/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# AProVE Commit ID: 48fb2092695e11cc9f56e44b17a92a5f88ffb256 marcel 20180622 unpublished dirty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 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, 0 ms]
        (9) QDP
        (10) PisEmptyProof [EQUIVALENT, 0 ms]
        (11) YES
    (12) QDP
        (13) UsableRulesProof [EQUIVALENT, 0 ms]
        (14) QDP
        (15) MNOCProof [EQUIVALENT, 0 ms]
        (16) QDP
        (17) MRRProof [EQUIVALENT, 0 ms]
        (18) QDP
        (19) DependencyGraphProof [EQUIVALENT, 0 ms]
        (20) TRUE
    (21) QDP
        (22) UsableRulesProof [EQUIVALENT, 0 ms]
        (23) QDP
        (24) TransformationProof [EQUIVALENT, 0 ms]
        (25) QDP
        (26) DependencyGraphProof [EQUIVALENT, 0 ms]
        (27) TRUE


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

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

   active(f(x)) -> mark(f(f(x)))
   chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
   mat(f(x), f(y)) -> f(mat(x, y))
   chk(no(c)) -> active(c)
   mat(f(x), c) -> no(c)
   f(active(x)) -> active(f(x))
   f(no(x)) -> no(f(x))
   f(mark(x)) -> mark(f(x))
   tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), 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:

   ACTIVE(f(x)) -> F(f(x))
   CHK(no(f(x))) -> F(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
   CHK(no(f(x))) -> CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
   CHK(no(f(x))) -> MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
   CHK(no(f(x))) -> F(f(f(f(f(f(f(f(f(f(X))))))))))
   CHK(no(f(x))) -> F(f(f(f(f(f(f(f(f(X)))))))))
   CHK(no(f(x))) -> F(f(f(f(f(f(f(f(X))))))))
   CHK(no(f(x))) -> F(f(f(f(f(f(f(X)))))))
   CHK(no(f(x))) -> F(f(f(f(f(f(X))))))
   CHK(no(f(x))) -> F(f(f(f(f(X)))))
   CHK(no(f(x))) -> F(f(f(f(X))))
   CHK(no(f(x))) -> F(f(f(X)))
   CHK(no(f(x))) -> F(f(X))
   CHK(no(f(x))) -> F(X)
   MAT(f(x), f(y)) -> F(mat(x, y))
   MAT(f(x), f(y)) -> MAT(x, y)
   CHK(no(c)) -> ACTIVE(c)
   F(active(x)) -> ACTIVE(f(x))
   F(active(x)) -> F(x)
   F(no(x)) -> F(x)
   F(mark(x)) -> F(x)
   TP(mark(x)) -> TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
   TP(mark(x)) -> CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))
   TP(mark(x)) -> MAT(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)
   TP(mark(x)) -> F(f(f(f(f(f(f(f(f(f(X))))))))))
   TP(mark(x)) -> F(f(f(f(f(f(f(f(f(X)))))))))
   TP(mark(x)) -> F(f(f(f(f(f(f(f(X))))))))
   TP(mark(x)) -> F(f(f(f(f(f(f(X)))))))
   TP(mark(x)) -> F(f(f(f(f(f(X))))))
   TP(mark(x)) -> F(f(f(f(f(X)))))
   TP(mark(x)) -> F(f(f(f(X))))
   TP(mark(x)) -> F(f(f(X)))
   TP(mark(x)) -> F(f(X))
   TP(mark(x)) -> F(X)

The TRS R consists of the following rules:

   active(f(x)) -> mark(f(f(x)))
   chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
   mat(f(x), f(y)) -> f(mat(x, y))
   chk(no(c)) -> active(c)
   mat(f(x), c) -> no(c)
   f(active(x)) -> active(f(x))
   f(no(x)) -> no(f(x))
   f(mark(x)) -> mark(f(x))
   tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), 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 3 SCCs with 27 less nodes.
----------------------------------------

(4)
Complex Obligation (AND)

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

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

   F(active(x)) -> ACTIVE(f(x))
   ACTIVE(f(x)) -> F(f(x))
   F(active(x)) -> F(x)
   F(no(x)) -> F(x)
   F(mark(x)) -> F(x)

The TRS R consists of the following rules:

   active(f(x)) -> mark(f(f(x)))
   chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
   mat(f(x), f(y)) -> f(mat(x, y))
   chk(no(c)) -> active(c)
   mat(f(x), c) -> no(c)
   f(active(x)) -> active(f(x))
   f(no(x)) -> no(f(x))
   f(mark(x)) -> mark(f(x))
   tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), 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(active(x)) -> ACTIVE(f(x))
   ACTIVE(f(x)) -> F(f(x))
   F(active(x)) -> F(x)
   F(no(x)) -> F(x)
   F(mark(x)) -> F(x)

The TRS R consists of the following rules:

   f(active(x)) -> active(f(x))
   f(no(x)) -> no(f(x))
   f(mark(x)) -> mark(f(x))
   active(f(x)) -> mark(f(f(x)))

