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

(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) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) 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)))

The set Q consists of the following terms:

quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
plus(0, x0)
plus(s(x0), x1)
quot(x0, 0, s(x1))

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) 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)))

The set Q consists of the following terms:

quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
plus(0, x0)
plus(s(x0), x1)
quot(x0, 0, s(x1))

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(6) Complex Obligation (AND)

(7) 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)))

The set Q consists of the following terms:

quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
plus(0, x0)
plus(s(x0), x1)
quot(x0, 0, s(x1))

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

(8) 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: Recursive path order with status [RPO].
Precedence:

s₁ > PLUS₂

Status:

PLUS₂: [2,1]
s₁: multiset

The following usable rules [FROCOS05] were oriented: none

(9) 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)))

The set Q consists of the following terms:

quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
plus(0, x0)
plus(s(x0), x1)
quot(x0, 0, s(x1))

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

(10) PisEmptyProof (EQUIVALENT transformation)

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

(11) TRUE

(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))
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)))

The set Q consists of the following terms:

quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
plus(0, x0)
plus(s(x0), x1)
quot(x0, 0, s(x1))

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(s(x), s(y), z) → QUOT(x, y, z)

The remaining pairs can at least be oriented weakly.
Used ordering: 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].
Precedence:

s₁ > 0

Status:

plus: multiset
s₁: [1]
0: multiset

The following usable rules [FROCOS05] were oriented: none

(14) 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)))

The set Q consists of the following terms:

quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
plus(0, x0)
plus(s(x0), x1)
quot(x0, 0, s(x1))

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

(15) 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: Combined order from the following AFS and order.
QUOT(x1, x2, x3)  =  x2
0  =  0
s(x1)  =  s
plus(x1, x2)  =  plus(x2)

Recursive path order with status [RPO].
Precedence:

0 > plus₁ > s

Status:

plus₁: multiset
s: multiset
0: multiset

The following usable rules [FROCOS05] were oriented:

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

(16) 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)))

The set Q consists of the following terms:

quot(0, s(x0), s(x1))
quot(s(x0), s(x1), x2)
plus(0, x0)
plus(s(x0), x1)
quot(x0, 0, s(x1))

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

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE