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

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 QDPOrderProof (⇔)
↳4 QDP
↳5 Instantiation (⇔)
↳6 QDP
↳7 DependencyGraphProof (⇔)
↳8 TRUE

(0) Obligation:

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

div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))

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:

DIV(x, y) → QUOT(x, y, y)
QUOT(s(x), s(y), z) → QUOT(x, y, z)
QUOT(x, 0, s(z)) → DIV(x, s(z))

The TRS R consists of the following rules:

div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))

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.


QUOT(s(x), s(y), z) → QUOT(x, y, z)
The remaining pairs can at least be oriented weakly.
Used ordering: Matrix interpretation [MATRO]:

POL(DIV(x1, x2)) =
/1\
\0/
 +
/10\
\00/
·x1 +
/01\
\00/
·x2

POL(QUOT(x1, x2, x3)) =
/1\
\0/
 +
/10\
\00/
·x1 +
/00\
\00/
·x2 +
/00\
\00/
·x3

POL(s(x1)) =
/1\
\0/
 +
/10\
\00/
·x1

POL(0) =
/0\
\0/

The following usable rules [FROCOS05] were oriented: none

(4) Obligation:

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

DIV(x, y) → QUOT(x, y, y)
QUOT(x, 0, s(z)) → DIV(x, s(z))

The TRS R consists of the following rules:

div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))

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

(5) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule DIV(x, y) → QUOT(x, y, y) we obtained the following new rules [LPAR04]:

DIV(y_0, s(y_1)) → QUOT(y_0, s(y_1), s(y_1))

(6) Obligation:

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

QUOT(x, 0, s(z)) → DIV(x, s(z))
DIV(y_0, s(y_1)) → QUOT(y_0, s(y_1), s(y_1))

The TRS R consists of the following rules:

div(0, y) → 0
div(x, y) → quot(x, y, y)
quot(0, s(y), z) → 0
quot(s(x), s(y), z) → quot(x, y, z)
quot(x, 0, s(z)) → s(div(x, s(z)))

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

(7) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 2 less nodes.

(8) TRUE