.00/0.29	    prod : [b] --> c 
0.00/0.29	    sum : [b] --> c 
0.00/0.29	
0.00/0.29	  Rules:
0.00/0.29	
0.00/0.29	    fold(/\x./\y.X(x, y), nil, Y) => Y 
0.00/0.29	    fold(/\x./\y.X(x, y), cons(Y, Z), U) => fold(/\z./\u.X(z, u), Z, X(U, Y)) 
0.00/0.29	    sum(X) => fold(/\x./\y.add(x, y), X, 0) 
0.00/0.29	    fold(/\x./\y.mul(x, y), X, 1) => prod(X) 
0.00/0.29	
0.00/0.29	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.29	
0.00/0.29	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.29	
0.00/0.29	  fold(/\x./\y.X(x, y), nil, Y) >? Y 
0.00/0.29	  fold(/\x./\y.X(x, y), cons(Y, Z), U) >? fold(/\z./\u.X(z, u), Z, X(U, Y)) 
0.00/0.29	  sum(X) >? fold(/\x./\y.add(x, y), X, 0) 
0.00/0.29	  fold(/\x./\y.mul(x, y), X, 1) >? prod(X) 
0.00/0.29	
0.00/0.29	We use a recursive path ordering as defined in [Kop12, Chapter 5].
0.00/0.29	
0.00/0.29	Argument functions:
0.00/0.29	
0.00/0.29	  [[0]] = _|_ 
0.00/0.29	  [[prod(x_1)]] = x_1 
0.00/0.29	
0.00/0.29	We choose Lex = {fold} and Mul = {1, add, cons, mul, nil, sum}, and the following precedence: 1 > cons > mul > nil > sum > add > fold
0.00/0.29	
0.00/0.29	Taking the argument function into account, and fixing the greater / greater equal choices, the constraints can be denoted as follows:
0.00/0.29	
0.00/0.29	  fold(/\x./\y.X(x, y), nil, Y) >= Y 
0.00/0.29	  fold(/\x./\y.X(x, y), cons(Y, Z), U) > fold(/\x./\y.X(x, y), Z, X(U, Y)) 
0.00/0.29	  sum(X) >= fold(/\x./\y.add(x, y), X, _|_) 
0.00/0.29	  fold(/\x./\y.mul(x, y), X, 1) >= X 
0.00/0.29	
0.00/0.29	With these choices, we have:
0.00/0.29	
0.00/0.29	  1] fold(/\x./\y.X(x, y), nil, Y) >= Y  because [2], by (Star) 
0.00/0.29	  2] fold*(/\x./\y.X(x, y), nil, Y) >= Y  because [3], by (Select) 
0.00/0.29	  3] Y >= Y  by (Meta)

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