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 DependencyGraphProof (⇔)
        ↳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:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
minus(0, Y) → 0
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) → 0
ifMinus(false, s(X), Y) → s(minus(X, Y))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(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:

LE(s(X), s(Y)) → LE(X, Y)
MINUS(s(X), Y) → IFMINUS(le(s(X), Y), s(X), Y)
MINUS(s(X), Y) → LE(s(X), Y)
IFMINUS(false, s(X), Y) → MINUS(X, Y)
QUOT(s(X), s(Y)) → QUOT(minus(X, Y), s(Y))
QUOT(s(X), s(Y)) → MINUS(X, Y)

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
minus(0, Y) → 0
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) → 0
ifMinus(false, s(X), Y) → s(minus(X, Y))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(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:

LE(s(X), s(Y)) → LE(X, Y)

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
minus(0, Y) → 0
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) → 0
ifMinus(false, s(X), Y) → s(minus(X, Y))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(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.


LE(s(X), s(Y)) → LE(X, Y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LE(x1, x2)  =  LE(x2)
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:
trivial

Status:
LE1: [1]
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:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
minus(0, Y) → 0
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) → 0
ifMinus(false, s(X), Y) → s(minus(X, Y))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(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:

MINUS(s(X), Y) → IFMINUS(le(s(X), Y), s(X), Y)
IFMINUS(false, s(X), Y) → MINUS(X, Y)

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
minus(0, Y) → 0
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) → 0
ifMinus(false, s(X), Y) → s(minus(X, Y))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(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.


IFMINUS(false, s(X), Y) → MINUS(X, Y)
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2)  =  MINUS(x1, x2)
s(x1)  =  s(x1)
IFMINUS(x1, x2, x3)  =  IFMINUS(x2, x3)
le(x1, x2)  =  le(x1, x2)
false  =  false
0  =  0
true  =  true

Lexicographic path order with status [LPO].
Quasi-Precedence:
[s1, le2, false, 0] > [MINUS2, IFMINUS2]
true > [MINUS2, IFMINUS2]

Status:
MINUS2: [2,1]
s1: [1]
IFMINUS2: [2,1]
le2: [2,1]
false: []
0: []
true: []


The following usable rules [FROCOS05] were oriented: none

(12) Obligation:

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

MINUS(s(X), Y) → IFMINUS(le(s(X), Y), s(X), Y)

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
minus(0, Y) → 0
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) → 0
ifMinus(false, s(X), Y) → s(minus(X, Y))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(X, Y), s(Y)))

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

(13) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(14) TRUE

(15) Obligation:

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

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

The TRS R consists of the following rules:

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
minus(0, Y) → 0
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) → 0
ifMinus(false, s(X), Y) → s(minus(X, Y))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(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(minus(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, x2)
s(x1)  =  s(x1)
minus(x1, x2)  =  x1
0  =  0
ifMinus(x1, x2, x3)  =  x2
le(x1, x2)  =  le
false  =  false
true  =  true

Lexicographic path order with status [LPO].
Quasi-Precedence:
QUOT2 > [s1, 0, true] > false > le

Status:
QUOT2: [2,1]
s1: [1]
0: []
le: []
false: []
true: []


The following usable rules [FROCOS05] were oriented:

minus(0, Y) → 0
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y)
ifMinus(false, s(X), Y) → s(minus(X, Y))
ifMinus(true, s(X), Y) → 0

(17) Obligation:

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

le(0, Y) → true
le(s(X), 0) → false
le(s(X), s(Y)) → le(X, Y)
minus(0, Y) → 0
minus(s(X), Y) → ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) → 0
ifMinus(false, s(X), Y) → s(minus(X, Y))
quot(0, s(Y)) → 0
quot(s(X), s(Y)) → s(quot(minus(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