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:

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) QDPSizeChangeProof (EQUIVALENT transformation)

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

Order:Polynomial interpretation [POLO]:


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

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

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

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

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

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

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

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

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

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

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

  • MINUS(*(x, y)) → MINUS(minus(y)) (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].


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))))

(7) TRUE

(8) 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.

(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)

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

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

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

(10) TRUE