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HRS union beta 16688 pair #381734291
details
property
value
status
complete
benchmark
lambda_prod.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n109.star.cs.uiowa.edu
space
Hamana_17
run statistics
property
value
solver
Wanda
configuration
HigherOrder
runtime (wallclock)
0.689041852951 seconds
cpu usage
1.343351536
max memory
9.4130176E7
stage attributes
key
value
output-size
6057
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: app : [] --> arrab -> a -> b lam : [] --> a -> b -> arrab pair : [] --> a -> b -> prodab pia : [] --> prodab -> a pib : [] --> prodab -> b Rules: app (lam (/\x.f x)) y => f y lam (/\x.app y x) => y pia (pair x y) => x pib (pair x y) => y pair (pia x) (pib x) => x Using the transformations described in [Kop11], this system can be brought in a form without leading free variables in the left-hand side, and where the left-hand side of a variable is always a functional term or application headed by a functional term. We now transform the resulting AFS into an AFSM by replacing all free variables by meta-variables (with arity 0). This leads to the following AFSM: Alphabet: app : [arrab * a] --> b lam : [a -> b] --> arrab pair : [a * b] --> prodab pia : [prodab] --> a pib : [prodab] --> b ~AP1 : [a -> b * a] --> b Rules: app(lam(/\x.~AP1(F, x)), X) => ~AP1(F, X) lam(/\x.app(X, x)) => X pia(pair(X, Y)) => X pib(pair(X, Y)) => Y pair(pia(X), pib(X)) => X app(lam(/\x.app(X, x)), Y) => app(X, Y) ~AP1(F, X) => F X We observe that the rules contain a first-order subset: pia(pair(X, Y)) => X pib(pair(X, Y)) => Y pair(pia(X), pib(X)) => X 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: || || pia(pair(%X, %Y)) -> %X || pib(pair(%X, %Y)) -> %Y || pair(pia(%X), pib(%X)) -> %X || ~PAIR(%X, %Y) -> %X || ~PAIR(%X, %Y) -> %Y || || Q is empty. || || ---------------------------------------- || || (1) QTRSRRRProof (EQUIVALENT) || Used ordering: || Polynomial interpretation [POLO]: || || POL(pair(x_1, x_2)) = 1 + x_1 + x_2 || POL(pia(x_1)) = 1 + 2*x_1 || POL(pib(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: || || pia(pair(%X, %Y)) -> %X || pib(pair(%X, %Y)) -> %Y
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