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 QDPSizeChangeProof (⇔)
        ↳7 TRUE
    ↳8 QDP
        ↳9 QDPSizeChangeProof (⇔)
        ↳10 TRUE
    ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔)
        ↳13 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) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
s(x1)  =  s(x1)

From the DPs we obtained the following set of size-change graphs:

  • LE(s(X), s(Y)) → LE(X, Y) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 > 1, 2 > 2

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(7) TRUE

(8) 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.

(9) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order and afs together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Homeomorphic Embedding Order

AFS:
false  =  false
s(x1)  =  s(x1)

From the DPs we obtained the following set of size-change graphs:

  • IFMINUS(false, s(X), Y) → MINUS(X, Y) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 2 > 1, 3 >= 2

  • MINUS(s(X), Y) → IFMINUS(le(s(X), Y), s(X), Y) (allowed arguments on rhs = {2, 3})
    The graph contains the following edges 1 >= 2, 2 >= 3

We oriented the following set of usable rules [AAECC05,FROCOS05]. none

(10) TRUE

(11) 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.

(12) QDPSizeChangeProof (EQUIVALENT transformation)

We used the following order together with the size-change analysis [AAECC05] to show that there are no infinite chains for this DP problem.

Order:Polynomial interpretation [POLO]:


POL(0) = 0   
POL(false) = 0   
POL(ifMinus(x1, x2, x3)) = x2   
POL(le(x1, x2)) = 0   
POL(minus(x1, x2)) = x1   
POL(s(x1)) = 1 + x1   
POL(true) = 0   

From the DPs we obtained the following set of size-change graphs:

  • QUOT(s(X), s(Y)) → QUOT(minus(X, Y), s(Y)) (allowed arguments on rhs = {1, 2})
    The graph contains the following edges 1 > 1, 2 >= 2

We oriented the following set of usable rules [AAECC05,FROCOS05].


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

(13) TRUE