SRS Standard pair #487083142

details

property	value
status	complete
benchmark	03.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n143.star.cs.uiowa.edu
space	Zantema_06

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	45.0454 seconds
cpu usage	174.182
user time	170.9
system time	3.28214
max virtual memory	8.2065016E7
max residence set size	5381572.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) DependencyPairsProof [EQUIVALENT, 14 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 1 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) QDPOrderProof [EQUIVALENT, 39 ms]
        (12) QDP
        (13) QDPOrderProof [EQUIVALENT, 2491 ms]
        (14) QDP
        (15) QDPOrderProof [EQUIVALENT, 3672 ms]
        (16) QDP
        (17) PisEmptyProof [EQUIVALENT, 0 ms]
        (18) YES


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

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

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

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

The TRS R consists of the following rules:

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

   C(b(x1)) -> C(x1)

The TRS R consists of the following rules:

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

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

(6) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(7)

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