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HRS union beta 16688 pair #381734351
details
property
value
status
complete
benchmark
eval.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n023.star.cs.uiowa.edu
space
Mixed_HO_10
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
0.707914113998 seconds
cpu usage
1.486470506
max memory
9.9729408E7
stage attributes
key
value
output-size
6119
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: dom : [N * N * N] --> N eval : [N * N] --> N fun : [N -> N * N * N] --> N o : [] --> N s : [N] --> N Rules: eval(fun(f, x, y), z) => f dom(x, y, z) dom(s(x), s(y), s(z)) => s(dom(x, y, z)) dom(o, s(x), s(y)) => s(dom(o, x, y)) dom(x, y, o) => x dom(o, o, x) => o 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: dom(s(X), s(Y), s(Z)) => s(dom(X, Y, Z)) dom(o, s(X), s(Y)) => s(dom(o, X, Y)) dom(X, Y, o) => X dom(o, o, X) => o Moreover, the system is finitely branching. 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 Ce-terminating when seen as a many-sorted first-order TRS. According to the external first-order termination prover, this system is indeed Ce-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) QTRSRRRProof [EQUIVALENT] || (2) QTRS || (3) RisEmptyProof [EQUIVALENT] || (4) YES || || || ---------------------------------------- || || (0) || Obligation: || Q restricted rewrite system: || The TRS R consists of the following rules: || || dom(s(%X), s(%Y), s(%Z)) -> s(dom(%X, %Y, %Z)) || dom(o, s(%X), s(%Y)) -> s(dom(o, %X, %Y)) || dom(%X, %Y, o) -> %X || dom(o, o, %X) -> o || ~PAIR(%X, %Y) -> %X || ~PAIR(%X, %Y) -> %Y || || Q is empty. || || ---------------------------------------- || || (1) QTRSRRRProof (EQUIVALENT) || Used ordering: || Polynomial interpretation [POLO]: || || POL(dom(x_1, x_2, x_3)) = 1 + 2*x_1 + 2*x_2 + x_3 || POL(o) = 2 || POL(s(x_1)) = 1 + x_1 || POL(~PAIR(x_1, x_2)) = 2 + x_1 + x_2 || With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: || || dom(s(%X), s(%Y), s(%Z)) -> s(dom(%X, %Y, %Z)) || dom(o, s(%X), s(%Y)) -> s(dom(o, %X, %Y)) || dom(%X, %Y, o) -> %X || dom(o, o, %X) -> o || ~PAIR(%X, %Y) -> %X || ~PAIR(%X, %Y) -> %Y || || || || || ---------------------------------------- || || (2) || Obligation: || Q restricted rewrite system: || R is empty. || Q is empty. || || ---------------------------------------- || || (3) RisEmptyProof (EQUIVALENT)
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