SRS Standard pair #487090900

details

property	value
status	complete
benchmark	multum6.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n023.star.cs.uiowa.edu
space	Secret_06_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	5.20893 seconds
cpu usage	17.5682
user time	16.8016
system time	0.766655
max virtual memory	6.0599676E7
max residence set size	2442536.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 10 ms]
(4) QDP
(5) MRRProof [EQUIVALENT, 66 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) AND
    (9) QDP
        (10) QDPOrderProof [EQUIVALENT, 16 ms]
        (11) QDP
        (12) DependencyGraphProof [EQUIVALENT, 0 ms]
        (13) TRUE
    (14) QDP
        (15) QDPOrderProof [EQUIVALENT, 409 ms]
        (16) QDP
        (17) PisEmptyProof [EQUIVALENT, 0 ms]
        (18) YES


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

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

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

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(5) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

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