/export/starexec/sandbox2/solver/bin/starexec_run_standard /export/starexec/sandbox2/benchmark/theBenchmark.xml /export/starexec/sandbox2/output/output_files


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YES
proof of /export/starexec/sandbox2/benchmark/theBenchmark.xml
# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty


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

(0) QTRS
(1) DependencyPairsProof [EQUIVALENT, 0 ms]
(2) QDP
(3) DependencyGraphProof [EQUIVALENT, 0 ms]
(4) AND
    (5) QDP
        (6) UsableRulesProof [EQUIVALENT, 0 ms]
        (7) QDP
        (8) MRRProof [EQUIVALENT, 47 ms]
        (9) QDP
        (10) DependencyGraphProof [EQUIVALENT, 0 ms]
        (11) QDP
        (12) UsableRulesProof [EQUIVALENT, 0 ms]
        (13) QDP
        (14) MRRProof [EQUIVALENT, 0 ms]
        (15) QDP
        (16) PisEmptyProof [EQUIVALENT, 0 ms]
        (17) YES
    (18) QDP
        (19) UsableRulesProof [EQUIVALENT, 0 ms]
        (20) QDP
        (21) QDPSizeChangeProof [EQUIVALENT, 0 ms]
        (22) YES


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

(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)
   fold(t, x, 0) -> t
   fold(t, x, s(n)) -> f(fold(t, x, n), x)

Q is empty.

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

(1) DependencyPairsProof (EQUIVALENT)
Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.
----------------------------------------

(2)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FOLDB(t, s(n)) -> F(foldB(t, n), B)
   FOLDB(t, s(n)) -> FOLDB(t, n)
   FOLDC(t, s(n)) -> F(foldC(t, n), C)
   FOLDC(t, s(n)) -> FOLDC(t, n)
   F(t, x) -> F'(t, g(x))
   F(t, x) -> G(x)
   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), A) -> FOLDB(triple(s(a), 0, c), b)
   F''(triple(a, b, c)) -> FOLDC(triple(a, b, 0), c)
   FOLD(t, x, s(n)) -> F(fold(t, x, n), x)
   FOLD(t, x, s(n)) -> FOLD(t, x, n)

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)
   fold(t, x, 0) -> t
   fold(t, x, s(n)) -> f(fold(t, x, n), x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(3) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 2 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

(5)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

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)
   fold(t, x, 0) -> t
   fold(t, x, s(n)) -> f(fold(t, x, n), x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(6) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(7)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(8) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   F''(triple(a, b, c)) -> FOLDC(triple(a, b, 0), c)
   FOLDC(t, s(n)) -> FOLDC(t, n)
   F'(triple(a, b, c), A) -> FOLDB(triple(s(a), 0, c), b)
   FOLDB(t, s(n)) -> F(foldB(t, n), B)
   FOLDB(t, s(n)) -> FOLDB(t, n)

Strictly oriented rules of the TRS R:

   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)

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)) = 2 + x_1
   POL(FOLDB(x_1, x_2)) = 1 + x_1 + 2*x_2
   POL(FOLDC(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_2)) = x_1 + 2*x_2
   POL(f''(x_1)) = 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


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

(9)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F(t, x) -> F'(t, g(x))
   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))
   FOLDC(t, s(n)) -> F(foldC(t, n), C)

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(10) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.
----------------------------------------

(11)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F'(triple(a, b, c), B) -> F(triple(a, b, c), A)
   F(t, x) -> F'(t, g(x))

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(12) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(13)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   F'(triple(a, b, c), B) -> F(triple(a, b, c), A)
   F(t, x) -> F'(t, g(x))

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(14) MRRProof (EQUIVALENT)
By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.

Strictly oriented dependency pairs:

   F'(triple(a, b, c), B) -> F(triple(a, b, c), A)
   F(t, x) -> F'(t, g(x))

Strictly oriented rules of the TRS R:

   g(A) -> A
   g(B) -> A
   g(B) -> B
   g(C) -> A
   g(C) -> B
   g(C) -> C

Used ordering: Knuth-Bendix order [KBO] with precedence:g_1 > F_2 > triple_3 > F'_2 > C > B > A

and weight map:

   A=1
   B=2
   C=2
   g_1=0
   F'_2=0
   F_2=0
   triple_3=0

The variable weight is 1

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

(15)
Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(16) PisEmptyProof (EQUIVALENT)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
----------------------------------------

(17)
YES

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

(18)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FOLD(t, x, s(n)) -> FOLD(t, x, n)

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)
   fold(t, x, 0) -> t
   fold(t, x, s(n)) -> f(fold(t, x, n), x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(19) UsableRulesProof (EQUIVALENT)
We can use the usable rules and reduction pair processor [LPAR04] with the Ce-compatible extension of the polynomial order that maps every function symbol to the sum of its arguments. Then, we can delete all non-usable rules [FROCOS05] from R.
----------------------------------------

(20)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   FOLD(t, x, s(n)) -> FOLD(t, x, n)

R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

(21) QDPSizeChangeProof (EQUIVALENT)
By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem. 

From the DPs we obtained the following set of size-change graphs:
*FOLD(t, x, s(n)) -> FOLD(t, x, n)
The graph contains the following edges 1 >= 1, 2 >= 2, 3 > 3


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

(22)
YES