Termination w.r.t. Q proof of TRS/secret05aprove4.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) → s(s(0))
p(s(x)) → x
p(p(s(x))) → p(x)
le(p(s(x)), x) → le(x, x)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
minus(x, y) → if(le(x, y), x, y)
if(true, x, y) → 0
if(false, x, y) → s(minus(p(x), y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

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

Q is empty.

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

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

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

The TRS R consists of the following rules:

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

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

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

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

The TRS R consists of the following rules:

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

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

LE(p(s(x)), x) → LE(x, x)
LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

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

Q is empty.
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(p(s(x)), x) → LE(x, x)
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(x1)
p(x1) = p(x1)
s(x1) = s(x1)

Recursive path order with status [2].
Precedence:

p₁ > LE₁
s₁ > LE₁

Status:

LE₁: multiset
s₁: multiset
p₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

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

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

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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

              ↳ QDP

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

P(p(s(x))) → P(x)

The TRS R consists of the following rules:

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

Q is empty.
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.

P(p(s(x))) → P(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
P(x1) = P(x1)
p(x1) = p(x1)
s(x1) = x1

Recursive path order with status [2].
Precedence:

p₁ > P₁

Status:

P₁: multiset
p₁: multiset

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

              ↳ QDP

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

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

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

      ↳ EdgeDeletionProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

MINUS(x, y) → IF(le(x, y), x, y)
IF(false, x, y) → MINUS(p(x), y)

The TRS R consists of the following rules:

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

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