Spaces
Explore
Communities
Statistics
Reports
Cluster
Status
Help
HRS union beta 16688 pair #381734866
details
property
value
status
complete
benchmark
Applicative_first_order_05__#3.38.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n114.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
1.0218269825 seconds
cpu usage
2.191906997
max memory
9.814016E7
stage attributes
key
value
output-size
16328
starexec-result
YES
output
/export/starexec/sandbox/solver/bin/starexec_run_HigherOrder /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files -------------------------------------------------------------------------------- YES We consider the system theBenchmark. Alphabet: 0 : [] --> c cons : [c * d] --> d false : [] --> b filter : [c -> b * d] --> d filter2 : [b * c -> b * c * d] --> d map : [c -> c * d] --> d nil : [] --> d rev : [d] --> d rev1 : [c * d] --> c rev2 : [c * d] --> d s : [a] --> c true : [] --> b Rules: rev(nil) => nil rev(cons(x, y)) => cons(rev1(x, y), rev2(x, y)) rev1(0, nil) => 0 rev1(s(x), nil) => s(x) rev1(x, cons(y, z)) => rev1(y, z) rev2(x, nil) => nil rev2(x, cons(y, z)) => rev(cons(x, rev2(y, z))) map(f, nil) => nil map(f, cons(x, y)) => cons(f x, map(f, y)) filter(f, nil) => nil filter(f, cons(x, y)) => filter2(f x, f, x, y) filter2(true, f, x, y) => cons(x, filter(f, y)) filter2(false, f, x, y) => filter(f, 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: rev(nil) => nil rev(cons(X, Y)) => cons(rev1(X, Y), rev2(X, Y)) rev1(0, nil) => 0 rev1(s(X), nil) => s(X) rev1(X, cons(Y, Z)) => rev1(Y, Z) rev2(X, nil) => nil rev2(X, cons(Y, Z)) => rev(cons(X, rev2(Y, 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: || 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) AND || (7) QDP || (8) UsableRulesProof [EQUIVALENT] || (9) QDP || (10) QReductionProof [EQUIVALENT] || (11) QDP || (12) QDPSizeChangeProof [EQUIVALENT] || (13) YES || (14) QDP || (15) UsableRulesProof [EQUIVALENT] || (16) QDP || (17) QDPOrderProof [EQUIVALENT] || (18) QDP || (19) DependencyGraphProof [EQUIVALENT] || (20) TRUE || || || ---------------------------------------- || || (0) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || rev(nil) -> nil || rev(cons(%X, %Y)) -> cons(rev1(%X, %Y), rev2(%X, %Y)) || rev1(0, nil) -> 0 || rev1(s(%X), nil) -> s(%X) || rev1(%X, cons(%Y, %Z)) -> rev1(%Y, %Z) || rev2(%X, nil) -> nil || rev2(%X, cons(%Y, %Z)) -> rev(cons(%X, rev2(%Y, %Z))) ||
popout
output may be truncated. 'popout' for the full output.
job log
popout
actions
all output
return to HRS union beta 16688