Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270341

details

property	value
status	complete
benchmark	kripke.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n160.star.cs.uiowa.edu
space	Hamana_17

run statistics

property	value
solver	Wanda 2.1c
configuration	default
runtime (wallclock)	0.643143 seconds
cpu usage	1.4536
user time	1.34165
system time	0.111945
max virtual memory	6748936.0
max residence set size	105820.0

stage attributes

key	value
starexec-result	YES

output

1.40/0.63	YES
1.40/0.64	We consider the system theBenchmark.
1.40/0.64	
1.40/0.64	  Alphabet:
1.40/0.64	
1.40/0.64	    abs : [] --> (a -> b) -> Arrab 
1.40/0.64	    app : [] --> Arrab -> a -> b 
1.40/0.64	    box : [] --> a -> Boxa 
1.40/0.64	    unbox : [] --> Boxa -> a 
1.40/0.64	
1.40/0.64	  Rules:
1.40/0.64	
1.40/0.64	    app (abs (/\x.f x)) y => f y 
1.40/0.64	    abs (/\x.app y x) => y 
1.40/0.64	    unbox (box x) => x 
1.40/0.64	    box (unbox x) => x 
1.40/0.64	
1.40/0.64	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.
1.40/0.64	
1.40/0.64	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:
1.40/0.64	
1.40/0.64	  Alphabet:
1.40/0.64	
1.40/0.64	    abs : [a -> b] --> Arrab 
1.40/0.64	    app : [Arrab * a] --> b 
1.40/0.64	    box : [a] --> Boxa 
1.40/0.64	    unbox : [Boxa] --> a 
1.40/0.64	    ~AP1 : [a -> b * a] --> b 
1.40/0.64	
1.40/0.64	  Rules:
1.40/0.64	
1.40/0.64	    app(abs(/\x.~AP1(F, x)), X) => ~AP1(F, X) 
1.40/0.64	    abs(/\x.app(X, x)) => X 
1.40/0.64	    unbox(box(X)) => X 
1.40/0.64	    box(unbox(X)) => X 
1.40/0.64	    app(abs(/\x.app(X, x)), Y) => app(X, Y) 
1.40/0.64	    ~AP1(F, X) => F X 
1.40/0.64	
1.40/0.64	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:
1.40/0.64	
1.40/0.64	  Alphabet:
1.40/0.64	
1.40/0.64	    abs : [a -> b] --> Arrab 
1.40/0.64	    app : [Arrab * a] --> b 
1.40/0.64	    box : [a] --> Boxa 
1.40/0.64	    unbox : [Boxa] --> a 
1.40/0.64	
1.40/0.64	  Rules:
1.40/0.64	
1.40/0.64	    app(abs(/\x.X(x)), Y) => X(Y) 
1.40/0.64	    abs(/\x.app(X, x)) => X 
1.40/0.64	    unbox(box(X)) => X 
1.40/0.64	    box(unbox(X)) => X 
1.40/0.64	
1.40/0.64	We observe that the rules contain a first-order subset:
1.40/0.64	
1.40/0.64	  unbox(box(X)) => X 
1.40/0.64	  box(unbox(X)) => X 
1.40/0.64	
1.40/0.64	Moreover, the system is finitely branching.  Thus, by [Kop12, Thm. 7.55], we may omit all first-order dependency pairs from the dependency pair problem (DP(R), R) if this first-order part is Ce-terminating when seen as a many-sorted first-order TRS.
1.40/0.64	
1.40/0.64	According to the external first-order termination prover, this system is indeed Ce-terminating:
1.40/0.64	
1.40/0.64	 || proof of resources/system.trs
1.40/0.64	 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616
1.40/0.64	 || 
1.40/0.64	 || 
1.40/0.64	 || Termination w.r.t. Q of the given QTRS could be proven:
1.40/0.64	 || 
1.40/0.64	 || (0) QTRS
1.40/0.64	 || (1) QTRSRRRProof [EQUIVALENT]
1.40/0.64	 || (2) QTRS
1.40/0.64	 || (3) RisEmptyProof [EQUIVALENT]
1.40/0.64	 || (4) YES
1.40/0.64	 || 
1.40/0.64	 || 
1.40/0.64	 || ----------------------------------------
1.40/0.64	 || 
1.40/0.64	 || (0)
1.40/0.64	 || Obligation:
1.40/0.64	 || Q restricted rewrite system:
1.40/0.64	 || The TRS R consists of the following rules:
1.40/0.64	 || 
1.40/0.64	 ||    unbox(box(%X)) -> %X
1.40/0.64	 ||    box(unbox(%X)) -> %X
1.40/0.64	 ||    ~PAIR(%X, %Y) -> %X
1.40/0.64	 ||    ~PAIR(%X, %Y) -> %Y
1.40/0.64	 || 
1.40/0.64	 || Q is empty.
1.40/0.64	 || 
1.40/0.64	 || ----------------------------------------
1.40/0.64	 || 
1.40/0.64	 || (1) QTRSRRRProof (EQUIVALENT)
1.40/0.64	 || Used ordering:
1.40/0.64	 || Polynomial interpretation [POLO]:
1.40/0.64	 || 
1.40/0.64	 ||    POL(box(x_1)) = 2 + x_1
1.40/0.64	 ||    POL(unbox(x_1)) = 1 + 2*x_1
1.40/0.64	 ||    POL(~PAIR(x_1, x_2)) = 2 + x_1 + x_2
1.40/0.64	 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

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