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SRS Standard pair #487092148
details
property
value
status
complete
benchmark
size-11-alpha-3-num-9.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n146.star.cs.uiowa.edu
space
Waldmann_07_size11
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
3.953 seconds
cpu usage
12.5824
user time
11.989
system time
0.5934
max virtual memory
1.00254788E8
max residence set size
1515244.0
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) QTRS Reverse [EQUIVALENT, 0 ms] (2) QTRS (3) DependencyPairsProof [EQUIVALENT, 0 ms] (4) QDP (5) DependencyGraphProof [EQUIVALENT, 0 ms] (6) QDP (7) QDPOrderProof [EQUIVALENT, 116 ms] (8) QDP (9) UsableRulesProof [EQUIVALENT, 0 ms] (10) QDP (11) QDPSizeChangeProof [EQUIVALENT, 1 ms] (12) YES ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(x1) -> b(x1) a(c(x1)) -> c(c(a(a(b(x1))))) b(b(x1)) -> 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) -> b(x1) c(a(x1)) -> b(a(a(c(c(x1))))) b(b(x1)) -> 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) -> B(x1) C(a(x1)) -> B(a(a(c(c(x1))))) C(a(x1)) -> A(a(c(c(x1)))) C(a(x1)) -> A(c(c(x1))) C(a(x1)) -> C(c(x1)) C(a(x1)) -> C(x1) The TRS R consists of the following rules: a(x1) -> b(x1) c(a(x1)) -> b(a(a(c(c(x1))))) b(b(x1)) -> 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 4 less nodes. ---------------------------------------- (6) Obligation: Q DP problem: The TRS P consists of the following rules: C(a(x1)) -> C(x1) C(a(x1)) -> C(c(x1)) The TRS R consists of the following rules: a(x1) -> b(x1) c(a(x1)) -> b(a(a(c(c(x1))))) b(b(x1)) -> x1 Q is empty. We have to consider all minimal (P,Q,R)-chains.
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