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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 QDPOrderProof (⇔)
↳4 QDP
↳5 PisEmptyProof (⇔)
↳6 TRUE

(0) Obligation:

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

-(-(neg(x), neg(x)), -(neg(y), neg(y))) → -(-(x, y), -(x, y))

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:

-1(-(neg(x), neg(x)), -(neg(y), neg(y))) → -1(-(x, y), -(x, y))
-1(-(neg(x), neg(x)), -(neg(y), neg(y))) → -1(x, y)

The TRS R consists of the following rules:

-(-(neg(x), neg(x)), -(neg(y), neg(y))) → -(-(x, y), -(x, y))

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

(3) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


-1(-(neg(x), neg(x)), -(neg(y), neg(y))) → -1(-(x, y), -(x, y))
-1(-(neg(x), neg(x)), -(neg(y), neg(y))) → -1(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(-(x1, x2)) = 1 + x1 + x2   
POL(-1(x1, x2)) = x1 + x2   
POL(neg(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented:

-(-(neg(x), neg(x)), -(neg(y), neg(y))) → -(-(x, y), -(x, y))

(4) Obligation:

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

-(-(neg(x), neg(x)), -(neg(y), neg(y))) → -(-(x, y), -(x, y))

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

(5) PisEmptyProof (EQUIVALENT transformation)

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

(6) TRUE