SRS Standard pair #516976474

details

property	value
status	complete
benchmark	random-55.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n081.star.cs.uiowa.edu
space	Waldmann_19

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	7.53080201149 seconds
cpu usage	26.589005737
max memory	1.514254336E9

stage attributes

key	value
output-size	10246
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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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, 3 ms]
(4) QDP
(5) QDPOrderProof [EQUIVALENT, 40 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) AND
    (9) QDP
        (10) QDPOrderProof [EQUIVALENT, 63 ms]
        (11) QDP
        (12) QDPOrderProof [EQUIVALENT, 184 ms]
        (13) QDP
        (14) PisEmptyProof [EQUIVALENT, 0 ms]
        (15) YES
    (16) QDP
        (17) QDPOrderProof [EQUIVALENT, 9 ms]
        (18) QDP
        (19) QDPOrderProof [EQUIVALENT, 162 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(b(b(a(x1)))) -> b(a(a(a(x1))))
   b(b(b(b(x1)))) -> b(a(a(b(x1))))
   a(a(b(a(x1)))) -> b(a(b(a(x1))))
   b(a(a(b(x1)))) -> a(a(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:

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

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

The TRS R consists of the following rules:

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

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