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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
  ↳5 QDP
    ↳6 QDPOrderProof (⇔)
    ↳7 QDP
    ↳8 PisEmptyProof (⇔)
    ↳9 TRUE
  ↳10 QDP
    ↳11 QDPOrderProof (⇔)
    ↳12 QDP
    ↳13 PisEmptyProof (⇔)
    ↳14 TRUE

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

minus(minus(x)) → x
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))
f(minus(x)) → minus(minus(minus(f(x))))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

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:

MINUS(+(x, y)) → MINUS(minus(minus(x)))
MINUS(+(x, y)) → MINUS(minus(x))
MINUS(+(x, y)) → MINUS(x)
MINUS(+(x, y)) → MINUS(minus(minus(y)))
MINUS(+(x, y)) → MINUS(minus(y))
MINUS(+(x, y)) → MINUS(y)
MINUS(*(x, y)) → MINUS(minus(minus(x)))
MINUS(*(x, y)) → MINUS(minus(x))
MINUS(*(x, y)) → MINUS(x)
MINUS(*(x, y)) → MINUS(minus(minus(y)))
MINUS(*(x, y)) → MINUS(minus(y))
MINUS(*(x, y)) → MINUS(y)
F(minus(x)) → MINUS(minus(minus(f(x))))
F(minus(x)) → MINUS(minus(f(x)))
F(minus(x)) → MINUS(f(x))
F(minus(x)) → F(x)

The TRS R consists of the following rules:

minus(minus(x)) → x
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))
f(minus(x)) → minus(minus(minus(f(x))))

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

(3) DependencyGraphProof (EQUIVALENT transformation)

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

(4) Complex Obligation (AND)

(5) Obligation:

Q DP problem:
The TRS P consists of the following rules:

MINUS(+(x, y)) → MINUS(minus(x))
MINUS(+(x, y)) → MINUS(minus(minus(x)))
MINUS(+(x, y)) → MINUS(x)
MINUS(+(x, y)) → MINUS(minus(minus(y)))
MINUS(+(x, y)) → MINUS(minus(y))
MINUS(+(x, y)) → MINUS(y)
MINUS(*(x, y)) → MINUS(minus(minus(x)))
MINUS(*(x, y)) → MINUS(minus(x))
MINUS(*(x, y)) → MINUS(x)
MINUS(*(x, y)) → MINUS(minus(minus(y)))
MINUS(*(x, y)) → MINUS(minus(y))
MINUS(*(x, y)) → MINUS(y)

The TRS R consists of the following rules:

minus(minus(x)) → x
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))
f(minus(x)) → minus(minus(minus(f(x))))

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


MINUS(+(x, y)) → MINUS(minus(x))
MINUS(+(x, y)) → MINUS(minus(minus(x)))
MINUS(+(x, y)) → MINUS(x)
MINUS(+(x, y)) → MINUS(minus(minus(y)))
MINUS(+(x, y)) → MINUS(minus(y))
MINUS(+(x, y)) → MINUS(y)
MINUS(*(x, y)) → MINUS(minus(minus(x)))
MINUS(*(x, y)) → MINUS(minus(x))
MINUS(*(x, y)) → MINUS(x)
MINUS(*(x, y)) → MINUS(minus(minus(y)))
MINUS(*(x, y)) → MINUS(minus(y))
MINUS(*(x, y)) → MINUS(y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1)  =  x1
+(x1, x2)  =  +(x1, x2)
minus(x1)  =  x1
*(x1, x2)  =  *(x1, x2)

Recursive path order with status [RPO].
Quasi-Precedence:
[+2, *2]

Status:
*2: multiset
+2: multiset


The following usable rules [FROCOS05] were oriented:

minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(minus(x)) → x

(7) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

minus(minus(x)) → x
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))
f(minus(x)) → minus(minus(minus(f(x))))

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

(8) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(9) TRUE

(10) Obligation:

Q DP problem:
The TRS P consists of the following rules:

F(minus(x)) → F(x)

The TRS R consists of the following rules:

minus(minus(x)) → x
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))
f(minus(x)) → minus(minus(minus(f(x))))

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

(11) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


F(minus(x)) → F(x)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
F(x1)  =  x1
minus(x1)  =  minus(x1)

Recursive path order with status [RPO].
Quasi-Precedence:
trivial

Status:
trivial


The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

minus(minus(x)) → x
minus(+(x, y)) → *(minus(minus(minus(x))), minus(minus(minus(y))))
minus(*(x, y)) → +(minus(minus(minus(x))), minus(minus(minus(y))))
f(minus(x)) → minus(minus(minus(f(x))))

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

(13) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(14) TRUE