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 QDPOrderProof (⇔)
        ↳7 QDP
        ↳8 PisEmptyProof (⇔)
        ↳9 TRUE
    ↳10 QDP
        ↳11 QDPOrderProof (⇔)
        ↳12 QDP
        ↳13 QDPOrderProof (⇔)
        ↳14 QDP
        ↳15 PisEmptyProof (⇔)
        ↳16 TRUE

(0) Obligation:

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

quot(0, s(y), s(z)) → 0
quot(s(x), s(y), z) → quot(x, y, z)
plus(0, y) → y
plus(s(x), y) → s(plus(x, y))
quot(x, 0, s(z)) → s(quot(x, plus(z, s(0)), 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:

QUOT(s(x), s(y), z) → QUOT(x, y, z)
PLUS(s(x), y) → PLUS(x, y)
QUOT(x, 0, s(z)) → QUOT(x, plus(z, s(0)), s(z))
QUOT(x, 0, s(z)) → PLUS(z, s(0))

The TRS R consists of the following rules:

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

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 1 less node.

(4) Complex Obligation (AND)

(5) Obligation:

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

PLUS(s(x), y) → PLUS(x, y)

The TRS R consists of the following rules:

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

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

(6) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PLUS(s(x), y) → PLUS(x, y)
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
PLUS(x0, x1, x2)  =  PLUS(x1)

Tags:
PLUS has argument tags [1,2,1] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
PLUS(x1, x2)  =  PLUS
s(x1)  =  s(x1)

Recursive path order with status [RPO].
Quasi-Precedence:
trivial

Status:
PLUS: multiset
s1: [1]


The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

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

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

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) Obligation:

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

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

The TRS R consists of the following rules:

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

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

(11) 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: SCNP Order with the following components:
Level mapping:
Top level AFS:
QUOT(x0, x1, x2, x3)  =  QUOT(x0)

Tags:
QUOT has argument tags [0,2,3,0] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
QUOT(x1, x2, x3)  =  x1
0  =  0
s(x1)  =  s(x1)
plus(x1, x2)  =  plus

Recursive path order with status [RPO].
Quasi-Precedence:
plus > s1 > 0

Status:
0: multiset
s1: multiset
plus: multiset


The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

QUOT(x, 0, s(z)) → QUOT(x, plus(z, s(0)), s(z))

The TRS R consists of the following rules:

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

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

(13) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


QUOT(x, 0, s(z)) → QUOT(x, plus(z, s(0)), s(z))
The remaining pairs can at least be oriented weakly.
Used ordering: SCNP Order with the following components:
Level mapping:
Top level AFS:
QUOT(x0, x1, x2, x3)  =  QUOT(x0)

Tags:
QUOT has argument tags [0,0,0,3] and root tag 0

Comparison: MAX
Underlying order for the size change arcs and the rules of R:
Combined order from the following AFS and order.
QUOT(x1, x2, x3)  =  QUOT(x1, x2, x3)
0  =  0
s(x1)  =  s
plus(x1, x2)  =  plus(x2)

Recursive path order with status [RPO].
Quasi-Precedence:
0 > [QUOT3, s, plus1]

Status:
QUOT3: [3,2,1]
0: multiset
s: multiset
plus1: multiset


The following usable rules [FROCOS05] were oriented:

plus(0, y) → y
plus(s(x), y) → s(plus(x, y))

(14) Obligation:

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

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

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

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE