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 QDPSizeChangeProof (⇔)
        ↳7 TRUE
    ↳8 QDP
        ↳9 QDPSizeChangeProof (⇔)
        ↳10 TRUE

(0) Obligation:

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

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(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:

MINUS(+(x, y)) → +1(minus(y), minus(x))
MINUS(+(x, y)) → MINUS(y)
MINUS(+(x, y)) → MINUS(x)
*1(x, +(y, z)) → +1(*(x, y), *(x, z))
*1(x, +(y, z)) → *1(x, y)
*1(x, +(y, z)) → *1(x, z)
*1(x, minus(y)) → MINUS(*(x, y))
*1(x, minus(y)) → *1(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, y))

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(x)
MINUS(+(x, y)) → MINUS(y)

The TRS R consists of the following rules:

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, y))

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

(6) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
+(x1, x2)  =  +(x1, x2)

From the DPs we obtained the following set of size-change graphs:

  • MINUS(+(x, y)) → MINUS(x) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

  • MINUS(+(x, y)) → MINUS(y) (allowed arguments on rhs = {1})
    The graph contains the following edges 1 > 1

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(7) TRUE

(8) Obligation:

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

*1(x, +(y, z)) → *1(x, z)
*1(x, +(y, z)) → *1(x, y)
*1(x, minus(y)) → *1(x, y)

The TRS R consists of the following rules:

+(x, 0) → x
+(minus(x), x) → 0
minus(0) → 0
minus(minus(x)) → x
minus(+(x, y)) → +(minus(y), minus(x))
*(x, 1) → x
*(x, 0) → 0
*(x, +(y, z)) → +(*(x, y), *(x, z))
*(x, minus(y)) → minus(*(x, y))

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

(9) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
minus(x1)  =  minus(x1)
+(x1, x2)  =  +(x1, x2)

From the DPs we obtained the following set of size-change graphs:

  • *1(x, +(y, z)) → *1(x, z) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 >= 1, 2 > 2

  • *1(x, +(y, z)) → *1(x, y) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 >= 1, 2 > 2

  • *1(x, minus(y)) → *1(x, y) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 >= 1, 2 > 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(10) TRUE