Higher_Order_Rewriting_Union_Beta 2019-03-28 22.10 pair #432270243

details

property	value
status	complete
benchmark	prefixshuffle.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n141.star.cs.uiowa.edu
space	Mixed_HO_12

run statistics

property	value
solver	Wanda 2.1c
configuration	default
runtime (wallclock)	2.16221 seconds
cpu usage	3.71934
user time	3.48377
system time	0.235573
max virtual memory	7030500.0
max residence set size	173980.0

stage attributes

key	value
starexec-result	YES

output

3.62/2.14	YES
3.62/2.15	We consider the system theBenchmark.
3.62/2.15	
3.62/2.15	  Alphabet:
3.62/2.15	
3.62/2.15	    0 : [] --> nat 
3.62/2.15	    app : [natlist * natlist] --> natlist 
3.62/2.15	    apply2 : [pair -> nat -> pair * pair * nat] --> pair 
3.62/2.15	    cons : [nat * natlist] --> natlist 
3.62/2.15	    fst : [pair] --> natlist 
3.62/2.15	    nil : [] --> natlist 
3.62/2.15	    p : [natlist * natlist] --> pair 
3.62/2.15	    pcons : [pair * plist] --> plist 
3.62/2.15	    pnil : [] --> plist 
3.62/2.15	    pps : [natlist] --> plist 
3.62/2.15	    prefixshuffle : [pair * natlist] --> plist 
3.62/2.15	    pshuffle : [natlist] --> pair 
3.62/2.15	    reverse : [natlist] --> natlist 
3.62/2.15	    s : [nat] --> nat 
3.62/2.15	    shuffle : [natlist] --> natlist 
3.62/2.15	
3.62/2.15	  Rules:
3.62/2.15	
3.62/2.15	    app(nil, x) => x 
3.62/2.15	    app(cons(x, y), z) => cons(x, app(y, z)) 
3.62/2.15	    reverse(nil) => nil 
3.62/2.15	    reverse(cons(x, y)) => app(reverse(y), cons(x, nil)) 
3.62/2.15	    shuffle(nil) => nil 
3.62/2.15	    shuffle(cons(x, y)) => cons(x, shuffle(reverse(y))) 
3.62/2.15	    fst(p(x, y)) => x 
3.62/2.15	    pshuffle(x) => p(x, shuffle(x)) 
3.62/2.15	    prefixshuffle(x, nil) => pcons(x, pnil) 
3.62/2.15	    prefixshuffle(x, cons(y, z)) => pcons(x, prefixshuffle(apply2(/\u./\v.pshuffle(app(fst(u), cons(v, nil))), x, y), reverse(z))) 
3.62/2.16	    apply2(f, x, 0) => x 
3.62/2.16	    apply2(f, x, s(y)) => f x s(y) 
3.62/2.16	    pps(x) => prefixshuffle(p(nil, nil), x) 
3.62/2.16	
3.62/2.16	This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0).
3.62/2.16	
3.62/2.16	We observe that the rules contain a first-order subset:
3.62/2.16	
3.62/2.16	  app(nil, X) => X 
3.62/2.16	  app(cons(X, Y), Z) => cons(X, app(Y, Z)) 
3.62/2.16	  reverse(nil) => nil 
3.62/2.16	  reverse(cons(X, Y)) => app(reverse(Y), cons(X, nil)) 
3.62/2.16	  shuffle(nil) => nil 
3.62/2.16	  shuffle(cons(X, Y)) => cons(X, shuffle(reverse(Y))) 
3.62/2.16	  fst(p(X, Y)) => X 
3.62/2.16	  pshuffle(X) => p(X, shuffle(X)) 
3.62/2.16	
3.62/2.16	Moreover, the system is orthogonal.  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 terminating when seen as a many-sorted first-order TRS.
3.62/2.16	
3.62/2.16	According to the external first-order termination prover, this system is indeed terminating:
3.62/2.16	
3.62/2.16	 || proof of resources/system.trs
3.62/2.16	 || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616
3.62/2.16	 || 
3.62/2.16	 || 
3.62/2.16	 || Termination w.r.t. Q of the given QTRS could be proven:
3.62/2.16	 || 
3.62/2.16	 || (0) QTRS
3.62/2.16	 || (1) QTRSRRRProof [EQUIVALENT]
3.62/2.16	 || (2) QTRS
3.62/2.16	 || (3) QTRSRRRProof [EQUIVALENT]
3.62/2.16	 || (4) QTRS
3.62/2.16	 || (5) QTRSRRRProof [EQUIVALENT]
3.62/2.16	 || (6) QTRS
3.62/2.16	 || (7) QTRSRRRProof [EQUIVALENT]
3.62/2.16	 || (8) QTRS
3.62/2.16	 || (9) QTRSRRRProof [EQUIVALENT]
3.62/2.16	 || (10) QTRS
3.62/2.16	 || (11) QTRSRRRProof [EQUIVALENT]
3.62/2.16	 || (12) QTRS
3.62/2.16	 || (13) QTRSRRRProof [EQUIVALENT]
3.62/2.16	 || (14) QTRS
3.62/2.16	 || (15) RisEmptyProof [EQUIVALENT]
3.62/2.16	 || (16) YES
3.62/2.16	 || 
3.62/2.16	 || 
3.62/2.16	 || ----------------------------------------
3.62/2.16	 || 
3.62/2.16	 || (0)
3.62/2.16	 || Obligation:
3.62/2.16	 || Q restricted rewrite system:
3.62/2.16	 || The TRS R consists of the following rules:
3.62/2.16	 || 
3.62/2.16	 ||    app(nil, %X) -> %X
3.62/2.16	 ||    app(cons(%X, %Y), %Z) -> cons(%X, app(%Y, %Z))
3.62/2.16	 ||    reverse(nil) -> nil
3.62/2.16	 ||    reverse(cons(%X, %Y)) -> app(reverse(%Y), cons(%X, nil))
3.62/2.16	 ||    shuffle(nil) -> nil
3.62/2.16	 ||    shuffle(cons(%X, %Y)) -> cons(%X, shuffle(reverse(%Y)))
3.62/2.16	 ||    fst(p(%X, %Y)) -> %X
3.62/2.16	 ||    pshuffle(%X) -> p(%X, shuffle(%X))
3.62/2.16	 || 
3.62/2.16	 || Q is empty.
3.62/2.16	 || 
3.62/2.16	 || ----------------------------------------
3.62/2.16	 || 
3.62/2.16	 || (1) QTRSRRRProof (EQUIVALENT)

popout

output may be truncated. 'popout' for the full output.

job log

popout

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