details

property	value
status	complete
benchmark	Applicative_first_order_05__perfect.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n010.star.cs.uiowa.edu
space	Uncurried_Applicative_11

run statistics

property	value
solver	Wanda
configuration	HigherOrder
runtime (wallclock)	0.963912963867 seconds
cpu usage	1.833493767
max memory	1.13983488E8

stage attributes

key	value
output-size	12398
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. Alphabet: 0 : [] --> b cons : [d * e] --> e f : [b * b * b * b] --> c false : [] --> c filter : [d -> c * e] --> e filter2 : [c * d -> c * d * e] --> e if : [a * c * c] --> c le : [b * b] --> a map : [d -> d * e] --> e minus : [b * b] --> b nil : [] --> e perfectp : [b] --> c s : [b] --> b true : [] --> c Rules: 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: 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: || proof of resources/system.trs || # AProVE Commit ID: d84c10301d352dfd14de2104819581f4682260f5 fuhs 20130616 || || || Termination w.r.t. Q of the given QTRS could be proven: || || (0) QTRS || (1) Overlay + Local Confluence [EQUIVALENT] || (2) QTRS || (3) DependencyPairsProof [EQUIVALENT] || (4) QDP || (5) DependencyGraphProof [EQUIVALENT] || (6) QDP || (7) UsableRulesProof [EQUIVALENT] || (8) QDP || (9) QReductionProof [EQUIVALENT] || (10) QDP || (11) QDPSizeChangeProof [EQUIVALENT] || (12) YES || || || ---------------------------------------- || || (0) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || 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)) || || Q is empty. || || ---------------------------------------- || || (1) Overlay + Local Confluence (EQUIVALENT) || The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. || ---------------------------------------- || || (2) popout

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

job log

popout

actions

all output return to HRS union beta 16688