/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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, 143 ms]
(4) RelTRS
(5) RelTRSRRRProof [EQUIVALENT, 630 ms]
(6) RelTRS
(7) RelTRSRRRProof [EQUIVALENT, 0 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:

   b(q(b(x1))) -> b(p(b(x1)))

The relative TRS consists of the following S rules:

   0(p(0(x1))) -> q(x1)
   1(p(1(x1))) -> q(x1)
   0(q(0(x1))) -> q(x1)
   1(q(1(x1))) -> q(x1)
   p(x1) -> 1(p(1(0(1(x1)))))


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

The set of rules S is 
   0(p(0(x1))) -> q(x1)
   1(p(1(x1))) -> q(x1)
   0(q(0(x1))) -> q(x1)
   1(q(1(x1))) -> q(x1)
   p(x1) -> 1(p(1(0(1(x1)))))

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

The set of rules S is 
   0(p(0(x1))) -> q(x1)
   1(p(1(x1))) -> q(x1)
   0(q(0(x1))) -> q(x1)
   1(q(1(x1))) -> q(x1)
   p(x1) -> 1(0(1(p(1(x1)))))


----------------------------------------

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

   b(q(b(x1))) -> b(p(b(x1)))

The relative TRS consists of the following S rules:

   0(p(0(x1))) -> q(x1)
   1(p(1(x1))) -> q(x1)
   0(q(0(x1))) -> q(x1)
   1(q(1(x1))) -> q(x1)
   p(x1) -> 1(0(1(p(1(x1)))))


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

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

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

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

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

   <<<
 POL(1(x_1)) =  	[[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:
none
Rules from S:

   0(p(0(x1))) -> q(x1)




----------------------------------------

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

   b(q(b(x1))) -> b(p(b(x1)))

The relative TRS consists of the following S rules:

   1(p(1(x1))) -> q(x1)
   0(q(0(x1))) -> q(x1)
   1(q(1(x1))) -> q(x1)
   p(x1) -> 1(0(1(p(1(x1)))))


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

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

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

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

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

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

With this ordering the following rules can be removed [MATRO] because they are oriented strictly:
Rules from R:
none
Rules from S:

   1(p(1(x1))) -> q(x1)




----------------------------------------

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

   b(q(b(x1))) -> b(p(b(x1)))

The relative TRS consists of the following S rules:

   0(q(0(x1))) -> q(x1)
   1(q(1(x1))) -> q(x1)
   p(x1) -> 1(0(1(p(1(x1)))))


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

   POL(0(x_1)) = x_1
   POL(1(x_1)) = x_1
   POL(b(x_1)) = x_1
   POL(p(x_1)) = x_1
   POL(q(x_1)) = 1 + x_1
With this ordering the following rules can be removed [MATRO] because they are oriented strictly:
Rules from R:

   b(q(b(x1))) -> b(p(b(x1)))
Rules from S:
none




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

   0(q(0(x1))) -> q(x1)
   1(q(1(x1))) -> q(x1)
   p(x1) -> 1(0(1(p(1(x1)))))


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