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 UsableRulesProof (⇔)
    ↳7 QDP
    ↳8 MRRProof (⇔)
    ↳9 QDP
    ↳10 PisEmptyProof (⇔)
    ↳11 TRUE
  ↳12 QDP
    ↳13 UsableRulesProof (⇔)
    ↳14 QDP
    ↳15 QDPSizeChangeProof (⇔)
    ↳16 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) UsableRulesProof (EQUIVALENT transformation)

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:

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))))

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

(8) MRRProof (EQUIVALENT transformation)

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:

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)


Used ordering: Polynomial interpretation [POLO]:

POL(*(x1, x2)) = 1 + x1 + x2   
POL(+(x1, x2)) = 1 + x1 + x2   
POL(MINUS(x1)) = x1   
POL(minus(x1)) = x1   

(9) 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))))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(12) 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.

(13) UsableRulesProof (EQUIVALENT transformation)

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:

F(minus(x)) → F(x)

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

(15) QDPSizeChangeProof (EQUIVALENT transformation)

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:

  • F(minus(x)) → F(x)
    The graph contains the following edges 1 > 1

(16) TRUE