details

property	value
status	complete
benchmark	size-12-alpha-3-num-536.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n139.star.cs.uiowa.edu
space	Waldmann_07_size12

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	4.27495 seconds
cpu usage	13.2179
user time	12.5912
system time	0.626716
max virtual memory	4.0331716E7
max residence set size	1579140.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) Overlay + Local Confluence [EQUIVALENT, 0 ms] (4) QTRS (5) DependencyPairsProof [EQUIVALENT, 8 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 172 ms] (8) QDP (9) DependencyGraphProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 113 ms] (12) QDP (13) PisEmptyProof [EQUIVALENT, 0 ms] (14) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(a(b(x1))) -> c(b(a(a(a(x1))))) a(c(x1)) -> b(a(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: b(a(a(x1))) -> a(a(a(b(c(x1))))) c(a(x1)) -> a(b(x1)) Q is empty. ---------------------------------------- (3) Overlay + Local Confluence (EQUIVALENT) The TRS is overlay and locally confluent. By [NOC] we can switch to innermost. ---------------------------------------- (4) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(a(a(x1))) -> a(a(a(b(c(x1))))) c(a(x1)) -> a(b(x1)) The set Q consists of the following terms: b(a(a(x0))) c(a(x0)) ---------------------------------------- (5) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: B(a(a(x1))) -> B(c(x1)) B(a(a(x1))) -> C(x1) C(a(x1)) -> B(x1) The TRS R consists of the following rules: b(a(a(x1))) -> a(a(a(b(c(x1))))) c(a(x1)) -> a(b(x1)) The set Q consists of the following terms: b(a(a(x0))) c(a(x0)) We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- popout

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all output return to SRS Standard