SRS Standard pair #487083286

details

property	value
status	complete
benchmark	jw3.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n146.star.cs.uiowa.edu
space	Waldmann_06_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	4.21418 seconds
cpu usage	13.4178
user time	12.7242
system time	0.693553
max virtual memory	6.0602908E7
max residence set size	1890472.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) QDPOrderProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (9) YES
    (10) QDP
        (11) QDPOrderProof [EQUIVALENT, 33 ms]
        (12) QDP
        (13) QDPOrderProof [EQUIVALENT, 376 ms]
        (14) QDP
        (15) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (16) YES


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

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

   a(b(b(x1))) -> P(a(b(x1)))
   a(P(x1)) -> P(a(x(x1)))
   a(x(x1)) -> x(a(x1))
   b(P(x1)) -> b(Q(x1))
   Q(x(x1)) -> a(Q(x1))
   Q(a(x1)) -> b(b(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:

   A(b(b(x1))) -> A(b(x1))
   A(P(x1)) -> A(x(x1))
   A(x(x1)) -> A(x1)
   B(P(x1)) -> B(Q(x1))
   B(P(x1)) -> Q^1(x1)
   Q^1(x(x1)) -> A(Q(x1))
   Q^1(x(x1)) -> Q^1(x1)
   Q^1(a(x1)) -> B(b(a(x1)))
   Q^1(a(x1)) -> B(a(x1))

The TRS R consists of the following rules:

   a(b(b(x1))) -> P(a(b(x1)))
   a(P(x1)) -> P(a(x(x1)))
   a(x(x1)) -> x(a(x1))
   b(P(x1)) -> b(Q(x1))
   Q(x(x1)) -> a(Q(x1))
   Q(a(x1)) -> b(b(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:

   A(P(x1)) -> A(x(x1))
   A(x(x1)) -> A(x1)
   A(b(b(x1))) -> A(b(x1))

The TRS R consists of the following rules:

   a(b(b(x1))) -> P(a(b(x1)))
   a(P(x1)) -> P(a(x(x1)))
   a(x(x1)) -> x(a(x1))
   b(P(x1)) -> b(Q(x1))

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