details

property	value
status	complete
benchmark	findzero.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n073.star.cs.uiowa.edu
space	Mixed_HO_10

run statistics

property	value
solver	Wanda
configuration	HigherOrder
runtime (wallclock)	1.01620388031 seconds
cpu usage	1.688754742
max memory	9.3720576E7

stage attributes

key	value
output-size	18558
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 : [] --> nat find0 : [nat -> nat * nat * nat] --> nat if : [nat * nat * nat] --> nat min : [nat * nat] --> nat nul : [nat -> nat * nat] --> nat s : [nat] --> nat Rules: min(s(x), s(y)) => min(x, y) min(x, 0) => 0 min(0, x) => 0 min(nul(f, x), y) => nul(f, min(x, y)) nul(f, x) => find0(f, 0, x) find0(f, x, 0) => x find0(f, x, s(y)) => if(f x, find0(f, s(x), y), x) if(s(x), y, z) => y if(0, x, y) => 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: if(s(X), Y, Z) => Y if(0, X, Y) => Y 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: || || if(s(%X), %Y, %Z) -> %Y || if(0, %X, %Y) -> %Y || ~PAIR(%X, %Y) -> %X || ~PAIR(%X, %Y) -> %Y || || Q is empty. || || ---------------------------------------- || || (1) QTRSRRRProof (EQUIVALENT) || Used ordering: || Polynomial interpretation [POLO]: || || POL(0) = 2 || POL(if(x_1, x_2, x_3)) = 2 + 2*x_1 + x_2 + x_3 || POL(s(x_1)) = 2 + 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: || || if(s(%X), %Y, %Z) -> %Y || if(0, %X, %Y) -> %Y || ~PAIR(%X, %Y) -> %X || ~PAIR(%X, %Y) -> %Y || || || || || ---------------------------------------- || || (2) || Obligation: || Q restricted rewrite system: || R is empty. || Q is empty. || || ---------------------------------------- || || (3) RisEmptyProof (EQUIVALENT) || The TRS R is empty. Hence, termination is trivially proven. popout

output may be truncated. 'popout' for the full output.

job log

popout

actions

all output return to HRS union beta 16688