Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270483

details

property	value
status	complete
benchmark	h60.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n128.star.cs.uiowa.edu
space	Hamana_Kikuchi_18

run statistics

property	value
solver	Wanda 2.1c
configuration	default
runtime (wallclock)	9.347 seconds
cpu usage	9.34798
user time	9.18878
system time	0.159199
max virtual memory	387728.0
max residence set size	115904.0

stage attributes

key	value
starexec-result	YES

output

9.27/9.32	YES
9.27/9.33	We consider the system theBenchmark.
9.27/9.33	
9.27/9.33	  Alphabet:
9.27/9.33	
9.27/9.33	    0 : [] --> nat 
9.27/9.33	    rec : [nat * a * nat -> a -> a] --> a 
9.27/9.33	    s : [nat] --> nat 
9.27/9.33	    xap : [nat -> a -> a * nat] --> a -> a 
9.27/9.33	    yap : [a -> a * a] --> a 
9.27/9.33	
9.27/9.33	  Rules:
9.27/9.33	
9.27/9.33	    rec(0, x, /\y./\z.yap(xap(f, y), z)) => x 
9.27/9.33	    rec(s(x), y, /\z./\u.yap(xap(f, z), u)) => yap(xap(f, x), rec(x, y, /\v./\w.yap(xap(f, v), w))) 
9.27/9.33	    xap(f, x) => f x 
9.27/9.33	    yap(f, x) => f x 
9.27/9.33	
9.27/9.33	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
9.27/9.33	
9.27/9.33	Symbol xap 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:
9.27/9.33	
9.27/9.33	  Alphabet:
9.27/9.33	
9.27/9.33	    0 : [] --> nat 
9.27/9.33	    rec : [nat * a * nat -> a -> a] --> a 
9.27/9.33	    s : [nat] --> nat 
9.27/9.33	    yap : [a -> a * a] --> a 
9.27/9.33	
9.27/9.33	  Rules:
9.27/9.33	
9.27/9.33	    rec(0, X, /\x./\y.yap(F(x), y)) => X 
9.27/9.33	    rec(s(X), Y, /\x./\y.yap(F(x), y)) => yap(F(X), rec(X, Y, /\z./\u.yap(F(z), u))) 
9.27/9.33	    yap(F, X) => F X 
9.27/9.33	
9.27/9.33	We use rule removal, following [Kop12, Theorem 2.23].
9.27/9.33	
9.27/9.33	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
9.27/9.33	
9.27/9.33	  rec(0, X, /\x./\y.yap(F(x), y)) >? X 
9.27/9.33	  rec(s(X), Y, /\x./\y.yap(F(x), y)) >? yap(F(X), rec(X, Y, /\z./\u.yap(F(z), u))) 
9.27/9.33	  yap(F, X) >? F X 
9.27/9.33	
9.27/9.33	We use a recursive path ordering as defined in [Kop12, Chapter 5].
9.27/9.33	
9.27/9.33	We choose Lex = {} and Mul = {0, @_{o -> o}, rec, s, yap}, and the following precedence: 0 > rec > s > yap > @_{o -> o}
9.27/9.33	
9.27/9.33	With these choices, we have:
9.27/9.33	
9.27/9.33	  1] rec(0, X, /\x./\y.yap(F(x), y)) > X  because [2], by definition 
9.27/9.33	  2] rec*(0, X, /\x./\y.yap(F(x), y)) >= X  because [3], by (Select) 
9.27/9.33	  3] X >= X  by (Meta) 
9.27/9.33	
9.27/9.33	  4] rec(s(X), Y, /\x./\y.yap(F(x), y)) >= yap(F(X), rec(X, Y, /\x./\y.yap(F(x), y)))  because [5], by (Star) 
9.27/9.33	  5] rec*(s(X), Y, /\x./\y.yap(F(x), y)) >= yap(F(X), rec(X, Y, /\x./\y.yap(F(x), y)))  because rec > yap, [6] and [13], by (Copy) 
