/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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 S Cleaner [EQUIVALENT, 0 ms]
(2) RelTRS
(3) RelTRSRRRProof [EQUIVALENT, 73 ms]
(4) RelTRS
(5) RelTRSRRRProof [EQUIVALENT, 18 ms]
(6) RelTRS
(7) RelTRSRRRProof [EQUIVALENT, 9 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:

   topB(i, N1(x), y) -> topA(1, T1(x), y)
   topA(i, x, N2(y)) -> topB(0, x, T2(y))
   topB(i, S1(x), y) -> topA(i, N1(x), y)
   topA(i, x, S2(y)) -> topB(i, x, N2(y))
   topA(i, N1(x), T2(y)) -> topB(i, N1(x), S2(y))
   topA(1, T1(x), T2(y)) -> topB(1, T1(x), S2(y))

The relative TRS consists of the following S rules:

   topA(i, N1(x), y) -> topA(1, T1(x), y)
   topB(i, x, N2(y)) -> topB(0, x, T2(y))
   topA(i, S1(x), y) -> topA(i, N1(x), y)
   topB(i, x, S2(y)) -> topB(i, x, N2(y))
   topB(i, N1(x), T2(y)) -> topB(i, N1(x), S2(y))
   topB(1, T1(x), T2(y)) -> topB(1, T1(x), S2(y))
   topA(i, N1(x), y) -> topA(i, N1(C(x)), y)
   topB(i, x, N2(y)) -> topB(i, x, N2(C(y)))
   topA(i, T1(x), y) -> topA(i, T1(x), y)
   topB(i, x, T2(y)) -> topB(i, x, T2(y))
   topB(i, x, S2(y)) -> topB(i, x, S2(D(y)))


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

   topA(i, T1(x), y) -> topA(i, T1(x), y)
   topB(i, x, T2(y)) -> topB(i, x, T2(y))


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

   topB(i, N1(x), y) -> topA(1, T1(x), y)
   topA(i, x, N2(y)) -> topB(0, x, T2(y))
   topB(i, S1(x), y) -> topA(i, N1(x), y)
   topA(i, x, S2(y)) -> topB(i, x, N2(y))
   topA(i, N1(x), T2(y)) -> topB(i, N1(x), S2(y))
   topA(1, T1(x), T2(y)) -> topB(1, T1(x), S2(y))

The relative TRS consists of the following S rules:

   topA(i, N1(x), y) -> topA(1, T1(x), y)
   topB(i, x, N2(y)) -> topB(0, x, T2(y))
   topA(i, S1(x), y) -> topA(i, N1(x), y)
   topB(i, x, S2(y)) -> topB(i, x, N2(y))
   topB(i, N1(x), T2(y)) -> topB(i, N1(x), S2(y))
   topB(1, T1(x), T2(y)) -> topB(1, T1(x), S2(y))
   topA(i, N1(x), y) -> topA(i, N1(C(x)), y)
   topB(i, x, N2(y)) -> topB(i, x, N2(C(y)))
   topB(i, x, S2(y)) -> topB(i, x, S2(D(y)))


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

   POL(0) = 0
   POL(1) = 0
   POL(C(x_1)) = x_1
   POL(D(x_1)) = x_1
   POL(N1(x_1)) = 1 + x_1
   POL(N2(x_1)) = x_1
   POL(S1(x_1)) = 1 + x_1
   POL(S2(x_1)) = x_1
   POL(T1(x_1)) = x_1
   POL(T2(x_1)) = x_1
   POL(topA(x_1, x_2, x_3)) = x_1 + x_2 + x_3
   POL(topB(x_1, x_2, x_3)) = x_1 + x_2 + x_3
With this ordering the following rules can be removed [MATRO] because they are oriented strictly:
Rules from R:

   topB(i, N1(x), y) -> topA(1, T1(x), y)
Rules from S:

   topA(i, N1(x), y) -> topA(1, T1(x), y)




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

   topA(i, x, N2(y)) -> topB(0, x, T2(y))
   topB(i, S1(x), y) -> topA(i, N1(x), y)
   topA(i, x, S2(y)) -> topB(i, x, N2(y))
   topA(i, N1(x), T2(y)) -> topB(i, N1(x), S2(y))
   topA(1, T1(x), T2(y)) -> topB(1, T1(x), S2(y))

