details

property	value
status	complete
benchmark	z29.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n056.star.cs.uiowa.edu
space	Zantema_05

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.23605298996 seconds
cpu usage	5.817101805
max memory	4.74513408E8

stage attributes

key	value
output-size	3828
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files -------------------------------------------------------------------------------- YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRSRRRProof [EQUIVALENT, 71 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 3 ms] (4) QDP (5) MRRProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPSizeChangeProof [EQUIVALENT, 0 ms] (8) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(lambda(x), y) -> lambda(a(x, 1)) a(lambda(x), y) -> lambda(a(x, a(y, t))) a(a(x, y), z) -> a(x, a(y, z)) lambda(x) -> x a(x, y) -> x a(x, y) -> y Q is empty. ---------------------------------------- (1) QTRSRRRProof (EQUIVALENT) Used ordering: Polynomial interpretation [POLO]: POL(1) = 0 POL(a(x_1, x_2)) = x_1 + x_2 POL(lambda(x_1)) = 1 + x_1 POL(t) = 0 With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly: lambda(x) -> x ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(lambda(x), y) -> lambda(a(x, 1)) a(lambda(x), y) -> lambda(a(x, a(y, t))) a(a(x, y), z) -> a(x, a(y, z)) a(x, y) -> x a(x, y) -> y Q is empty. ---------------------------------------- (3) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: A(lambda(x), y) -> A(x, 1) A(lambda(x), y) -> A(x, a(y, t)) A(lambda(x), y) -> A(y, t) A(a(x, y), z) -> A(x, a(y, z)) A(a(x, y), z) -> A(y, z) The TRS R consists of the following rules: a(lambda(x), y) -> lambda(a(x, 1)) a(lambda(x), y) -> lambda(a(x, a(y, t))) a(a(x, y), z) -> a(x, a(y, z)) a(x, y) -> x a(x, y) -> y Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- popout

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job log

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actions

all output return to TRS Standard