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# AProVE Commit ID: 794c25de1cacf0d048858bcd21c9a779e1221865 marcel 20200619 unpublished dirty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 Reverse [SOUND, 0 ms]
(2) RelTRS
(3) RelTRSRRRProof [EQUIVALENT, 16 ms]
(4) RelTRS
(5) RelTRSRRRProof [EQUIVALENT, 0 ms]
(6) RelTRS
(7) RelTRSRRRProof [EQUIVALENT, 19 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:

   g(s(x)) -> f(x)
   f(0) -> s(0)
   f(s(x)) -> s(s(g(x)))
   g(0) -> 0

The relative TRS consists of the following S rules:

   rand(x) -> x
   rand(x) -> rand(s(x))


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(1) RelTRS Reverse (SOUND)
We have reversed the following relative TRS [REVERSE]:
The set of rules R is 
   g(s(x)) -> f(x)
   f(0) -> s(0)
   f(s(x)) -> s(s(g(x)))
   g(0) -> 0

The set of rules S is 
   rand(x) -> x
   rand(x) -> rand(s(x))

We have obtained the following relative TRS:
The set of rules R is 
   s(g(x)) -> f(x)
   0'(f(x)) -> 0'(s(x))
   s(f(x)) -> g(s(s(x)))
   0'(g(x)) -> 0'(x)

The set of rules S is 
   rand(x) -> x
   rand(x) -> s(rand(x))


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

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

   s(g(x)) -> f(x)
   0'(f(x)) -> 0'(s(x))
   s(f(x)) -> g(s(s(x)))
   0'(g(x)) -> 0'(x)

The relative TRS consists of the following S rules:

   rand(x) -> x
   rand(x) -> s(rand(x))


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

   POL(0'(x_1)) = x_1
   POL(f(x_1)) = x_1
   POL(g(x_1)) = x_1
   POL(rand(x_1)) = 1 + x_1
   POL(s(x_1)) = x_1
With this ordering the following rules can be removed [MATRO] because they are oriented strictly:
Rules from R:
none
Rules from S:

   rand(x) -> x




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

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

   s(g(x)) -> f(x)
   0'(f(x)) -> 0'(s(x))
   s(f(x)) -> g(s(s(x)))
   0'(g(x)) -> 0'(x)

The relative TRS consists of the following S rules:

   rand(x) -> s(rand(x))


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

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

   POL(0'(x_1)) = x_1
   POL(f(x_1)) = 1 + x_1
   POL(g(x_1)) = 1 + x_1
   POL(rand(x_1)) = x_1
   POL(s(x_1)) = x_1
With this ordering the following rules can be removed [MATRO] because they are oriented strictly:
Rules from R:

   0'(f(x)) -> 0'(s(x))
   0'(g(x)) -> 0'(x)
Rules from S:
none




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

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

   s(g(x)) -> f(x)
   s(f(x)) -> g(s(s(x)))

The relative TRS consists of the following S rules:

   rand(x) -> s(rand(x))


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

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

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

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

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

   <<<
 POL(rand(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:

   s(g(x)) -> f(x)
   s(f(x)) -> g(s(s(x)))
Rules from S:
none




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

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

   rand(x) -> s(rand(x))


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