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 PisEmptyProof (⇔)
        ↳14 TRUE
    ↳15 QDP
        ↳16 QDPOrderProof (⇔)
        ↳17 QDP
        ↳18 PisEmptyProof (⇔)
        ↳19 TRUE

(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: Polynomial interpretation [POLO]:

POL(PLUS(x1, x2)) = x1   
POL(s(x1)) = 1 + x1   

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: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(MIN(x1, x2)) = x1   
POL(Z) = 0   
POL(min(x1, x2)) = 1 + x1 + x2   
POL(plus(x1, x2)) = 0   
POL(s(x1)) = 1 + x1   

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: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(QUOT(x1, x2)) = x1   
POL(Z) = 0   
POL(min(x1, x2)) = x1   
POL(plus(x1, x2)) = 0   
POL(s(x1)) = 1 + x1   

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