SRS Stand 10685 pair #381723543

details

property	value
status	complete
benchmark	size-12-alpha-3-num-82.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n080.star.cs.uiowa.edu
space	Waldmann_07_size12

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	5.8391430378 seconds
cpu usage	18.685914461
max memory	2.300817408E9

stage attributes

key	value
output-size	7037
starexec-result	YES

output

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


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


YES
proof of /export/starexec/sandbox/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 1 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) QDP
(5) QDPOrderProof [EQUIVALENT, 167 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) AND
    (9) QDP
        (10) QDPOrderProof [EQUIVALENT, 47 ms]
        (11) QDP
        (12) PisEmptyProof [EQUIVALENT, 0 ms]
        (13) YES
    (14) QDP
        (15) UsableRulesProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (18) YES


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

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

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

The TRS R consists of the following rules:

   a(x1) -> x1
   a(a(x1)) -> b(a(b(c(x1))))
   b(x1) -> x1
   c(b(x1)) -> a(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 1 SCC with 2 less nodes.
----------------------------------------

(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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

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