SRS Standard pair #487090918

details

property	value
status	complete
benchmark	secr2.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n144.star.cs.uiowa.edu
space	Secret_06_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	4.64808 seconds
cpu usage	15.1105
user time	14.3229
system time	0.787622
max virtual memory	8.1559124E7
max residence set size	2373980.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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


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

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

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

The TRS R consists of the following rules:

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