/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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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 of the given RelTRS could be proven:

(0) RelTRS
(1) RelTRS S Cleaner [EQUIVALENT, 0 ms]
(2) RelTRS
(3) RelTRS Reverse [EQUIVALENT, 0 ms]
(4) RelTRS
(5) RelTRSRRRProof [EQUIVALENT, 169 ms]
(6) RelTRS
(7) RelTRSRRRProof [EQUIVALENT, 3 ms]
(8) RelTRS
(9) RIsEmptyProof [EQUIVALENT, 0 ms]
(10) YES


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

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

The relative TRS consists of the following S rules:

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


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(1) RelTRS S Cleaner (EQUIVALENT)
We have deleted all rules from S that have the shape t -> t:

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


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

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

The relative TRS consists of the following S rules:

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


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

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

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

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


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

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

The relative TRS consists of the following S rules:

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


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(5) RelTRSRRRProof (EQUIVALENT)
We used the following monotonic ordering for rule removal:
Matrix interpretation [MATRO] to (N^2, +, *, >=, >) :

   <<<
 POL(b(x_1)) =  	[[0], [0]] 	 +  	[[2, 0], [0, 0]] 	* 	x_1
>>>

   <<<
 POL(a(x_1)) =  	[[0], [0]] 	 +  	[[2, 1], [0, 0]] 	* 	x_1
>>>

   <<<
 POL(c(x_1)) =  	[[0], [1]] 	 +  	[[2, 0], [0, 2]] 	* 	x_1
>>>

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

   b(a(c(x1))) -> a(c(a(x1)))
Rules from S:

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




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(6)
Obligation:
Relative term rewrite system:
The relative TRS consists of the following R rules:

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

The relative TRS consists of the following S rules:

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


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(7) RelTRSRRRProof (EQUIVALENT)
We used the following monotonic ordering for rule removal:
Polynomial interpretation [POLO]:

   POL(a(x_1)) = 1 + x_1
   POL(b(x_1)) = x_1
   POL(c(x_1)) = 1 + x_1
With this ordering the following rules can be removed [MATRO] because they are oriented strictly:
Rules from R:

   c(c(c(x1))) -> b(b(b(x1)))
Rules from S:

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




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(8)
Obligation:
Relative term rewrite system:
R is empty.
The relative TRS consists of the following S rules:

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


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(9) RIsEmptyProof (EQUIVALENT)
The TRS R is empty. Hence, termination is trivially proven.
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(10)
YES