Termination w.r.t. Q of the given QTRS could not be shown:

(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)
quot(x, 0) → quotZeroErro
quot(x, s(y)) → quotIter(x, s(y), 0, 0, 0)
quotIter(x, s(y), z, u, v) → if(le(x, z), x, s(y), z, u, v)
if(true, x, y, z, u, v) → v
if(false, x, y, z, u, v) → if2(le(y, s(u)), x, y, s(z), s(u), v)
if2(false, x, y, z, u, v) → quotIter(x, y, z, u, v)
if2(true, x, y, z, u, v) → quotIter(x, y, z, 0, s(v))

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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
quot(x, 0) → quotZeroErro
quot(x, s(y)) → quotIter(x, s(y), 0, 0, 0)
quotIter(x, s(y), z, u, v) → if(le(x, z), x, s(y), z, u, v)
if(true, x, y, z, u, v) → v
if(false, x, y, z, u, v) → if2(le(y, s(u)), x, y, s(z), s(u), v)
if2(false, x, y, z, u, v) → quotIter(x, y, z, u, v)
if2(true, x, y, z, u, v) → quotIter(x, y, z, 0, s(v))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, 0)
quot(x0, s(x1))
quotIter(x0, s(x1), x2, x3, x4)
if(true, x0, x1, x2, x3, x4)
if(false, x0, x1, x2, x3, x4)
if2(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, x3, x4)

(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:

LE(s(x), s(y)) → LE(x, y)
QUOT(x, s(y)) → QUOTITER(x, s(y), 0, 0, 0)
QUOTITER(x, s(y), z, u, v) → IF(le(x, z), x, s(y), z, u, v)
QUOTITER(x, s(y), z, u, v) → LE(x, z)
IF(false, x, y, z, u, v) → IF2(le(y, s(u)), x, y, s(z), s(u), v)
IF(false, x, y, z, u, v) → LE(y, s(u))
IF2(false, x, y, z, u, v) → QUOTITER(x, y, z, u, v)
IF2(true, x, y, z, u, v) → QUOTITER(x, y, z, 0, s(v))

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)
quot(x, 0) → quotZeroErro
quot(x, s(y)) → quotIter(x, s(y), 0, 0, 0)
quotIter(x, s(y), z, u, v) → if(le(x, z), x, s(y), z, u, v)
if(true, x, y, z, u, v) → v
if(false, x, y, z, u, v) → if2(le(y, s(u)), x, y, s(z), s(u), v)
if2(false, x, y, z, u, v) → quotIter(x, y, z, u, v)
if2(true, x, y, z, u, v) → quotIter(x, y, z, 0, s(v))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, 0)
quot(x0, s(x1))
quotIter(x0, s(x1), x2, x3, x4)
if(true, x0, x1, x2, x3, x4)
if(false, x0, x1, x2, x3, x4)
if2(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, x3, x4)

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 3 less nodes.

(6) Complex Obligation (AND)

(7) 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)
quot(x, 0) → quotZeroErro
quot(x, s(y)) → quotIter(x, s(y), 0, 0, 0)
quotIter(x, s(y), z, u, v) → if(le(x, z), x, s(y), z, u, v)
if(true, x, y, z, u, v) → v
if(false, x, y, z, u, v) → if2(le(y, s(u)), x, y, s(z), s(u), v)
if2(false, x, y, z, u, v) → quotIter(x, y, z, u, v)
if2(true, x, y, z, u, v) → quotIter(x, y, z, 0, s(v))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, 0)
quot(x0, s(x1))
quotIter(x0, s(x1), x2, x3, x4)
if(true, x0, x1, x2, x3, x4)
if(false, x0, x1, x2, x3, x4)
if2(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, x3, x4)

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.

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(x1)
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Precedence:

s₁ > LE₁

Status:

s₁: [1]
LE₁: [1]

The following usable rules [FROCOS05] were oriented: none

(9) 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)
quot(x, 0) → quotZeroErro
quot(x, s(y)) → quotIter(x, s(y), 0, 0, 0)
quotIter(x, s(y), z, u, v) → if(le(x, z), x, s(y), z, u, v)
if(true, x, y, z, u, v) → v
if(false, x, y, z, u, v) → if2(le(y, s(u)), x, y, s(z), s(u), v)
if2(false, x, y, z, u, v) → quotIter(x, y, z, u, v)
if2(true, x, y, z, u, v) → quotIter(x, y, z, 0, s(v))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, 0)
quot(x0, s(x1))
quotIter(x0, s(x1), x2, x3, x4)
if(true, x0, x1, x2, x3, x4)
if(false, x0, x1, x2, x3, x4)
if2(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, x3, x4)

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:

QUOTITER(x, s(y), z, u, v) → IF(le(x, z), x, s(y), z, u, v)
IF(false, x, y, z, u, v) → IF2(le(y, s(u)), x, y, s(z), s(u), v)
IF2(false, x, y, z, u, v) → QUOTITER(x, y, z, u, v)
IF2(true, x, y, z, u, v) → QUOTITER(x, y, z, 0, s(v))

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)
quot(x, 0) → quotZeroErro
quot(x, s(y)) → quotIter(x, s(y), 0, 0, 0)
quotIter(x, s(y), z, u, v) → if(le(x, z), x, s(y), z, u, v)
if(true, x, y, z, u, v) → v
if(false, x, y, z, u, v) → if2(le(y, s(u)), x, y, s(z), s(u), v)
if2(false, x, y, z, u, v) → quotIter(x, y, z, u, v)
if2(true, x, y, z, u, v) → quotIter(x, y, z, 0, s(v))

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
quot(x0, 0)
quot(x0, s(x1))
quotIter(x0, s(x1), x2, x3, x4)
if(true, x0, x1, x2, x3, x4)
if(false, x0, x1, x2, x3, x4)
if2(false, x0, x1, x2, x3, x4)
if2(true, x0, x1, x2, x3, x4)

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