Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270659

details

property	value
status	complete
benchmark	Applicative_first_order_05__#3.22.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n074.star.cs.uiowa.edu
space	Uncurried_Applicative_11

run statistics

property	value
solver	Wanda 2.1c
configuration	default
runtime (wallclock)	1.15517 seconds
cpu usage	2.36336
user time	2.19089
system time	0.172471
max virtual memory	7023584.0
max residence set size	143904.0

stage attributes

key	value
starexec-result	YES

output

2.26/1.14	YES
2.26/1.15	We consider the system theBenchmark.
2.26/1.15	
2.26/1.15	  Alphabet:
2.26/1.15	
2.26/1.15	    0 : [] --> a 
2.26/1.15	    cons : [c * d] --> d 
2.26/1.15	    false : [] --> b 
2.26/1.15	    filter : [c -> b * d] --> d 
2.26/1.15	    filter2 : [b * c -> b * c * d] --> d 
2.26/1.15	    map : [c -> c * d] --> d 
2.26/1.15	    nil : [] --> d 
2.26/1.15	    plus : [a * a] --> a 
2.26/1.15	    s : [a] --> a 
2.26/1.15	    times : [a * a] --> a 
2.26/1.15	    true : [] --> b 
2.26/1.15	
2.26/1.15	  Rules:
2.26/1.15	
2.26/1.15	    times(x, plus(y, s(z))) => plus(times(x, plus(y, times(s(z), 0))), times(x, s(z))) 
2.26/1.15	    times(x, 0) => 0 
2.26/1.15	    times(x, s(y)) => plus(times(x, y), x) 
2.26/1.15	    plus(x, 0) => x 
2.26/1.15	    plus(x, s(y)) => s(plus(x, y)) 
2.26/1.15	    map(f, nil) => nil 
2.26/1.15	    map(f, cons(x, y)) => cons(f x, map(f, y)) 
2.26/1.15	    filter(f, nil) => nil 
2.26/1.15	    filter(f, cons(x, y)) => filter2(f x, f, x, y) 
2.26/1.15	    filter2(true, f, x, y) => cons(x, filter(f, y)) 
2.26/1.15	    filter2(false, f, x, y) => filter(f, y) 
2.26/1.15	
2.26/1.15	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
2.26/1.15	
2.26/1.15	We observe that the rules contain a first-order subset:
2.26/1.15	
2.26/1.15	  times(X, plus(Y, s(Z))) => plus(times(X, plus(Y, times(s(Z), 0))), times(X, s(Z))) 
2.26/1.15	  times(X, 0) => 0 
2.26/1.15	  times(X, s(Y)) => plus(times(X, Y), X) 
2.26/1.15	  plus(X, 0) => X 
2.26/1.15	  plus(X, s(Y)) => s(plus(X, Y)) 
2.26/1.15	
2.26/1.15	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.
2.26/1.15	
2.26/1.15	According to the external first-order termination prover, this system is indeed Ce-terminating:
2.26/1.15	
2.26/1.15	 || proof of resources/system.trs
2.26/1.15	 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616
2.26/1.15	 || 
2.26/1.15	 || 
2.26/1.15	 || Termination w.r.t. Q of the given QTRS could be proven:
2.26/1.15	 || 
2.26/1.15	 || (0) QTRS
2.26/1.15	 || (1) DependencyPairsProof [EQUIVALENT]
2.26/1.15	 || (2) QDP
2.26/1.15	 || (3) DependencyGraphProof [EQUIVALENT]
2.26/1.15	 || (4) AND
2.26/1.15	 ||     (5) QDP
2.26/1.15	 ||         (6) UsableRulesProof [EQUIVALENT]
2.26/1.15	 ||         (7) QDP
2.26/1.15	 ||         (8) QDPSizeChangeProof [EQUIVALENT]
2.26/1.15	 ||         (9) YES
2.26/1.15	 ||     (10) QDP
2.26/1.15	 ||         (11) UsableRulesProof [EQUIVALENT]
2.26/1.15	 ||         (12) QDP
2.26/1.15	 ||         (13) MRRProof [EQUIVALENT]
2.26/1.15	 ||         (14) QDP
2.26/1.15	 ||         (15) MRRProof [EQUIVALENT]
2.26/1.15	 ||         (16) QDP
2.26/1.15	 ||         (17) MRRProof [EQUIVALENT]
2.26/1.15	 ||         (18) QDP
2.26/1.15	 ||         (19) DependencyGraphProof [EQUIVALENT]
2.26/1.15	 ||         (20) TRUE
2.26/1.15	 || 
2.26/1.15	 || 
2.26/1.15	 || ----------------------------------------
2.26/1.15	 || 
2.26/1.15	 || (0)
2.26/1.15	 || Obligation:
2.26/1.15	 || Q restricted rewrite system:
2.26/1.15	 || The TRS R consists of the following rules:
2.26/1.15	 || 
2.26/1.15	 ||    times(%X, plus(%Y, s(%Z))) -> plus(times(%X, plus(%Y, times(s(%Z), 0))), times(%X, s(%Z)))
2.26/1.15	 ||    times(%X, 0) -> 0
2.26/1.15	 ||    times(%X, s(%Y)) -> plus(times(%X, %Y), %X)
2.26/1.15	 ||    plus(%X, 0) -> %X
2.26/1.15	 ||    plus(%X, s(%Y)) -> s(plus(%X, %Y))
2.26/1.15	 ||    ~PAIR(%X, %Y) -> %X
2.26/1.15	 ||    ~PAIR(%X, %Y) -> %Y
2.26/1.15	 || 
2.26/1.15	 || Q is empty.
2.26/1.15	 || 
2.26/1.15	 || ----------------------------------------
2.26/1.15	 || 
2.26/1.15	 || (1) DependencyPairsProof (EQUIVALENT)
2.26/1.15	 || Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
2.26/1.15	 || ----------------------------------------
2.26/1.15	 || 
2.26/1.15	 || (2)
2.26/1.15	 || Obligation:
2.26/1.15	 || Q DP problem:

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