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:

plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
min(min(X, Y), Z) → min(X, plus(Y, Z))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(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:

PLUS(s(X), Y) → PLUS(X, Y)
MIN(s(X), s(Y)) → MIN(X, Y)
MIN(min(X, Y), Z) → MIN(X, plus(Y, Z))
MIN(min(X, Y), Z) → PLUS(Y, Z)
QUOT(s(X), s(Y)) → QUOT(min(X, Y), s(Y))
QUOT(s(X), s(Y)) → MIN(X, Y)

The TRS R consists of the following rules:

plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
min(min(X, Y), Z) → min(X, plus(Y, Z))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(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 2 less nodes.

(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:

plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
min(min(X, Y), Z) → min(X, plus(Y, Z))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))

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: Combined order from the following AFS and order.
PLUS(x1, x2)  =  x1
s(x1)  =  s(x1)

Lexicographic Path Order [LPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(7) Obligation:

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

plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
min(min(X, Y), Z) → min(X, plus(Y, Z))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))

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:

MIN(min(X, Y), Z) → MIN(X, plus(Y, Z))
MIN(s(X), s(Y)) → MIN(X, Y)

The TRS R consists of the following rules:

plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
min(min(X, Y), Z) → min(X, plus(Y, Z))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))

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.

MIN(min(X, Y), Z) → MIN(X, plus(Y, Z))
MIN(s(X), s(Y)) → MIN(X, Y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MIN(x1, x2)  =  MIN(x1)
min(x1, x2)  =  min(x1)
Z  =  Z
plus(x1, x2)  =  plus
s(x1)  =  s(x1)
0  =  0

Lexicographic Path Order [LPO].
Precedence:

min₁ > [MIN₁, Z, plus, s₁]
0 > [MIN₁, Z, plus, s₁]

The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
min(min(X, Y), Z) → min(X, plus(Y, Z))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))

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

(13) PisEmptyProof (EQUIVALENT transformation)

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

(14) TRUE

(15) Obligation:

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

QUOT(s(X), s(Y)) → QUOT(min(X, Y), s(Y))

The TRS R consists of the following rules:

plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
min(min(X, Y), Z) → min(X, plus(Y, Z))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))

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

(16) 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)) → QUOT(min(X, Y), s(Y))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
QUOT(x1, x2)  =  QUOT(x1)
s(x1)  =  s(x1)
min(x1, x2)  =  x1
0  =  0
Z  =  Z
plus(x1, x2)  =  x1

Lexicographic Path Order [LPO].
Precedence:

s₁ > QUOT₁
0 > QUOT₁
Z > QUOT₁

The following usable rules [FROCOS05] were oriented:

min(X, 0) → X
min(min(X, Y), Z) → min(X, plus(Y, Z))
min(s(X), s(Y)) → min(X, Y)

(17) Obligation:

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

plus(0, Y) → Y
plus(s(X), Y) → s(plus(X, Y))
min(X, 0) → X
min(s(X), s(Y)) → min(X, Y)
min(min(X, Y), Z) → min(X, plus(Y, Z))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(min(X, Y), s(Y)))

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

(18) PisEmptyProof (EQUIVALENT transformation)

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

(19) TRUE