SRS Stand 10685 pair #381717460

details

property	value
status	complete
benchmark	beans3.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n071.star.cs.uiowa.edu
space	Zantema_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.34508895874 seconds
cpu usage	9.442193365
max memory	7.86870272E8

stage attributes

key	value
output-size	10933
starexec-result	YES

output

/export/starexec/sandbox/solver/bin/starexec_run_standard /export/starexec/sandbox/benchmark/theBenchmark.xml /export/starexec/sandbox/output/output_files


--------------------------------------------------------------------------------


YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 2 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 3 ms]
        (7) QDP
        (8) QDPOrderProof [EQUIVALENT, 23 ms]
        (9) QDP
        (10) DependencyGraphProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) QDPOrderProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) DependencyGraphProof [EQUIVALENT, 0 ms]
        (15) TRUE
    (16) QDP
        (17) UsableRulesProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QDPOrderProof [EQUIVALENT, 67 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))
   b(c(a(x1))) -> a(b(c(x1)))
   c(b(x1)) -> d(x1)
   a(d(x1)) -> d(a(x1))
   d(x1) -> b(a(x1))
   L(a(a(x1))) -> L(a(b(c(x1))))
   c(R(x1)) -> c(b(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))) -> A(b(c(x1)))
   B(a(a(x1))) -> B(c(x1))
   B(a(a(x1))) -> C(x1)
   C(a(x1)) -> A(c(x1))
   C(a(x1)) -> C(x1)
   B(c(a(x1))) -> A(b(c(x1)))
   B(c(a(x1))) -> B(c(x1))
   B(c(a(x1))) -> C(x1)
   C(b(x1)) -> D(x1)
   A(d(x1)) -> D(a(x1))
   A(d(x1)) -> A(x1)
   D(x1) -> B(a(x1))
   D(x1) -> A(x1)
   L^1(a(a(x1))) -> L^1(a(b(c(x1))))
   L^1(a(a(x1))) -> A(b(c(x1)))
   L^1(a(a(x1))) -> B(c(x1))
   L^1(a(a(x1))) -> C(x1)
   C(R(x1)) -> C(b(R(x1)))
   C(R(x1)) -> B(R(x1))

The TRS R consists of the following rules:

   b(a(a(x1))) -> a(b(c(x1)))
   c(a(x1)) -> a(c(x1))
   b(c(a(x1))) -> a(b(c(x1)))
   c(b(x1)) -> d(x1)
   a(d(x1)) -> d(a(x1))
   d(x1) -> b(a(x1))
   L(a(a(x1))) -> L(a(b(c(x1))))
   c(R(x1)) -> c(b(R(x1)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

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