SRS Standard pair #516968890

details

property	value
status	complete
benchmark	z022.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n092.star.cs.uiowa.edu
space	Zantema_04

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	5.14955997467 seconds
cpu usage	17.329131929
max memory	1.290428416E9

stage attributes

key	value
output-size	8268
starexec-result	YES

output

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


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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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished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) QTRS Reverse [EQUIVALENT, 0 ms]
(2) QTRS
(3) DependencyPairsProof [EQUIVALENT, 31 ms]
(4) QDP
(5) MRRProof [EQUIVALENT, 59 ms]
(6) QDP
(7) DependencyGraphProof [EQUIVALENT, 0 ms]
(8) AND
    (9) QDP
        (10) QDPOrderProof [EQUIVALENT, 11 ms]
        (11) QDP
        (12) DependencyGraphProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) UsableRulesProof [EQUIVALENT, 0 ms]
        (15) QDP
        (16) MRRProof [EQUIVALENT, 2 ms]
        (17) QDP
        (18) PisEmptyProof [EQUIVALENT, 0 ms]
        (19) YES
    (20) QDP
        (21) UsableRulesProof [EQUIVALENT, 0 ms]
        (22) QDP
        (23) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (24) YES


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

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

Q is empty.

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

Q is empty.

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(3) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
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(4)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

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