Termination w.r.t. Q proof of TRS/nontermin/AG01#4.27.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:

p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(x, s(y)) → if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
if(true, x, y) → x
if(false, x, y) → y

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

Q restricted rewrite system:
The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(x, s(y)) → if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
if(true, x, y) → x
if(false, x, y) → y

Q is empty.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

Q restricted rewrite system:
The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(x, s(y)) → if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
if(true, x, y) → x
if(false, x, y) → y

The set Q consists of the following terms:

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(x0, s(x1))
if(true, x0, x1)
if(false, x0, x1)

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

MINUS(x, s(y)) → MINUS(x, p(s(y)))
MINUS(x, s(y)) → P(minus(x, p(s(y))))
MINUS(x, s(y)) → LE(x, s(y))
MINUS(x, s(y)) → IF(le(x, s(y)), 0, p(minus(x, p(s(y)))))
LE(s(x), s(y)) → LE(x, y)
MINUS(x, s(y)) → P(s(y))

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(x, s(y)) → if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
if(true, x, y) → x
if(false, x, y) → y

The set Q consists of the following terms:

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(x0, s(x1))
if(true, x0, x1)
if(false, x0, x1)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

MINUS(x, s(y)) → MINUS(x, p(s(y)))
MINUS(x, s(y)) → P(minus(x, p(s(y))))
MINUS(x, s(y)) → LE(x, s(y))
MINUS(x, s(y)) → IF(le(x, s(y)), 0, p(minus(x, p(s(y)))))
LE(s(x), s(y)) → LE(x, y)
MINUS(x, s(y)) → P(s(y))

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(x, s(y)) → if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
if(true, x, y) → x
if(false, x, y) → y

The set Q consists of the following terms:

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(x0, s(x1))
if(true, x0, x1)
if(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

MINUS(x, s(y)) → MINUS(x, p(s(y)))
MINUS(x, s(y)) → LE(x, s(y))
MINUS(x, s(y)) → P(minus(x, p(s(y))))
MINUS(x, s(y)) → IF(le(x, s(y)), 0, p(minus(x, p(s(y)))))
LE(s(x), s(y)) → LE(x, y)
MINUS(x, s(y)) → P(s(y))

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(x, s(y)) → if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
if(true, x, y) → x
if(false, x, y) → y

The set Q consists of the following terms:

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(x0, s(x1))
if(true, x0, x1)
if(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 4 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ 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:

p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(x, s(y)) → if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
if(true, x, y) → x
if(false, x, y) → y

The set Q consists of the following terms:

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(x0, s(x1))
if(true, x0, x1)
if(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

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: Combined order from the following AFS and order.
LE(x1, x2) = LE(x2)
s(x1) = s(x1)

Recursive Path Order [2].
Precedence:

s₁ > LE₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(x, s(y)) → if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
if(true, x, y) → x
if(false, x, y) → y

The set Q consists of the following terms:

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(x0, s(x1))
if(true, x0, x1)
if(false, x0, x1)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

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

MINUS(x, s(y)) → MINUS(x, p(s(y)))

The TRS R consists of the following rules:

p(0) → 0
p(s(x)) → x
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, 0) → x
minus(x, s(y)) → if(le(x, s(y)), 0, p(minus(x, p(s(y)))))
if(true, x, y) → x
if(false, x, y) → y

The set Q consists of the following terms:

p(0)
p(s(x0))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
minus(x0, 0)
minus(x0, s(x1))
if(true, x0, x1)
if(false, x0, x1)

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