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
    ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔)
        ↳13 TRUE

(0) Obligation:

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

minus(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → 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(s(X), s(Y)) → MINUS(X, Y)
GEQ(s(X), s(Y)) → GEQ(X, Y)
DIV(s(X), s(Y)) → IF(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
DIV(s(X), s(Y)) → GEQ(X, Y)
DIV(s(X), s(Y)) → DIV(minus(X, Y), s(Y))
DIV(s(X), s(Y)) → MINUS(X, Y)

The TRS R consists of the following rules:

minus(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → 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 3 SCCs with 3 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

GEQ(s(X), s(Y)) → GEQ(X, Y)

The TRS R consists of the following rules:

minus(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → 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:
s(x1)  =  s(x1)

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

  • GEQ(s(X), s(Y)) → GEQ(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

(7) TRUE

(8) Obligation:

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

MINUS(s(X), s(Y)) → MINUS(X, Y)

The TRS R consists of the following rules:

minus(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → 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:
s(x1)  =  s(x1)

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

  • MINUS(s(X), s(Y)) → MINUS(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

(11) Obligation:

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

DIV(s(X), s(Y)) → DIV(minus(X, Y), s(Y))

The TRS R consists of the following rules:

minus(0, Y) → 0
minus(s(X), s(Y)) → minus(X, Y)
geq(X, 0) → true
geq(0, s(Y)) → false
geq(s(X), s(Y)) → geq(X, Y)
div(0, s(Y)) → 0
div(s(X), s(Y)) → if(geq(X, Y), s(div(minus(X, Y), s(Y))), 0)
if(true, X, Y) → X
if(false, X, Y) → Y

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

(12) 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(0) = 0   
POL(minus(x1, x2)) = x1   
POL(s(x1)) = 1 + x1   

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

  • DIV(s(X), s(Y)) → DIV(minus(X, Y), s(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].


minus(s(X), s(Y)) → minus(X, Y)
minus(0, Y) → 0

(13) TRUE