/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# AProVE Commit ID: c69e44bd14796315568835c1ffa2502984884775 mhark 20210624 unpublished


Termination w.r.t. Q of the given QTRS could be proven:

(0) QTRS
(1) QTRSRRRProof [EQUIVALENT, 112 ms]
(2) QTRS
(3) QTRSRRRProof [EQUIVALENT, 24 ms]
(4) QTRS
(5) QTRSRRRProof [EQUIVALENT, 0 ms]
(6) QTRS
(7) RisEmptyProof [EQUIVALENT, 0 ms]
(8) YES


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(0)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   g(A) -> A
   g(B) -> A
   g(B) -> B
   g(C) -> A
   g(C) -> B
   g(C) -> C
   foldB(t, 0) -> t
   foldB(t, s(n)) -> f(foldB(t, n), B)
   foldC(t, 0) -> t
   foldC(t, s(n)) -> f(foldC(t, n), C)
   f(t, x) -> f'(t, g(x))
   f'(triple(a, b, c), C) -> triple(a, b, s(c))
   f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
   f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))
   f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)

Q is empty.

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(1) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(A) = 2
   POL(B) = 2
   POL(C) = 2
   POL(f(x_1, x_2)) = x_1 + 2*x_2
   POL(f'(x_1, x_2)) = x_1 + 2*x_2
   POL(f''(x_1)) = x_1
   POL(foldB(x_1, x_2)) = x_1 + 2*x_2
   POL(foldC(x_1, x_2)) = x_1 + 2*x_2
   POL(g(x_1)) = x_1
   POL(s(x_1)) = 2 + x_1
   POL(triple(x_1, x_2, x_3)) = x_1 + 2*x_2 + 2*x_3
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   f'(triple(a, b, c), A) -> f''(foldB(triple(s(a), 0, c), b))




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(2)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   g(A) -> A
   g(B) -> A
   g(B) -> B
   g(C) -> A
   g(C) -> B
   g(C) -> C
   foldB(t, 0) -> t
   foldB(t, s(n)) -> f(foldB(t, n), B)
   foldC(t, 0) -> t
   foldC(t, s(n)) -> f(foldC(t, n), C)
   f(t, x) -> f'(t, g(x))
   f'(triple(a, b, c), C) -> triple(a, b, s(c))
   f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
   f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)

Q is empty.

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(3) QTRSRRRProof (EQUIVALENT)
Used ordering:
Polynomial interpretation [POLO]:

   POL(0) = 0
   POL(A) = 0
   POL(B) = 2
   POL(C) = 2
   POL(f(x_1, x_2)) = x_1 + 2*x_2
   POL(f'(x_1, x_2)) = x_1 + 2*x_2
   POL(f''(x_1)) = 1 + 2*x_1
   POL(foldB(x_1, x_2)) = 2*x_1 + 2*x_2
   POL(foldC(x_1, x_2)) = x_1 + 2*x_2
   POL(g(x_1)) = x_1
   POL(s(x_1)) = 2 + x_1
   POL(triple(x_1, x_2, x_3)) = 2 + 2*x_1 + 2*x_2 + x_3
With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   g(B) -> A
   g(C) -> A
   f'(triple(a, b, c), C) -> triple(a, b, s(c))
   f'(triple(a, b, c), B) -> f(triple(a, b, c), A)
   f''(triple(a, b, c)) -> foldC(triple(a, b, 0), c)




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(4)
Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:

   g(A) -> A
   g(B) -> B
   g(C) -> B
   g(C) -> C
   foldB(t, 0) -> t
   foldB(t, s(n)) -> f(foldB(t, n), B)
   foldC(t, 0) -> t
   foldC(t, s(n)) -> f(foldC(t, n), C)
   f(t, x) -> f'(t, g(x))

Q is empty.

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(5) QTRSRRRProof (EQUIVALENT)
Used ordering:
Knuth-Bendix order [KBO] with precedence:foldB_2 > f'_2 > foldC_2 > f_2 > 0 > s_1 > C > A > B > g_1

and weight map:

   A=1
   B=2
   C=1
   0=1
   g_1=2
   s_1=5
   foldB_2=0
   f_2=3
   foldC_2=0
   f'_2=0

The variable weight is 1With this ordering the following rules can be removed by the rule removal processor [LPAR04] because they are oriented strictly:

   g(A) -> A
   g(B) -> B
   g(C) -> B
   g(C) -> C
   foldB(t, 0) -> t
   foldB(t, s(n)) -> f(foldB(t, n), B)
   foldC(t, 0) -> t
   foldC(t, s(n)) -> f(foldC(t, n), C)
   f(t, x) -> f'(t, g(x))




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(6)
Obligation:
Q restricted rewrite system:
R is empty.
Q is empty.

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(7) RisEmptyProof (EQUIVALENT)
The TRS R is empty. Hence, termination is trivially proven.
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(8)
YES