details

property	value
status	complete
benchmark	size-12-alpha-3-num-456.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n141.star.cs.uiowa.edu
space	Waldmann_07_size12

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	4.32763 seconds
cpu usage	14.0679
user time	13.4266
system time	0.641336
max virtual memory	9.8916088E7
max residence set size	1950180.0

stage attributes

key	value
starexec-result	YES

output

YES proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml # AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 115 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 51 ms] (12) QDP (13) UsableRulesProof [EQUIVALENT, 2 ms] (14) QDP (15) QDPSizeChangeProof [EQUIVALENT, 1 ms] (16) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(x1) -> b(c(x1)) b(b(x1)) -> x1 b(c(c(x1))) -> c(c(a(b(x1)))) Q is empty. ---------------------------------------- (1) QTRS Reverse (EQUIVALENT) We applied the QTRS Reverse Processor [REVERSE]. ---------------------------------------- (2) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(x1) -> c(b(x1)) b(b(x1)) -> x1 c(c(b(x1))) -> b(a(c(c(x1)))) 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(x1) -> C(b(x1)) A(x1) -> B(x1) C(c(b(x1))) -> B(a(c(c(x1)))) C(c(b(x1))) -> A(c(c(x1))) C(c(b(x1))) -> C(c(x1)) C(c(b(x1))) -> C(x1) The TRS R consists of the following rules: a(x1) -> c(b(x1)) b(b(x1)) -> x1 c(c(b(x1))) -> b(a(c(c(x1)))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (5) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: C(c(b(x1))) -> A(c(c(x1))) A(x1) -> C(b(x1)) C(c(b(x1))) -> C(c(x1)) C(c(b(x1))) -> C(x1) The TRS R consists of the following rules: popout

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

job log

popout

actions

all output return to SRS Standard