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	n147.star.cs.uiowa.edu
space	Mixed_HO_12

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	1.31718 seconds
cpu usage	1.27717
user time	1.1995
system time	0.07767
max virtual memory	135472.0
max residence set size	19900.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: 0 : [] --> nat app : [natlist * natlist] --> natlist apply2 : [pair -> nat -> pair * pair * nat] --> pair cons : [nat * natlist] --> natlist fst : [pair] --> natlist nil : [] --> natlist p : [natlist * natlist] --> pair pcons : [pair * plist] --> plist pnil : [] --> plist pps : [natlist] --> plist prefixshuffle : [pair * natlist] --> plist pshuffle : [natlist] --> pair reverse : [natlist] --> natlist s : [nat] --> nat shuffle : [natlist] --> natlist Rules: app(nil, x) => x app(cons(x, y), z) => cons(x, app(y, z)) reverse(nil) => nil reverse(cons(x, y)) => app(reverse(y), cons(x, nil)) shuffle(nil) => nil shuffle(cons(x, y)) => cons(x, shuffle(reverse(y))) fst(p(x, y)) => x pshuffle(x) => p(x, shuffle(x)) prefixshuffle(x, nil) => pcons(x, pnil) prefixshuffle(x, cons(y, z)) => pcons(x, prefixshuffle(apply2(/\u./\v.pshuffle(app(fst(u), cons(v, nil))), x, y), reverse(z))) apply2(f, x, 0) => x apply2(f, x, s(y)) => f x s(y) pps(x) => prefixshuffle(p(nil, nil), x) This AFS is converted to an AFSM simply by replacing all free variables by meta-variables (with arity 0). We observe that the rules contain a first-order subset: app(nil, X) => X app(cons(X, Y), Z) => cons(X, app(Y, Z)) reverse(nil) => nil reverse(cons(X, Y)) => app(reverse(Y), cons(X, nil)) shuffle(nil) => nil shuffle(cons(X, Y)) => cons(X, shuffle(reverse(Y))) fst(p(X, Y)) => X pshuffle(X) => p(X, shuffle(X)) 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. According to the external first-order termination prover, this system is indeed terminating: || Input TRS: || 1: app(nil(),PeRCenTX) -> PeRCenTX || 2: app(cons(PeRCenTX,PeRCenTY),PeRCenTZ) -> cons(PeRCenTX,app(PeRCenTY,PeRCenTZ)) || 3: reverse(nil()) -> nil() || 4: reverse(cons(PeRCenTX,PeRCenTY)) -> app(reverse(PeRCenTY),cons(PeRCenTX,nil())) || 5: shuffle(nil()) -> nil() || 6: shuffle(cons(PeRCenTX,PeRCenTY)) -> cons(PeRCenTX,shuffle(reverse(PeRCenTY))) || 7: fst(p(PeRCenTX,PeRCenTY)) -> PeRCenTX || 8: pshuffle(PeRCenTX) -> p(PeRCenTX,shuffle(PeRCenTX)) || Number of strict rules: 8 || Direct POLO(bPol) ... removes: 8 5 7 || fst w: 2 * x1 + 1 || reverse w: x1 || shuffle w: x1 + 1 || p w: x1 + x2 + 1 || pshuffle w: 2 * x1 + 3 || nil w: 0 || cons w: x1 + x2 + 3 || app w: x1 + x2 || Number of strict rules: 5 || Direct POLO(bPol) ... removes: 6 || fst w: 2 * x1 + 1 || reverse w: x1 || shuffle w: 2 * x1 + 1 || p w: x1 + x2 + 1 || pshuffle w: 2 * x1 + 3 || nil w: 0 || cons w: x1 + x2 + 1 || app w: x1 + x2 || Number of strict rules: 4 || Direct POLO(bPol) ... removes: 4 3 || fst w: 2 * x1 + 1 || reverse w: 2 * x1 + 1 || shuffle w: x1 + 1 || p w: x1 + x2 + 1 || pshuffle w: 2 * x1 + 3 || nil w: 0 || cons w: x1 + x2 + 1 || app w: x1 + x2 || Number of strict rules: 2 || Direct POLO(bPol) ... removes: 1 || fst w: 2 * x1 + 1 || reverse w: 2 * x1 + 1 || shuffle w: x1 + 1 || p w: x1 + x2 + 1 || pshuffle w: 2 * x1 + 3 popout

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actions

all output return to Higher Order Rewriting Union Beta