/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


Termination of the given RelTRS could be proven:

(0) RelTRS
(1) RelTRS Reverse [EQUIVALENT, 0 ms]
(2) RelTRS
(3) RelTRSRRRProof [EQUIVALENT, 67.1 s]
(4) RelTRS
(5) RIsEmptyProof [EQUIVALENT, 0 ms]
(6) YES


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(0)
Obligation:
Relative term rewrite system:
The relative TRS consists of the following R rules:

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

The relative TRS consists of the following S rules:

   b(x1) -> b(b(b(x1)))


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(1) RelTRS Reverse (EQUIVALENT)
We have reversed the following relative TRS [REVERSE]:
The set of rules R is 
   b(a(a(b(x1)))) -> a(a(a(a(x1))))
   a(b(b(a(x1)))) -> b(b(b(b(x1))))

The set of rules S is 
   b(x1) -> b(b(b(x1)))

We have obtained the following relative TRS:
The set of rules R is 
   b(a(a(b(x1)))) -> a(a(a(a(x1))))
   a(b(b(a(x1)))) -> b(b(b(b(x1))))

The set of rules S is 
   b(x1) -> b(b(b(x1)))


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(2)
Obligation:
Relative term rewrite system:
The relative TRS consists of the following R rules:

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

The relative TRS consists of the following S rules:

   b(x1) -> b(b(b(x1)))


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(3) RelTRSRRRProof (EQUIVALENT)
We used the following monotonic ordering for rule removal:
Matrix interpretation [MATRO] to (N^6, +, *, >=, >) :

   <<<
 POL(b(x_1)) =  	[[0], [0], [1], [0], [0], [0]] 	 +  	[[1, 0, 0, 1, 0, 0], [0, 0, 0, 1, 0, 1], [0, 1, 0, 0, 1, 0], [0, 0, 0, 0, 0, 0], [0, 0, 0, 1, 0, 0], [0, 0, 0, 0, 1, 0]] 	* 	x_1
>>>

   <<<
 POL(a(x_1)) =  	[[0], [0], [0], [0], [1], [0]] 	 +  	[[1, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 0], [0, 1, 0, 0, 0, 0], [0, 0, 0, 0, 0, 1], [0, 0, 0, 1, 0, 0], [0, 0, 1, 0, 0, 0]] 	* 	x_1
>>>

With this ordering the following rules can be removed [MATRO] because they are oriented strictly:
Rules from R:

   b(a(a(b(x1)))) -> a(a(a(a(x1))))
   a(b(b(a(x1)))) -> b(b(b(b(x1))))
Rules from S:
none




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(4)
Obligation:
Relative term rewrite system:
R is empty.
The relative TRS consists of the following S rules:

   b(x1) -> b(b(b(x1)))


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(5) RIsEmptyProof (EQUIVALENT)
The TRS R is empty. Hence, termination is trivially proven.
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(6)
YES