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HRS union beta 16688 pair #381734887
details
property
value
status
complete
benchmark
Applicative_first_order_05__#3.6.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n096.star.cs.uiowa.edu
space
Uncurried_Applicative_11
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
1.07931303978 seconds
cpu usage
2.401579535
max memory
1.44019456E8
stage attributes
key
value
output-size
23395
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 : [c * d] --> d false : [] --> a filter : [c -> a * d] --> d filter2 : [a * c -> a * c * d] --> d gcd : [b * b] --> b if!fac6220gcd : [a * b * b] --> b le : [b * b] --> a map : [c -> c * d] --> d minus : [b * b] --> b nil : [] --> d pred : [b] --> b s : [b] --> b true : [] --> a Rules: le(0, x) => true le(s(x), 0) => false le(s(x), s(y)) => le(x, y) pred(s(x)) => x minus(x, 0) => x minus(x, s(y)) => pred(minus(x, y)) gcd(0, x) => x gcd(s(x), 0) => s(x) gcd(s(x), s(y)) => if!fac6220gcd(le(y, x), s(x), s(y)) if!fac6220gcd(true, s(x), s(y)) => gcd(minus(x, y), s(y)) if!fac6220gcd(false, s(x), s(y)) => gcd(minus(y, x), s(x)) 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: le(0, X) => true le(s(X), 0) => false le(s(X), s(Y)) => le(X, Y) pred(s(X)) => X minus(X, 0) => X minus(X, s(Y)) => pred(minus(X, Y)) gcd(0, X) => X gcd(s(X), 0) => s(X) gcd(s(X), s(Y)) => if!fac6220gcd(le(Y, X), s(X), s(Y)) if!fac6220gcd(true, s(X), s(Y)) => gcd(minus(X, Y), s(Y)) if!fac6220gcd(false, s(X), s(Y)) => gcd(minus(Y, X), s(X)) 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 || (22) UsableRulesProof [EQUIVALENT] || (23) QDP || (24) QReductionProof [EQUIVALENT] || (25) QDP || (26) QDPOrderProof [EQUIVALENT] || (27) QDP
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