SRS Standard pair #487510643

details

property	value
status	complete
benchmark	random-397.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n125.star.cs.uiowa.edu
space	Waldmann_19

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	4.16409492493 seconds
cpu usage	12.96305915
max memory	1.664638976E9

stage attributes

key	value
output-size	7096
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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


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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 19 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 7 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) DependencyGraphProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QDPOrderProof [EQUIVALENT, 29 ms]
        (11) QDP
        (12) PisEmptyProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) QDPOrderProof [EQUIVALENT, 30 ms]
        (16) QDP
        (17) PisEmptyProof [EQUIVALENT, 0 ms]
        (18) YES


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

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

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

Q is empty.

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

(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:

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

The TRS R consists of the following rules:

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

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

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.
----------------------------------------

(4)
Complex Obligation (AND)

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

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

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

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