TRS Standard pair #615297170

details

property	value
status	complete
benchmark	z24.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	AProVE23
configuration	standard
runtime (wallclock)	1.8449151516 seconds
cpu usage	4.752919342
max memory	2.76877312E8

stage attributes

key	value
output-size	9304
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# AProVE Commit ID: 363435ff5b19173ec8a83b45cfdec073678dabcc jan-christoph 20230815 unpublished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, 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) MRRProof [EQUIVALENT, 0 ms]
(14) QDP
(15) DependencyGraphProof [EQUIVALENT, 0 ms]
(16) AND
    (17) QDP
        (18) UsableRulesProof [EQUIVALENT, 0 ms]
        (19) QDP
        (20) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (21) YES
    (22) QDP
        (23) UsableRulesProof [EQUIVALENT, 0 ms]
        (24) QDP
        (25) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (26) YES


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

Q is empty.

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

The set Q consists of the following terms:

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


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(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

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