/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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) RelTRSRRRProof [EQUIVALENT, 46 ms]
(2) RelTRS
(3) RelTRSRRRProof [EQUIVALENT, 6 ms]
(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:

   s(a(x)) -> s(b(x))
   b(b(x)) -> a(x)

The relative TRS consists of the following S rules:

   f(s(x), y) -> f(x, s(y))
   s(a(x)) -> a(s(x))
   s(b(x)) -> b(s(x))
   a(s(x)) -> s(a(x))
   b(s(x)) -> s(b(x))


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(1) 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)) = 1 + x_1
   POL(f(x_1, x_2)) = x_1 + x_2
   POL(s(x_1)) = x_1
With this ordering the following rules can be removed [MATRO] because they are oriented strictly:
Rules from R:

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




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

   s(a(x)) -> s(b(x))

The relative TRS consists of the following S rules:

   f(s(x), y) -> f(x, s(y))
   s(a(x)) -> a(s(x))
   s(b(x)) -> b(s(x))
   a(s(x)) -> s(a(x))
   b(s(x)) -> s(b(x))


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(3) 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(f(x_1, x_2)) = x_1 + x_2
   POL(s(x_1)) = x_1
With this ordering the following rules can be removed [MATRO] because they are oriented strictly:
Rules from R:

   s(a(x)) -> s(b(x))
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:

   f(s(x), y) -> f(x, s(y))
   s(a(x)) -> a(s(x))
   s(b(x)) -> b(s(x))
   a(s(x)) -> s(a(x))
   b(s(x)) -> s(b(x))


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