SRS Stand 10685 pair #381717180

details

property	value
status	complete
benchmark	03.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n024.star.cs.uiowa.edu
space	Zantema_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	41.8831391335 seconds
cpu usage	164.144794176
max memory	5.73329408E9

stage attributes

key	value
output-size	7715
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 17 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 1 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) QDPOrderProof [EQUIVALENT, 36 ms]
        (12) QDP
        (13) QDPOrderProof [EQUIVALENT, 2745 ms]
        (14) QDP
        (15) QDPOrderProof [EQUIVALENT, 3380 ms]
        (16) QDP
        (17) PisEmptyProof [EQUIVALENT, 0 ms]
        (18) YES


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(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   a(a(b(b(x1)))) -> b(b(b(a(a(a(x1))))))
   a(c(x1)) -> c(a(x1))
   c(b(x1)) -> b(c(x1))

Q is empty.

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(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(a(b(b(x1)))) -> A(a(a(x1)))
   A(a(b(b(x1)))) -> A(a(x1))
   A(a(b(b(x1)))) -> A(x1)
   A(c(x1)) -> C(a(x1))
   A(c(x1)) -> A(x1)
   C(b(x1)) -> C(x1)

The TRS R consists of the following rules:

   a(a(b(b(x1)))) -> b(b(b(a(a(a(x1))))))
   a(c(x1)) -> c(a(x1))
   c(b(x1)) -> b(c(x1))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
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(4)
Complex Obligation (AND)

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(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   C(b(x1)) -> C(x1)

The TRS R consists of the following rules:

   a(a(b(b(x1)))) -> b(b(b(a(a(a(x1))))))
   a(c(x1)) -> c(a(x1))
   c(b(x1)) -> b(c(x1))

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