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HRS union beta 16688 pair #381734947
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
n088.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
1.13749194145 seconds
cpu usage
2.548838189
max memory
1.57933568E8
stage attributes
key
value
output-size
25564
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 : [] --> 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: || 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) QReductionProof [EQUIVALENT] || (18) QDP || (19) QDPSizeChangeProof [EQUIVALENT] || (20) YES || (21) QDP
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