TRS Standard pair #487068693

details

property	value
status	complete
benchmark	z15.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n175.star.cs.uiowa.edu
space	Zantema_05

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	1.74434 seconds
cpu usage	3.90719
user time	3.75305
system time	0.154138
max virtual memory	1.8409712E7
max residence set size	240692.0

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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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) AAECC Innermost [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 0 ms]
(4) QDP
(5) UsableRulesProof [EQUIVALENT, 0 ms]
(6) QDP
(7) QReductionProof [EQUIVALENT, 0 ms]
(8) QDP
(9) QDPToSRSProof [SOUND, 0 ms]
(10) QTRS
(11) DependencyPairsProof [EQUIVALENT, 0 ms]
(12) QDP
(13) MNOCProof [EQUIVALENT, 0 ms]
(14) QDP
(15) UsableRulesProof [EQUIVALENT, 0 ms]
(16) QDP
(17) QReductionProof [EQUIVALENT, 0 ms]
(18) QDP
(19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
(20) YES


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

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

   f(x, a(b(y))) -> f(a(b(x)), y)
   f(a(x), y) -> f(x, a(y))
   f(b(x), y) -> f(x, b(y))

Q is empty.

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

(1) AAECC Innermost (EQUIVALENT)
We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is none

The TRS R 2 is 
   f(x, a(b(y))) -> f(a(b(x)), y)
   f(a(x), y) -> f(x, a(y))
   f(b(x), y) -> f(x, b(y))

The signature Sigma is {f_2}
----------------------------------------

(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   f(x, a(b(y))) -> f(a(b(x)), y)
   f(a(x), y) -> f(x, a(y))
   f(b(x), y) -> f(x, b(y))

The set Q consists of the following terms:

   f(x0, a(b(x1)))
   f(a(x0), x1)
   f(b(x0), x1)


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

(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:

   F(x, a(b(y))) -> F(a(b(x)), y)
   F(a(x), y) -> F(x, a(y))
   F(b(x), y) -> F(x, b(y))

The TRS R consists of the following rules:

   f(x, a(b(y))) -> f(a(b(x)), y)
   f(a(x), y) -> f(x, a(y))
   f(b(x), y) -> f(x, b(y))

The set Q consists of the following terms:

   f(x0, a(b(x1)))
   f(a(x0), x1)
   f(b(x0), x1)

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

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