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SRS Standard pair #487083016
details
property
value
status
complete
benchmark
beans2.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n142.star.cs.uiowa.edu
space
Zantema_06
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
3.18842 seconds
cpu usage
9.48226
user time
9.01349
system time
0.468769
max virtual memory
1.23895504E8
max residence set size
1126464.0
stage attributes
key
value
starexec-result
YES
output
YES proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 15 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) AND (5) QDP (6) UsableRulesProof [EQUIVALENT, 0 ms] (7) QDP (8) QDPOrderProof [EQUIVALENT, 31 ms] (9) QDP (10) DependencyGraphProof [EQUIVALENT, 0 ms] (11) TRUE (12) QDP (13) MNOCProof [EQUIVALENT, 0 ms] (14) QDP (15) UsableRulesProof [EQUIVALENT, 0 ms] (16) QDP (17) QReductionProof [EQUIVALENT, 0 ms] (18) QDP (19) QDPOrderProof [EQUIVALENT, 35 ms] (20) QDP (21) PisEmptyProof [EQUIVALENT, 0 ms] (22) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: b(a(a(x1))) -> a(b(c(x1))) c(a(x1)) -> a(c(x1)) c(b(x1)) -> b(a(x1)) L(a(a(x1))) -> L(a(b(c(x1)))) c(R(x1)) -> b(a(R(x1))) Q is empty. ---------------------------------------- (1) DependencyPairsProof (EQUIVALENT) Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem. ---------------------------------------- (2) 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)) -> C(x1) C(b(x1)) -> B(a(x1)) L^1(a(a(x1))) -> L^1(a(b(c(x1)))) L^1(a(a(x1))) -> B(c(x1)) L^1(a(a(x1))) -> C(x1) C(R(x1)) -> B(a(R(x1))) The TRS R consists of the following rules: b(a(a(x1))) -> a(b(c(x1))) c(a(x1)) -> a(c(x1)) c(b(x1)) -> b(a(x1)) L(a(a(x1))) -> L(a(b(c(x1)))) c(R(x1)) -> b(a(R(x1))) Q is empty. We have to consider all minimal (P,Q,R)-chains. ---------------------------------------- (3) DependencyGraphProof (EQUIVALENT) The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes. ---------------------------------------- (4) Complex Obligation (AND) ---------------------------------------- (5) Obligation: Q DP problem: The TRS P consists of the following rules: B(a(a(x1))) -> C(x1) C(a(x1)) -> C(x1) C(b(x1)) -> B(a(x1)) B(a(a(x1))) -> B(c(x1)) The TRS R consists of the following rules:
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