Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270719

details

property	value
status	complete
benchmark	Applicative_first_order_05__motivation.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n161.star.cs.uiowa.edu
space	Uncurried_Applicative_11

run statistics

property	value
solver	Wanda 2.1c
configuration	default
runtime (wallclock)	2.00439 seconds
cpu usage	2.15514
user time	1.99372
system time	0.161423
max virtual memory	7088588.0
max residence set size	126424.0

stage attributes

key	value
starexec-result	YES

output

2.12/1.99	YES
2.12/2.00	We consider the system theBenchmark.
2.12/2.00	
2.12/2.00	  Alphabet:
2.12/2.00	
2.12/2.00	    cons : [c * d] --> d 
2.12/2.00	    f : [a * a] --> a 
2.12/2.00	    false : [] --> b 
2.12/2.00	    filter : [c -> b * d] --> d 
2.12/2.00	    filter2 : [b * c -> b * c * d] --> d 
2.12/2.00	    g : [a] --> a 
2.12/2.00	    h : [a] --> a 
2.12/2.00	    map : [c -> c * d] --> d 
2.12/2.00	    nil : [] --> d 
2.12/2.00	    true : [] --> b 
2.12/2.00	
2.12/2.00	  Rules:
2.12/2.00	
2.12/2.00	    g(h(g(x))) => g(x) 
2.12/2.00	    g(g(x)) => g(h(g(x))) 
2.12/2.00	    h(h(x)) => h(f(h(x), x)) 
2.12/2.00	    map(i, nil) => nil 
2.12/2.00	    map(i, cons(x, y)) => cons(i x, map(i, y)) 
2.12/2.00	    filter(i, nil) => nil 
2.12/2.00	    filter(i, cons(x, y)) => filter2(i x, i, x, y) 
2.12/2.00	    filter2(true, i, x, y) => cons(x, filter(i, y)) 
2.12/2.00	    filter2(false, i, x, y) => filter(i, y) 
2.12/2.00	
2.12/2.00	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
2.12/2.00	
2.12/2.00	We observe that the rules contain a first-order subset:
2.12/2.00	
2.12/2.00	  g(h(g(X))) => g(X) 
2.12/2.00	  g(g(X)) => g(h(g(X))) 
2.12/2.00	  h(h(X)) => h(f(h(X), X)) 
2.12/2.00	
2.12/2.00	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.
2.12/2.00	
2.12/2.00	According to the external first-order termination prover, this system is indeed Ce-terminating:
2.12/2.00	
2.12/2.00	 || proof of resources/system.trs
2.12/2.00	 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616
2.12/2.00	 || 
2.12/2.00	 || 
2.12/2.00	 || Termination w.r.t. Q of the given QTRS could be proven:
2.12/2.00	 || 
2.12/2.00	 || (0) QTRS
2.12/2.00	 || (1) DependencyPairsProof [EQUIVALENT]
2.12/2.00	 || (2) QDP
2.12/2.00	 || (3) DependencyGraphProof [EQUIVALENT]
2.12/2.00	 || (4) QDP
2.12/2.00	 || (5) QDPOrderProof [EQUIVALENT]
2.12/2.00	 || (6) QDP
2.12/2.00	 || (7) PisEmptyProof [EQUIVALENT]
2.12/2.00	 || (8) YES
2.12/2.00	 || 
2.12/2.00	 || 
2.12/2.00	 || ----------------------------------------
2.12/2.00	 || 
2.12/2.00	 || (0)
2.12/2.00	 || Obligation:
2.12/2.00	 || Q restricted rewrite system:
2.12/2.00	 || The TRS R consists of the following rules:
2.12/2.00	 || 
2.12/2.00	 ||    g(h(g(%X))) -> g(%X)
2.12/2.00	 ||    g(g(%X)) -> g(h(g(%X)))
2.12/2.00	 ||    h(h(%X)) -> h(f(h(%X), %X))
2.12/2.00	 ||    ~PAIR(%X, %Y) -> %X
2.12/2.00	 ||    ~PAIR(%X, %Y) -> %Y
2.12/2.00	 || 
2.12/2.00	 || Q is empty.
2.12/2.00	 || 
2.12/2.00	 || ----------------------------------------
2.12/2.00	 || 
2.12/2.00	 || (1) DependencyPairsProof (EQUIVALENT)
2.12/2.00	 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
2.12/2.00	 || ----------------------------------------
2.12/2.00	 || 
2.12/2.00	 || (2)
2.12/2.00	 || Obligation:
2.12/2.00	 || Q DP problem:
2.12/2.00	 || The TRS P consists of the following rules:
2.12/2.00	 || 
2.12/2.00	 ||    G(g(%X)) -> G(h(g(%X)))
2.12/2.00	 ||    G(g(%X)) -> H(g(%X))
2.12/2.00	 ||    H(h(%X)) -> H(f(h(%X), %X))
2.12/2.00	 || 
2.12/2.00	 || The TRS R consists of the following rules:
2.12/2.00	 || 
2.12/2.00	 ||    g(h(g(%X))) -> g(%X)
2.12/2.00	 ||    g(g(%X)) -> g(h(g(%X)))
2.12/2.00	 ||    h(h(%X)) -> h(f(h(%X), %X))
2.12/2.00	 ||    ~PAIR(%X, %Y) -> %X
2.12/2.00	 ||    ~PAIR(%X, %Y) -> %Y
2.12/2.00	 || 
2.12/2.00	 || Q is empty.
2.12/2.00	 || We have to consider all minimal (P,Q,R)-chains.
2.12/2.00	 || ----------------------------------------
2.12/2.00	 || 
2.12/2.00	 || (3) DependencyGraphProof (EQUIVALENT)

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actions all output return to Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10