SRS Standard pair #516975700

details

property	value
status	complete
benchmark	secr3.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n032.star.cs.uiowa.edu
space	Secret_06_SRS

run statistics

property	value
solver	AProVE21
configuration	standard
runtime (wallclock)	10.8104000092 seconds
cpu usage	31.205575042
max memory	1.423388672E9

stage attributes

key	value
output-size	7644
starexec-result	YES

output

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


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YES
proof of /export/starexec/sandbox/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished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, 9 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) AND
    (7) QDP
        (8) UsableRulesProof [EQUIVALENT, 0 ms]
        (9) QDP
        (10) QDPOrderProof [EQUIVALENT, 43 ms]
        (11) QDP
        (12) QDPOrderProof [EQUIVALENT, 51 ms]
        (13) QDP
        (14) PisEmptyProof [EQUIVALENT, 0 ms]
        (15) YES
    (16) QDP
        (17) UsableRulesProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (20) YES


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

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

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

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

The TRS R consists of the following rules:

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

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