Termination w.r.t. Q proof of /tmp/tmpDSbu8L/thiemann36.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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.

↳ QTRS

  ↳ DependencyPairsProof

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.

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

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

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.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 2 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

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.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x₁)) = 5/4 + (15/4)x_1
POL(LE(x₁, x₂)) = x_1 + (13/4)x_2

The value of delta used in the strict ordering is 85/16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

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

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

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