Termination w.r.t. Q proof of TRS/nontermin/CSREx1

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
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → X
if(false, X, Y) → Y
diff(X, Y) → if(leq(X, Y), 0, s(diff(p(X), 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
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → X
if(false, X, Y) → Y
diff(X, Y) → if(leq(X, Y), 0, s(diff(p(X), 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
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → X
if(false, X, Y) → Y
diff(X, Y) → if(leq(X, Y), 0, s(diff(p(X), Y)))

The set Q consists of the following terms:

p(0)
p(s(x0))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
diff(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:

DIFF(X, Y) → LEQ(X, Y)
LEQ(s(X), s(Y)) → LEQ(X, Y)
DIFF(X, Y) → P(X)
DIFF(X, Y) → DIFF(p(X), Y)
DIFF(X, Y) → IF(leq(X, Y), 0, s(diff(p(X), Y)))

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → X
if(false, X, Y) → Y
diff(X, Y) → if(leq(X, Y), 0, s(diff(p(X), Y)))

The set Q consists of the following terms:

p(0)
p(s(x0))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
diff(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:

DIFF(X, Y) → LEQ(X, Y)
LEQ(s(X), s(Y)) → LEQ(X, Y)
DIFF(X, Y) → P(X)
DIFF(X, Y) → DIFF(p(X), Y)
DIFF(X, Y) → IF(leq(X, Y), 0, s(diff(p(X), Y)))

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → X
if(false, X, Y) → Y
diff(X, Y) → if(leq(X, Y), 0, s(diff(p(X), Y)))

The set Q consists of the following terms:

p(0)
p(s(x0))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
diff(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:

DIFF(X, Y) → LEQ(X, Y)
LEQ(s(X), s(Y)) → LEQ(X, Y)
DIFF(X, Y) → P(X)
DIFF(X, Y) → DIFF(p(X), Y)
DIFF(X, Y) → IF(leq(X, Y), 0, s(diff(p(X), Y)))

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → X
if(false, X, Y) → Y
diff(X, Y) → if(leq(X, Y), 0, s(diff(p(X), Y)))

The set Q consists of the following terms:

p(0)
p(s(x0))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
diff(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 3 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:

LEQ(s(X), s(Y)) → LEQ(X, Y)

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → X
if(false, X, Y) → Y
diff(X, Y) → if(leq(X, Y), 0, s(diff(p(X), Y)))

The set Q consists of the following terms:

p(0)
p(s(x0))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
diff(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.

LEQ(s(X), s(Y)) → LEQ(X, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
LEQ(x1, x2) = LEQ(x2)
s(x1) = s(x1)

Recursive Path Order [2].
Precedence:

s₁ > LEQ₁

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
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → X
if(false, X, Y) → Y
diff(X, Y) → if(leq(X, Y), 0, s(diff(p(X), Y)))

The set Q consists of the following terms:

p(0)
p(s(x0))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
diff(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:

DIFF(X, Y) → DIFF(p(X), Y)

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → X
if(false, X, Y) → Y
diff(X, Y) → if(leq(X, Y), 0, s(diff(p(X), Y)))

The set Q consists of the following terms:

p(0)
p(s(x0))
leq(0, x0)
leq(s(x0), 0)
leq(s(x0), s(x1))
if(true, x0, x1)
if(false, x0, x1)
diff(x0, x1)

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