TRS Standard pair #516965415

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	n143.star.cs.uiowa.edu
space	Zantema_05

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	2.19801497459 seconds
cpu usage	5.610119064
max memory	4.74451968E8

stage attributes

key	value
output-size	6159
starexec-result	YES

output

/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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


YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


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, 14 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:

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