9.27/9.33	  6] rec*(s(X), Y, /\x./\y.yap(F(x), y)) >= F(X)  because [7], by (Select) 
9.27/9.33	  7] /\x.yap(F(rec*(s(X), Y, /\y./\z.yap(F(y), z))), x) >= F(X)  because [8], by (Eta)[Kop13:2] 
9.27/9.33	  8] F(rec*(s(X), Y, /\x./\y.yap(F(x), y))) >= F(X)  because [9], by (Meta) 
9.27/9.33	  9] rec*(s(X), Y, /\x./\y.yap(F(x), y)) >= X  because [10], by (Select) 
9.27/9.33	  10] s(X) >= X  because [11], by (Star) 
9.27/9.33	  11] s*(X) >= X  because [12], by (Select) 
9.27/9.33	  12] X >= X  by (Meta) 
9.27/9.33	  13] rec*(s(X), Y, /\x./\y.yap(F(x), y)) >= rec(X, Y, /\x./\y.yap(F(x), y))  because rec in Mul, [14], [16] and [17], by (Stat) 
9.27/9.33	  14] s(X) > X  because [15], by definition 
9.27/9.33	  15] s*(X) >= X  because [12], by (Select) 
9.27/9.33	  16] Y >= Y  by (Meta) 
9.27/9.33	  17] /\x./\z.yap(F(x), z) >= /\x./\z.yap(F(x), z)  because [18], by (Abs) 
9.27/9.33	  18] /\z.yap(F(y), z) >= /\z.yap(F(y), z)  because [19], by (Abs) 
9.27/9.33	  19] yap(F(y), x) >= yap(F(y), x)  because yap in Mul, [20] and [22], by (Fun) 
9.27/9.33	  20] F(y) >= F(y)  because [21], by (Meta) 
9.27/9.33	  21] y >= y  by (Var) 
9.27/9.33	  22] x >= x  by (Var) 
9.27/9.33	
9.27/9.33	  23] yap(F, X) >= @_{o -> o}(F, X)  because [24], by (Star) 
9.27/9.33	  24] yap*(F, X) >= @_{o -> o}(F, X)  because yap > @_{o -> o}, [25] and [27], by (Copy) 
9.27/9.33	  25] yap*(F, X) >= F  because [26], by (Select) 
9.27/9.33	  26] F >= F  by (Meta) 
9.27/9.33	  27] yap*(F, X) >= X  because [28], by (Select) 
9.27/9.33	  28] X >= X  by (Meta) 
9.27/9.33	
9.27/9.33	We can thus remove the following rules:
9.27/9.33	
9.27/9.33	  rec(0, X, /\x./\y.yap(F(x), y)) => X 
9.27/9.33	
9.27/9.33	We use rule removal, following [Kop12, Theorem 2.23].
9.27/9.33	
9.27/9.33	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
9.27/9.33	
9.27/9.33	  rec(s(X), Y, /\x./\y.yap(F(x), y)) >? yap(F(X), rec(X, Y, /\z./\u.yap(F(z), u))) 
9.27/9.33	  yap(F, X) >? F X 
9.27/9.33	
9.27/9.33	We use a recursive path ordering as defined in [Kop12, Chapter 5].
9.27/9.33	
9.27/9.33	We choose Lex = {} and Mul = {@_{o -> o}, rec, s, yap}, and the following precedence: rec > s > yap > @_{o -> o}
9.27/9.33	
9.27/9.33	With these choices, we have:
9.27/9.33	
9.27/9.33	  1] rec(s(X), Y, /\x./\y.yap(F(x), y)) >= yap(F(X), rec(X, Y, /\x./\y.yap(F(x), y)))  because [2], by (Star) 
9.27/9.33	  2] rec*(s(X), Y, /\x./\y.yap(F(x), y)) >= yap(F(X), rec(X, Y, /\x./\y.yap(F(x), y)))  because rec > yap, [3] and [10], by (Copy) 
9.27/9.33	  3] rec*(s(X), Y, /\x./\y.yap(F(x), y)) >= F(X)  because [4], by (Select)

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