SRS Standard pair #487090894

details

property	value
status	complete
benchmark	secr3.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	Secret_06_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	3.33847 seconds
cpu usage	10.263
user time	9.80779
system time	0.455193
max virtual memory	4.070678E7
max residence set size	1155008.0

stage attributes

key	value
starexec-result	YES

output

YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 1 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) MRRProof [EQUIVALENT, 12 ms]
        (11) QDP
        (12) DependencyGraphProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) QDPOrderProof [EQUIVALENT, 46 ms]
        (15) QDP
        (16) PisEmptyProof [EQUIVALENT, 0 ms]
        (17) YES
    (18) QDP
        (19) UsableRulesProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (22) YES


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

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

Q is empty.

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

Q is empty.

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(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   A(b(a(x1))) -> C(b(x1))
   A(b(a(x1))) -> B(x1)
   B(b(b(x1))) -> B(c(x1))
   B(b(b(x1))) -> C(x1)
   C(x1) -> B(a(x1))
   C(x1) -> A(x1)
   D(c(x1)) -> A(b(c(d(x1))))
   D(c(x1)) -> B(c(d(x1)))
   D(c(x1)) -> C(d(x1))
   D(c(x1)) -> D(x1)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
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(5) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.
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