The relative TRS consists of the following S rules:

   topB(i, x, N2(y)) -> topB(0, x, T2(y))
   topA(i, S1(x), y) -> topA(i, N1(x), y)
   topB(i, x, S2(y)) -> topB(i, x, N2(y))
   topB(i, N1(x), T2(y)) -> topB(i, N1(x), S2(y))
   topB(1, T1(x), T2(y)) -> topB(1, T1(x), S2(y))
   topA(i, N1(x), y) -> topA(i, N1(C(x)), y)
   topB(i, x, N2(y)) -> topB(i, x, N2(C(y)))
   topB(i, x, S2(y)) -> topB(i, x, S2(D(y)))


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

   POL(0) = 0
   POL(1) = 0
   POL(C(x_1)) = x_1
   POL(D(x_1)) = x_1
   POL(N1(x_1)) = x_1
   POL(N2(x_1)) = x_1
   POL(S1(x_1)) = 1 + x_1
   POL(S2(x_1)) = x_1
   POL(T1(x_1)) = x_1
   POL(T2(x_1)) = x_1
   POL(topA(x_1, x_2, x_3)) = x_1 + x_2 + x_3
   POL(topB(x_1, x_2, x_3)) = x_1 + x_2 + x_3
With this ordering the following rules can be removed [MATRO] because they are oriented strictly:
Rules from R:

   topB(i, S1(x), y) -> topA(i, N1(x), y)
Rules from S:

   topA(i, S1(x), y) -> topA(i, N1(x), y)




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

   topA(i, x, N2(y)) -> topB(0, x, T2(y))
   topA(i, x, S2(y)) -> topB(i, x, N2(y))
   topA(i, N1(x), T2(y)) -> topB(i, N1(x), S2(y))
   topA(1, T1(x), T2(y)) -> topB(1, T1(x), S2(y))

The relative TRS consists of the following S rules:

   topB(i, x, N2(y)) -> topB(0, x, T2(y))
   topB(i, x, S2(y)) -> topB(i, x, N2(y))
   topB(i, N1(x), T2(y)) -> topB(i, N1(x), S2(y))
   topB(1, T1(x), T2(y)) -> topB(1, T1(x), S2(y))
   topA(i, N1(x), y) -> topA(i, N1(C(x)), y)
   topB(i, x, N2(y)) -> topB(i, x, N2(C(y)))
   topB(i, x, S2(y)) -> topB(i, x, S2(D(y)))


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

   POL(0) = 0
   POL(1) = 0
   POL(C(x_1)) = x_1
   POL(D(x_1)) = x_1
   POL(N1(x_1)) = x_1
   POL(N2(x_1)) = x_1
   POL(S2(x_1)) = x_1
   POL(T1(x_1)) = x_1
   POL(T2(x_1)) = x_1
   POL(topA(x_1, x_2, x_3)) = 1 + x_1 + x_2 + x_3
   POL(topB(x_1, x_2, x_3)) = x_1 + x_2 + x_3
With this ordering the following rules can be removed [MATRO] because they are oriented strictly:
Rules from R:

   topA(i, x, N2(y)) -> topB(0, x, T2(y))
   topA(i, x, S2(y)) -> topB(i, x, N2(y))
   topA(i, N1(x), T2(y)) -> topB(i, N1(x), S2(y))
   topA(1, T1(x), T2(y)) -> topB(1, T1(x), S2(y))
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:

   topB(i, x, N2(y)) -> topB(0, x, T2(y))
   topB(i, x, S2(y)) -> topB(i, x, N2(y))
   topB(i, N1(x), T2(y)) -> topB(i, N1(x), S2(y))
   topB(1, T1(x), T2(y)) -> topB(1, T1(x), S2(y))
   topA(i, N1(x), y) -> topA(i, N1(C(x)), y)
   topB(i, x, N2(y)) -> topB(i, x, N2(C(y)))
   topB(i, x, S2(y)) -> topB(i, x, S2(D(y)))


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