SRS Stand 10685 pair #381717804

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	n034.star.cs.uiowa.edu
space	Secret_06_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	7.78728914261 seconds
cpu usage	27.877375346
max memory	4.247625728E9

stage attributes

key	value
output-size	7817
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) QDPOrderProof [EQUIVALENT, 40 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) AND
    (9) QDP
        (10) QDPOrderProof [EQUIVALENT, 12 ms]
        (11) QDP
        (12) DependencyGraphProof [EQUIVALENT, 0 ms]
        (13) TRUE
    (14) QDP
        (15) QDPOrderProof [EQUIVALENT, 87 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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