SRS Standard pair #487511747

details

property	value
status	complete
benchmark	turing_add.xml
ran by	Akihisa Yamada
cpu timeout	1200 seconds
wallclock timeout	300 seconds
memory limit	137438953472 bytes
execution host	n077.star.cs.uiowa.edu
space	Mixed_SRS

run statistics

property	value
solver	AProVE
configuration	standard
runtime (wallclock)	2.63463711739 seconds
cpu usage	7.2539718
max memory	8.45344768E8

stage attributes

key	value
output-size	5378
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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty


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, 22 ms]
(4) QDP
(5) DependencyGraphProof [EQUIVALENT, 0 ms]
(6) QDP
(7) QDPOrderProof [EQUIVALENT, 40 ms]
(8) QDP
(9) DependencyGraphProof [EQUIVALENT, 0 ms]
(10) TRUE


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

   1(q0(1(x1))) -> 0(1(q1(x1)))
   1(q0(0(x1))) -> 0(0(q1(x1)))
   1(q1(1(x1))) -> 1(1(q1(x1)))
   1(q1(0(x1))) -> 1(0(q1(x1)))
   0(q1(x1)) -> q2(1(x1))
   1(q2(x1)) -> q2(1(x1))
   0(q2(x1)) -> 0(q0(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:

   1(q0(1(x1))) -> q1(1(0(x1)))
   0(q0(1(x1))) -> q1(0(0(x1)))
   1(q1(1(x1))) -> q1(1(1(x1)))
   0(q1(1(x1))) -> q1(0(1(x1)))
   q1(0(x1)) -> 1(q2(x1))
   q2(1(x1)) -> 1(q2(x1))
   q2(0(x1)) -> q0(0(x1))

Q is empty.

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(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:

   1^1(q0(1(x1))) -> Q1(1(0(x1)))
   1^1(q0(1(x1))) -> 1^1(0(x1))
   1^1(q0(1(x1))) -> 0^1(x1)
   0^1(q0(1(x1))) -> Q1(0(0(x1)))
   0^1(q0(1(x1))) -> 0^1(0(x1))
   0^1(q0(1(x1))) -> 0^1(x1)
   1^1(q1(1(x1))) -> Q1(1(1(x1)))
   1^1(q1(1(x1))) -> 1^1(1(x1))
   0^1(q1(1(x1))) -> Q1(0(1(x1)))
   0^1(q1(1(x1))) -> 0^1(1(x1))
   Q1(0(x1)) -> 1^1(q2(x1))
   Q1(0(x1)) -> Q2(x1)
   Q2(1(x1)) -> 1^1(q2(x1))
   Q2(1(x1)) -> Q2(x1)

The TRS R consists of the following rules:

   1(q0(1(x1))) -> q1(1(0(x1)))
   0(q0(1(x1))) -> q1(0(0(x1)))
   1(q1(1(x1))) -> q1(1(1(x1)))
   0(q1(1(x1))) -> q1(0(1(x1)))
   q1(0(x1)) -> 1(q2(x1))
   q2(1(x1)) -> 1(q2(x1))
   q2(0(x1)) -> q0(0(x1))

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