Termination w.r.t. Q proof of TRS/Rubiogcd.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:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

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:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(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:

IF(false, s(X), s(Y)) → MINUS(Y, X)
IF(true, s(X), s(Y)) → GCD(minus(X, Y), s(Y))
IF(false, s(X), s(Y)) → GCD(minus(Y, X), s(X))
MINUS(X, s(Y)) → PRED(minus(X, Y))
LE(s(X), s(Y)) → LE(X, Y)
GCD(s(X), s(Y)) → IF(le(Y, X), s(X), s(Y))
GCD(s(X), s(Y)) → LE(Y, X)
IF(true, s(X), s(Y)) → MINUS(X, Y)
MINUS(X, s(Y)) → MINUS(X, Y)

The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(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:

IF(false, s(X), s(Y)) → MINUS(Y, X)
IF(true, s(X), s(Y)) → GCD(minus(X, Y), s(Y))
IF(false, s(X), s(Y)) → GCD(minus(Y, X), s(X))
MINUS(X, s(Y)) → PRED(minus(X, Y))
LE(s(X), s(Y)) → LE(X, Y)
GCD(s(X), s(Y)) → IF(le(Y, X), s(X), s(Y))
GCD(s(X), s(Y)) → LE(Y, X)
IF(true, s(X), s(Y)) → MINUS(X, Y)
MINUS(X, s(Y)) → MINUS(X, Y)

The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(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:

IF(true, s(X), s(Y)) → GCD(minus(X, Y), s(Y))
IF(false, s(X), s(Y)) → MINUS(Y, X)
IF(false, s(X), s(Y)) → GCD(minus(Y, X), s(X))
MINUS(X, s(Y)) → PRED(minus(X, Y))
LE(s(X), s(Y)) → LE(X, Y)
GCD(s(X), s(Y)) → IF(le(Y, X), s(X), s(Y))
GCD(s(X), s(Y)) → LE(Y, X)
IF(true, s(X), s(Y)) → MINUS(X, Y)
MINUS(X, s(Y)) → MINUS(X, Y)

The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 4 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ 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:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(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) = x2
s(x1) = s(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

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

                    ↳ QDPOrderProof

                  ↳ QDP

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

MINUS(X, s(Y)) → MINUS(X, Y)

The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

minus(x0, s(x1))
minus(x0, 0)
pred(s(x0))
le(s(x0), s(x1))
le(s(x0), 0)
le(0, x0)
gcd(0, x0)
gcd(s(x0), 0)
gcd(s(x0), s(x1))
if(true, s(x0), s(x1))
if(false, s(x0), s(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.

MINUS(X, s(Y)) → MINUS(X, Y)

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

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ 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:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

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

                  ↳ QDP

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

IF(true, s(X), s(Y)) → GCD(minus(X, Y), s(Y))
IF(false, s(X), s(Y)) → GCD(minus(Y, X), s(X))
GCD(s(X), s(Y)) → IF(le(Y, X), s(X), s(Y))

The TRS R consists of the following rules:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X
le(s(X), s(Y)) → le(X, Y)
le(s(X), 0) → false
le(0, Y) → true
gcd(0, Y) → 0
gcd(s(X), 0) → s(X)
gcd(s(X), s(Y)) → if(le(Y, X), s(X), s(Y))
if(true, s(X), s(Y)) → gcd(minus(X, Y), s(Y))
if(false, s(X), s(Y)) → gcd(minus(Y, X), s(X))

The set Q consists of the following terms:

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

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