details

property	value
status	complete
benchmark	Applicative_first_order_05__perfect2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n149.star.cs.uiowa.edu
space	Uncurried_Applicative_11

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	0.546462 seconds
cpu usage	0.516283
user time	0.463422
system time	0.052861
max virtual memory	130700.0
max residence set size	15400.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: 0 : [] --> a cons : [c * d] --> d f : [a * a * a * a] --> b false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d if : [b * b * b] --> b le : [a * a] --> b map : [c -> c * d] --> d minus : [a * a] --> a nil : [] --> d perfectp : [a] --> b s : [a] --> a true : [] --> b Rules: minus(0, x) => 0 minus(s(x), 0) => s(x) minus(s(x), s(y)) => minus(x, y) le(0, x) => true le(s(x), 0) => false le(s(x), s(y)) => le(x, y) if(true, x, y) => x if(false, x, y) => y perfectp(0) => false perfectp(s(x)) => f(x, s(0), s(x), s(x)) f(0, x, 0, y) => true f(0, x, s(y), z) => false f(s(x), 0, y, z) => f(x, z, minus(y, s(x)), z) f(s(x), s(y), z, u) => if(le(x, y), f(s(x), minus(y, x), z, u), f(x, u, z, u)) map(g, nil) => nil map(g, cons(x, y)) => cons(g x, map(g, y)) filter(g, nil) => nil filter(g, cons(x, y)) => filter2(g x, g, x, y) filter2(true, g, x, y) => cons(x, filter(g, y)) filter2(false, g, x, y) => filter(g, y) 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: minus(0, X) => 0 minus(s(X), 0) => s(X) minus(s(X), s(Y)) => minus(X, Y) le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) if(true, X, Y) => X if(false, X, Y) => Y perfectp(0) => false perfectp(s(X)) => f(X, s(0), s(X), s(X)) f(0, X, 0, Y) => true f(0, X, s(Y), Z) => false f(s(X), 0, Y, Z) => f(X, Z, minus(Y, s(X)), Z) f(s(X), s(Y), Z, U) => if(le(X, Y), f(s(X), minus(Y, X), Z, U), f(X, U, Z, U)) 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: minus(0(),PeRCenTX) -> 0() || 2: minus(s(PeRCenTX),0()) -> s(PeRCenTX) || 3: minus(s(PeRCenTX),s(PeRCenTY)) -> minus(PeRCenTX,PeRCenTY) || 4: le(0(),PeRCenTX) -> true() || 5: le(s(PeRCenTX),0()) -> false() || 6: le(s(PeRCenTX),s(PeRCenTY)) -> le(PeRCenTX,PeRCenTY) || 7: if(true(),PeRCenTX,PeRCenTY) -> PeRCenTX || 8: if(false(),PeRCenTX,PeRCenTY) -> PeRCenTY || 9: perfectp(0()) -> false() || 10: perfectp(s(PeRCenTX)) -> f(PeRCenTX,s(0()),s(PeRCenTX),s(PeRCenTX)) || 11: f(0(),PeRCenTX,0(),PeRCenTY) -> true() || 12: f(0(),PeRCenTX,s(PeRCenTY),PeRCenTZ) -> false() || 13: f(s(PeRCenTX),0(),PeRCenTY,PeRCenTZ) -> f(PeRCenTX,PeRCenTZ,minus(PeRCenTY,s(PeRCenTX)),PeRCenTZ) || 14: f(s(PeRCenTX),s(PeRCenTY),PeRCenTZ,PeRCenTU) -> if(le(PeRCenTX,PeRCenTY),f(s(PeRCenTX),minus(PeRCenTY,PeRCenTX),PeRCenTZ,PeRCenTU),f(PeRCenTX,PeRCenTU,PeRCenTZ,PeRCenTU)) || Number of strict rules: 14 || Direct POLO(bPol) ... failed. || Uncurrying f || 1: minus(0(),PeRCenTX) -> 0() || 2: minus(s(PeRCenTX),0()) -> s(PeRCenTX) || 3: minus(s(PeRCenTX),s(PeRCenTY)) -> minus(PeRCenTX,PeRCenTY) || 4: le(0(),PeRCenTX) -> true() || 5: le(s(PeRCenTX),0()) -> false() || 6: le(s(PeRCenTX),s(PeRCenTY)) -> le(PeRCenTX,PeRCenTY) || 7: if(true(),PeRCenTX,PeRCenTY) -> PeRCenTX || 8: if(false(),PeRCenTX,PeRCenTY) -> PeRCenTY || 9: perfectp(0()) -> false() || 10: perfectp(s(PeRCenTX)) -> f(PeRCenTX,s(0()),s(PeRCenTX),s(PeRCenTX)) || 11: f^1_0(PeRCenTX,0(),PeRCenTY) -> true() || 12: f^1_0(PeRCenTX,s(PeRCenTY),PeRCenTZ) -> false() || 13: f^1_s(PeRCenTX,0(),PeRCenTY,PeRCenTZ) -> f(PeRCenTX,PeRCenTZ,minus(PeRCenTY,s(PeRCenTX)),PeRCenTZ) || 14: f^1_s(PeRCenTX,s(PeRCenTY),PeRCenTZ,PeRCenTU) -> if(le(PeRCenTX,PeRCenTY),f^1_s(PeRCenTX,minus(PeRCenTY,PeRCenTX),PeRCenTZ,PeRCenTU),f(PeRCenTX,PeRCenTU,PeRCenTZ,PeRCenTU)) || 15: f(0(),_3,_4,_5) ->= f^1_0(_3,_4,_5) || 16: f(s(_1),_4,_5,_6) ->= f^1_s(_1,_4,_5,_6) popout

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

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all output return to Higher Order Rewriting Union Beta