details

property	value
status	complete
benchmark	from.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n140.star.cs.uiowa.edu
space	Mixed_HO_10

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	0.68933 seconds
cpu usage	0.674832
user time	0.590361
system time	0.084471
max virtual memory	131872.0
max residence set size	17148.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: chain : [N -> N * list] --> list cons : [N * list] --> list false : [] --> B from : [N * list] --> list if : [B * list * list] --> list incch : [list] --> list lteq : [N * N] --> B nil : [] --> list o : [] --> N s : [N] --> N true : [] --> B Rules: if(true, x, y) => x if(false, x, y) => y lteq(s(x), o) => false lteq(o, x) => true lteq(s(x), s(y)) => lteq(x, y) from(x, nil) => nil from(x, cons(y, z)) => if(lteq(x, y), cons(y, z), from(x, z)) chain(f, nil) => nil chain(f, cons(x, y)) => cons(f x, chain(f, from(f x, y))) incch(x) => chain(/\y.s(y), 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: if(true, X, Y) => X if(false, X, Y) => Y lteq(s(X), o) => false lteq(o, X) => true lteq(s(X), s(Y)) => lteq(X, Y) from(X, nil) => nil from(X, cons(Y, Z)) => if(lteq(X, Y), cons(Y, Z), from(X, Z)) 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: if(true(),PeRCenTX,PeRCenTY) -> PeRCenTX || 2: if(false(),PeRCenTX,PeRCenTY) -> PeRCenTY || 3: lteq(s(PeRCenTX),o()) -> false() || 4: lteq(o(),PeRCenTX) -> true() || 5: lteq(s(PeRCenTX),s(PeRCenTY)) -> lteq(PeRCenTX,PeRCenTY) || 6: from(PeRCenTX,nil()) -> nil() || 7: from(PeRCenTX,cons(PeRCenTY,PeRCenTZ)) -> if(lteq(PeRCenTX,PeRCenTY),cons(PeRCenTY,PeRCenTZ),from(PeRCenTX,PeRCenTZ)) || Number of strict rules: 7 || Direct POLO(bPol) ... failed. || Uncurrying ... failed. || Dependency Pairs: || #1: #from(PeRCenTX,cons(PeRCenTY,PeRCenTZ)) -> #if(lteq(PeRCenTX,PeRCenTY),cons(PeRCenTY,PeRCenTZ),from(PeRCenTX,PeRCenTZ)) || #2: #from(PeRCenTX,cons(PeRCenTY,PeRCenTZ)) -> #lteq(PeRCenTX,PeRCenTY) || #3: #from(PeRCenTX,cons(PeRCenTY,PeRCenTZ)) -> #from(PeRCenTX,PeRCenTZ) || #4: #lteq(s(PeRCenTX),s(PeRCenTY)) -> #lteq(PeRCenTX,PeRCenTY) || Number of SCCs: 2, DPs: 2 || SCC { #3 } || POLO(Sum)... succeeded. || s w: 0 || #lteq w: 0 || false w: 0 || true w: 0 || o w: 0 || lteq w: 0 || if w: 0 || from w: 0 || nil w: 0 || #from w: x2 || cons w: x2 + 1 || #if w: 0 || USABLE RULES: { } || Removed DPs: #3 || Number of SCCs: 1, DPs: 1 || SCC { #4 } || POLO(Sum)... succeeded. || s w: x1 + 1 || #lteq w: x1 || false w: 0 || true w: 0 || o w: 0 || lteq w: 0 || if w: 0 || from w: 0 || nil w: 0 || #from w: 0 || cons w: 1 || #if w: 0 || USABLE RULES: { } || Removed DPs: #4 || Number of SCCs: 0, DPs: 0 || We use the dependency pair framework as described in [Kop12, Ch. 6/7], with static dependency pairs (see [KusIsoSakBla09] and the adaptation for AFSMs and accessible arguments in [FuhKop19]). popout

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

job log

popout

actions

all output return to Higher Order Rewriting Union Beta