details

property	value
status	complete
benchmark	Applicative_05__mapDivMinusHard.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n145.star.cs.uiowa.edu
space	Uncurried_Applicative_11

run statistics

property	value
solver	Wanda 2.2a
configuration	default
runtime (wallclock)	0.234396 seconds
cpu usage	0.212154
user time	0.164851
system time	0.047303
max virtual memory	113188.0
max residence set size	7280.0

stage attributes

key	value
starexec-result	YES

output

YES We consider the system theBenchmark. Alphabet: 0 : [] --> c cons : [a * b] --> b div : [c * c] --> c map : [a -> a * b] --> b minus : [c * c] --> c nil : [] --> b p : [c] --> c s : [c] --> c Rules: map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, y)) minus(x, 0) => x minus(s(x), s(y)) => minus(p(s(x)), p(s(y))) p(s(x)) => x div(0, s(x)) => 0 div(s(x), s(y)) => s(div(minus(x, y), s(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(X, 0) => X minus(s(X), s(Y)) => minus(p(s(X)), p(s(Y))) p(s(X)) => X div(0, s(X)) => 0 div(s(X), s(Y)) => s(div(minus(X, Y), s(Y))) 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(PeRCenTX,0()) -> PeRCenTX || 2: minus(s(PeRCenTX),s(PeRCenTY)) -> minus(p(s(PeRCenTX)),p(s(PeRCenTY))) || 3: p(s(PeRCenTX)) -> PeRCenTX || 4: div(0(),s(PeRCenTX)) -> 0() || 5: div(s(PeRCenTX),s(PeRCenTY)) -> s(div(minus(PeRCenTX,PeRCenTY),s(PeRCenTY))) || Number of strict rules: 5 || Direct POLO(bPol) ... failed. || Uncurrying p || 1: minus(PeRCenTX,0()) -> PeRCenTX || 2: minus(s(PeRCenTX),s(PeRCenTY)) -> minus(p^1_s(PeRCenTX),p^1_s(PeRCenTY)) || 3: p^1_s(PeRCenTX) -> PeRCenTX || 4: div(0(),s(PeRCenTX)) -> 0() || 5: div(s(PeRCenTX),s(PeRCenTY)) -> s(div(minus(PeRCenTX,PeRCenTY),s(PeRCenTY))) || 6: p(s(_1)) ->= p^1_s(_1) || Number of strict rules: 5 || Direct POLO(bPol) ... failed. || Dependency Pairs: || #1: #minus(s(PeRCenTX),s(PeRCenTY)) -> #minus(p^1_s(PeRCenTX),p^1_s(PeRCenTY)) || #2: #minus(s(PeRCenTX),s(PeRCenTY)) -> #p^1_s(PeRCenTX) || #3: #minus(s(PeRCenTX),s(PeRCenTY)) -> #p^1_s(PeRCenTY) || #4: #p(s(_1)) ->? #p^1_s(_1) || #5: #div(s(PeRCenTX),s(PeRCenTY)) -> #div(minus(PeRCenTX,PeRCenTY),s(PeRCenTY)) || #6: #div(s(PeRCenTX),s(PeRCenTY)) -> #minus(PeRCenTX,PeRCenTY) || Number of SCCs: 2, DPs: 2 || SCC { #1 } || POLO(Sum)... succeeded. || #div w: 0 || s w: x1 + 2 || #p^1_s w: 0 || minus w: 0 || div w: 0 || p^1_s w: x1 + 1 || #p w: 0 || p w: 0 || 0 w: 0 || #minus w: x1 + x2 || USABLE RULES: { 3 } || Removed DPs: #1 || Number of SCCs: 1, DPs: 1 || SCC { #5 } || POLO(Sum)... succeeded. || #div w: x1 || s w: x1 + 2 || #p^1_s w: 0 || minus w: x1 + 1 || div w: 0 || p^1_s w: x1 + 1 || #p w: 0 || p w: 0 || 0 w: 1 || #minus w: x1 + x2 || USABLE RULES: { 1..3 } || Removed DPs: #5 || 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]). We thus obtain the following dependency pair problem (P_0, R_0, computable, formative): Dependency Pairs P_0: popout

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all output return to Higher Order Rewriting Union Beta