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SRS Standard pair #487083100
details
property
value
status
complete
benchmark
12.xml
ran by
Akihisa Yamada
cpu timeout
1200 seconds
wallclock timeout
300 seconds
memory limit
137438953472 bytes
execution host
n140.star.cs.uiowa.edu
space
Zantema_06
run statistics
property
value
solver
AProVE
configuration
standard
runtime (wallclock)
64.0923 seconds
cpu usage
248.397
user time
243.363
system time
5.03348
max virtual memory
4.2159156E7
max residence set size
6219116.0
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty Termination w.r.t. Q of the given QTRS could be proven: (0) QTRS (1) DependencyPairsProof [EQUIVALENT, 10 ms] (2) QDP (3) DependencyGraphProof [EQUIVALENT, 0 ms] (4) QDP (5) QDPOrderProof [EQUIVALENT, 55 ms] (6) QDP (7) DependencyGraphProof [EQUIVALENT, 0 ms] (8) QDP (9) QDPOrderProof [EQUIVALENT, 19 ms] (10) QDP (11) QDPOrderProof [EQUIVALENT, 0 ms] (12) QDP (13) QDPOrderProof [EQUIVALENT, 483 ms] (14) QDP (15) DependencyGraphProof [EQUIVALENT, 0 ms] (16) TRUE ---------------------------------------- (0) Obligation: Q restricted rewrite system: The TRS R consists of the following rules: a(l(x1)) -> l(a(x1)) a(c(x1)) -> c(a(x1)) c(a(r(x1))) -> r(a(x1)) l(r(a(a(x1)))) -> a(a(l(c(c(c(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: A(l(x1)) -> L(a(x1)) A(l(x1)) -> A(x1) A(c(x1)) -> C(a(x1)) A(c(x1)) -> A(x1) C(a(r(x1))) -> A(x1) L(r(a(a(x1)))) -> A(a(l(c(c(c(r(x1))))))) L(r(a(a(x1)))) -> A(l(c(c(c(r(x1)))))) L(r(a(a(x1)))) -> L(c(c(c(r(x1))))) L(r(a(a(x1)))) -> C(c(c(r(x1)))) L(r(a(a(x1)))) -> C(c(r(x1))) L(r(a(a(x1)))) -> C(r(x1)) The TRS R consists of the following rules: a(l(x1)) -> l(a(x1)) a(c(x1)) -> c(a(x1)) c(a(r(x1))) -> r(a(x1)) l(r(a(a(x1)))) -> a(a(l(c(c(c(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 1 SCC with 4 less nodes. ---------------------------------------- (4) Obligation: Q DP problem: The TRS P consists of the following rules: L(r(a(a(x1)))) -> A(a(l(c(c(c(r(x1))))))) A(l(x1)) -> L(a(x1)) L(r(a(a(x1)))) -> A(l(c(c(c(r(x1)))))) A(l(x1)) -> A(x1) A(c(x1)) -> C(a(x1)) C(a(r(x1))) -> A(x1) A(c(x1)) -> A(x1) The TRS R consists of the following rules: a(l(x1)) -> l(a(x1)) a(c(x1)) -> c(a(x1)) c(a(r(x1))) -> r(a(x1)) l(r(a(a(x1)))) -> a(a(l(c(c(c(r(x1))))))) Q is empty. We have to consider all minimal (P,Q,R)-chains.
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