Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270565

details

property	value
status	complete
benchmark	Applicative_05__Ex2_6_1Composition.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n125.star.cs.uiowa.edu
space	Uncurried_Applicative_11

run statistics

property	value
solver	Wanda 2.1c
configuration	default
runtime (wallclock)	0.031133 seconds
cpu usage	0.031206
user time	0.019884
system time	0.011322
max virtual memory	113176.0
max residence set size	2580.0

stage attributes

key	value
starexec-result	YES

output

0.00/0.02	YES
0.00/0.03	We consider the system theBenchmark.
0.00/0.03	
0.00/0.03	  Alphabet:
0.00/0.03	
0.00/0.03	    compose : [b -> c * a -> b * a] --> c 
0.00/0.03	
0.00/0.03	  Rules:
0.00/0.03	
0.00/0.03	    compose(f, g, x) => f (g x) 
0.00/0.03	
0.00/0.03	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
0.00/0.03	
0.00/0.03	We use rule removal, following [Kop12, Theorem 2.23].
0.00/0.03	
0.00/0.03	This gives the following requirements (possibly using Theorems 2.25 and 2.26 in [Kop12]):
0.00/0.03	
0.00/0.03	  compose(F, G, X) >? F (G X) 
0.00/0.03	
0.00/0.03	We use a recursive path ordering as defined in [Kop12, Chapter 5].
0.00/0.03	
0.00/0.03	We choose Lex = {} and Mul = {@_{o -> o}, compose}, and the following precedence: compose > @_{o -> o}
0.00/0.03	
0.00/0.03	With these choices, we have:
0.00/0.03	
0.00/0.03	  1] compose(F, G, X) > @_{o -> o}(F, @_{o -> o}(G, X))  because [2], by definition 
0.00/0.03	  2] compose*(F, G, X) >= @_{o -> o}(F, @_{o -> o}(G, X))  because compose > @_{o -> o}, [3] and [5], by (Copy) 
0.00/0.03	  3] compose*(F, G, X) >= F  because [4], by (Select) 
0.00/0.03	  4] F >= F  by (Meta) 
0.00/0.03	  5] compose*(F, G, X) >= @_{o -> o}(G, X)  because compose > @_{o -> o}, [6] and [8], by (Copy) 
0.00/0.03	  6] compose*(F, G, X) >= G  because [7], by (Select) 
0.00/0.03	  7] G >= G  by (Meta) 
0.00/0.03	  8] compose*(F, G, X) >= X  because [9], by (Select) 
0.00/0.03	  9] X >= X  by (Meta) 
0.00/0.03	
0.00/0.03	We can thus remove the following rules:
0.00/0.03	
0.00/0.03	  compose(F, G, X) => F (G X) 
0.00/0.03	
0.00/0.03	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.
0.00/0.03	
0.00/0.03	
0.00/0.03	+++ Citations +++
0.00/0.03	
0.00/0.03	[Kop12]  C. Kop.  Higher Order Termination.  PhD Thesis, 2012.
0.00/0.03	EOF

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