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(active(x)) -> ACTIVE(f(x))
   ACTIVE(f(x)) -> F(f(x))
   F(active(x)) -> F(x)
   F(no(x)) -> F(x)
   F(mark(x)) -> F(x)

Strictly oriented rules of the TRS R:

   f(active(x)) -> active(f(x))
   f(no(x)) -> no(f(x))
   f(mark(x)) -> mark(f(x))
   active(f(x)) -> mark(f(f(x)))

Used ordering: Polynomial interpretation [POLO]:

   POL(ACTIVE(x_1)) = 2 + 2*x_1
   POL(F(x_1)) = 1 + 2*x_1
   POL(active(x_1)) = 2 + 2*x_1
   POL(f(x_1)) = 2*x_1
   POL(mark(x_1)) = 1 + x_1
   POL(no(x_1)) = 1 + 2*x_1


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

(9)
Obligation:
Q DP problem:
P is empty.
R is empty.
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
----------------------------------------

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

(11)
YES

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(12)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   CHK(no(f(x))) -> CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))

The TRS R consists of the following rules:

   active(f(x)) -> mark(f(f(x)))
   chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
   mat(f(x), f(y)) -> f(mat(x, y))
   chk(no(c)) -> active(c)
   mat(f(x), c) -> no(c)
   f(active(x)) -> active(f(x))
   f(no(x)) -> no(f(x))
   f(mark(x)) -> mark(f(x))
   tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

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

(13) 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.
----------------------------------------

(14)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   CHK(no(f(x))) -> CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))

The TRS R consists of the following rules:

   mat(f(x), f(y)) -> f(mat(x, y))
   mat(f(x), c) -> no(c)

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

(15) MNOCProof (EQUIVALENT)
We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.
----------------------------------------

(16)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   CHK(no(f(x))) -> CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))

The TRS R consists of the following rules:

   mat(f(x), f(y)) -> f(mat(x, y))
   mat(f(x), c) -> no(c)

The set Q consists of the following terms:

   mat(f(x0), f(y))
   mat(f(x0), c)

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

(17) 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 rules of the TRS R:

   mat(f(x), c) -> no(c)

Used ordering: Polynomial interpretation [POLO]:

   POL(CHK(x_1)) = 2*x_1
   POL(X) = 0
   POL(c) = 1
   POL(f(x_1)) = 2*x_1
   POL(mat(x_1, x_2)) = x_1 + 2*x_2
   POL(no(x_1)) = x_1
   POL(y) = 0


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

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

   CHK(no(f(x))) -> CHK(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))

The TRS R consists of the following rules:

   mat(f(x), f(y)) -> f(mat(x, y))

The set Q consists of the following terms:

   mat(f(x0), f(y))
   mat(f(x0), c)

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

(19) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.
----------------------------------------

(20)
TRUE

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

(21)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TP(mark(x)) -> TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

The TRS R consists of the following rules:

   active(f(x)) -> mark(f(f(x)))
   chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
   mat(f(x), f(y)) -> f(mat(x, y))
   chk(no(c)) -> active(c)
   mat(f(x), c) -> no(c)
   f(active(x)) -> active(f(x))
   f(no(x)) -> no(f(x))
   f(mark(x)) -> mark(f(x))
   tp(mark(x)) -> tp(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

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

(22) 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.
----------------------------------------

(23)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TP(mark(x)) -> TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))

The TRS R consists of the following rules:

   mat(f(x), f(y)) -> f(mat(x, y))
   mat(f(x), c) -> no(c)
   chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
   chk(no(c)) -> active(c)
   f(active(x)) -> active(f(x))
   f(no(x)) -> no(f(x))
   f(mark(x)) -> mark(f(x))
   active(f(x)) -> mark(f(f(x)))

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

(24) TransformationProof (EQUIVALENT)
By narrowing [LPAR04] the rule TP(mark(x)) -> TP(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x))) at position [0] we obtained the following new rules [LPAR04]:

   (TP(mark(f(y))) -> TP(chk(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y)))),TP(mark(f(y))) -> TP(chk(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y)))))
   (TP(mark(c)) -> TP(chk(no(c))),TP(mark(c)) -> TP(chk(no(c))))


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

(25)
Obligation:
Q DP problem:
The TRS P consists of the following rules:

   TP(mark(f(y))) -> TP(chk(f(mat(f(f(f(f(f(f(f(f(f(X))))))))), y))))
   TP(mark(c)) -> TP(chk(no(c)))

The TRS R consists of the following rules:

   mat(f(x), f(y)) -> f(mat(x, y))
   mat(f(x), c) -> no(c)
   chk(no(f(x))) -> f(chk(mat(f(f(f(f(f(f(f(f(f(f(X)))))))))), x)))
   chk(no(c)) -> active(c)
   f(active(x)) -> active(f(x))
   f(no(x)) -> no(f(x))
   f(mark(x)) -> mark(f(x))
   active(f(x)) -> mark(f(f(x)))

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

(26) DependencyGraphProof (EQUIVALENT)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.
----------------------------------------

(27)
TRUE