Termination w.r.t. Q proof of /tmp/tmpfnN7F7/#4.30c.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)
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, x, y) → gcd(minus(x, y), y)
if_gcd(false, x, y) → gcd(minus(y, x), x)

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)
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, x, y) → gcd(minus(x, y), y)
if_gcd(false, x, y) → gcd(minus(y, x), x)

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:

IF_GCD(false, x, y) → GCD(minus(y, x), x)
IF_GCD(true, x, y) → GCD(minus(x, y), y)
MINUS(s(x), s(y)) → MINUS(x, y)
GCD(s(x), s(y)) → IF_GCD(le(y, x), s(x), s(y))
IF_GCD(false, x, y) → MINUS(y, x)
IF_GCD(true, x, y) → MINUS(x, y)
LE(s(x), s(y)) → LE(x, y)
GCD(s(x), s(y)) → LE(y, x)

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)
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, x, y) → gcd(minus(x, y), y)
if_gcd(false, x, y) → gcd(minus(y, x), x)

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:

IF_GCD(false, x, y) → GCD(minus(y, x), x)
IF_GCD(true, x, y) → GCD(minus(x, y), y)
MINUS(s(x), s(y)) → MINUS(x, y)
GCD(s(x), s(y)) → IF_GCD(le(y, x), s(x), s(y))
IF_GCD(false, x, y) → MINUS(y, x)
IF_GCD(true, x, y) → MINUS(x, y)
LE(s(x), s(y)) → LE(x, y)
GCD(s(x), s(y)) → LE(y, x)

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)
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, x, y) → gcd(minus(x, y), y)
if_gcd(false, x, y) → gcd(minus(y, x), x)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

MINUS(s(x), s(y)) → MINUS(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)
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, x, y) → gcd(minus(x, y), y)
if_gcd(false, x, y) → gcd(minus(y, x), x)

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.

MINUS(s(x), s(y)) → MINUS(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(MINUS(x₁, x₂)) = (3)x_2
POL(s(x₁)) = 4 + (2)x_1

The value of delta used in the strict ordering is 12.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ 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)
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, x, y) → gcd(minus(x, y), y)
if_gcd(false, x, y) → gcd(minus(y, x), x)

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

            ↳ 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)
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, x, y) → gcd(minus(x, y), y)
if_gcd(false, x, y) → gcd(minus(y, x), x)

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)) = 4 + (2)x_1
POL(LE(x₁, x₂)) = (3)x_2

The value of delta used in the strict ordering is 12.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ 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)
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, x, y) → gcd(minus(x, y), y)
if_gcd(false, x, y) → gcd(minus(y, x), x)

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

          ↳ QDP

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

IF_GCD(false, x, y) → GCD(minus(y, x), x)
IF_GCD(true, x, y) → GCD(minus(x, y), y)
GCD(s(x), s(y)) → IF_GCD(le(y, x), s(x), s(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)
minus(x, 0) → x
minus(0, x) → 0
minus(s(x), s(y)) → minus(x, y)
gcd(0, y) → y
gcd(s(x), 0) → s(x)
gcd(s(x), s(y)) → if_gcd(le(y, x), s(x), s(y))
if_gcd(true, x, y) → gcd(minus(x, y), y)
if_gcd(false, x, y) → gcd(minus(y, x), x)

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