Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270237

details

property	value
status	complete
benchmark	twice.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n100.star.cs.uiowa.edu
space	Kop_11

run statistics

property	value
solver	Wanda 2.1c
configuration	default
runtime (wallclock)	0.181761 seconds
cpu usage	0.182196
user time	0.146178
system time	0.036018
max virtual memory	113176.0
max residence set size	5140.0

stage attributes

key	value
starexec-result	YES

output

0.00/0.15	YES
0.00/0.18	We consider the system theBenchmark.
0.00/0.18	
0.00/0.18	  Alphabet:
0.00/0.18	
0.00/0.18	    0 : [] --> nat 
0.00/0.18	    I : [nat] --> nat 
0.00/0.18	    s : [nat] --> nat 
0.00/0.18	    twice : [nat -> nat] --> nat -> nat 
0.00/0.18	
0.00/0.18	  Rules:
0.00/0.18	
0.00/0.18	    I(0) => 0 
0.00/0.18	    I(s(x)) => s(twice(/\y.I(y)) x) 
0.00/0.18	    twice(f) => /\x.f (f x) 
0.00/0.18	
0.00/0.18	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.18	
0.00/0.18	We use the dependency pair framework as described in [Kop12, Ch. 6/7], with dynamic dependency pairs.
0.00/0.18	
0.00/0.18	To start, the system is beta-saturated by adding the following rules:
0.00/0.18	
0.00/0.18	  twice(F) X => F (F X) 
0.00/0.18	
0.00/0.18	After applying [Kop12, Thm. 7.22] to denote collapsing dependency pairs in an extended form, we thus obtain the following dependency pair problem (P_0, R_0, minimal, formative):
0.00/0.18	
0.00/0.18	  Dependency Pairs P_0:
0.00/0.18	
0.00/0.18	    0] I#(s(X)) =#> twice(/\x.I(x)) X   
0.00/0.18	    1] I#(s(X)) =#> twice#(/\x.I(x))   
0.00/0.18	    2] I#(s(X)) =#> I#(x)   
0.00/0.18	    3] twice#(F) =#> F(F x)   
0.00/0.18	    4] twice#(F) =#> F(x)   
0.00/0.18	    5] twice(F) X =#> F(F X)   
0.00/0.18	    6] twice(F) X =#> F(X)   
0.00/0.18	
0.00/0.18	  Rules R_0:
0.00/0.18	
0.00/0.18	    I(0) => 0 
0.00/0.18	    I(s(X)) => s(twice(/\x.I(x)) X) 
0.00/0.18	    twice(F) => /\x.F (F x) 
0.00/0.18	    twice(F) X => F (F X) 
0.00/0.18	
0.00/0.18	Thus, the original system is terminating if (P_0, R_0, minimal, formative) is finite.
0.00/0.18	
0.00/0.18	We consider the dependency pair problem (P_0, R_0, minimal, formative).
0.00/0.18	
0.00/0.18	We place the elements of P in a dependency graph approximation G (see e.g. [Kop12, Thm. 7.27, 7.29], as follows:
0.00/0.18	
0.00/0.18	    * 0 : 5, 6 
0.00/0.18	    * 1 : 3, 4 
0.00/0.18	    * 2 :  
0.00/0.18	    * 3 : 0, 1, 2, 3, 4, 5, 6 
0.00/0.18	    * 4 : 0, 1, 2, 3, 4, 5, 6 
0.00/0.18	    * 5 : 0, 1, 2, 3, 4, 5, 6 
0.00/0.18	    * 6 : 0, 1, 2, 3, 4, 5, 6 
0.00/0.18	
0.00/0.18	This graph has the following strongly connected components:
0.00/0.18	
0.00/0.18	  P_1:
0.00/0.18	
0.00/0.18	    I#(s(X)) =#> twice(/\x.I(x)) X   
0.00/0.18	    I#(s(X)) =#> twice#(/\x.I(x))   
0.00/0.18	    twice#(F) =#> F(F x)   
0.00/0.18	    twice#(F) =#> F(x)   
0.00/0.18	    twice(F) X =#> F(F X)   
0.00/0.18	    twice(F) X =#> F(X)   
0.00/0.18	
0.00/0.18	By [Kop12, Thm. 7.31], we may replace any dependency pair problem (P_0, R_0, m, f) by (P_1, R_0, m, f).
0.00/0.18	
0.00/0.18	Thus, the original system is terminating if (P_1, R_0, minimal, formative) is finite.
0.00/0.18	
0.00/0.18	We consider the dependency pair problem (P_1, R_0, minimal, formative).
0.00/0.18	
0.00/0.18	The formative rules of (P_1, R_0) are R_1 ::=
0.00/0.18	
0.00/0.18	  I(s(X)) => s(twice(/\x.I(x)) X) 
0.00/0.18	  twice(F) X => F (F X) 
0.00/0.18	
0.00/0.18	By [Kop12, Thm. 7.17], we may replace the dependency pair problem (P_1, R_0, minimal, formative) by (P_1, R_1, minimal, formative).
0.00/0.18	
0.00/0.18	Thus, the original system is terminating if (P_1, R_1, minimal, formative) is finite.
0.00/0.18	
0.00/0.18	We consider the dependency pair problem (P_1, R_1, minimal, formative).
0.00/0.18	
0.00/0.18	We will use the reduction pair processor [Kop12, Thm. 7.16].  As the system is abstraction-simple and the formative flag is set, it suffices to find a tagged reduction pair [Kop12, Def. 6.70].  Thus, we must orient:
0.00/0.18	
0.00/0.18	  I#(s(X)) >? twice(/\x.I-(x), X) 
0.00/0.18	  I#(s(X)) >? twice#(/\x.I-(x)) ~c0 
0.00/0.18	  twice#(F) X >? F(F ~c1) 
0.00/0.18	  twice#(F) X >? F(~c2) 
0.00/0.18	  twice(F, X) >? F(F X) 
0.00/0.18	  twice(F, X) >? F(X) 
0.00/0.18	  I(s(X)) >= s(twice(/\x.I-(x), X)) 
0.00/0.18	  twice(F, X) >= F (F X) 
0.00/0.18	  I-(X) >= I(X) 
0.00/0.18	  I-(X) >= I#(X) 
0.00/0.18	
0.00/0.18	We apply [Kop12, Thm. 6.75] and use the following argument functions:
0.00/0.18

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