SRS Stand 10685 pair #381717942

details

property	value
status	complete
benchmark	08.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n003.star.cs.uiowa.edu
space	Mixed_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	49.3767440319 seconds
cpu usage	192.669101306
max memory	6.067798016E9

stage attributes

key	value
output-size	8128
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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QDP
        (13) QDPOrderProof [EQUIVALENT, 941 ms]
        (14) QDP
        (15) SemLabProof [SOUND, 49 ms]
        (16) QDP
        (17) DependencyGraphProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) UsableRulesReductionPairsProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) PisEmptyProof [EQUIVALENT, 0 ms]
        (22) YES


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

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

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

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

The TRS R consists of the following rules:

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

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

(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 4 less nodes.
----------------------------------------

(6)

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