Termination w.r.t. Q proof of /tmp/tmpiXuQ8U/gcd.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

  ↳ 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))

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:

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

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:

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

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 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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))

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)) = 1/2 + (13/4)x_1
POL(LE(x₁, x₂)) = (15/4)x_2

The value of delta used in the strict ordering is 15/8.
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:

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.
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:

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))

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(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)) = (2)x_2
POL(s(x₁)) = 1/4 + (7/2)x_1

The value of delta used in the strict ordering is 1/2.
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:

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.
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

            ↳ QDPOrderProof

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

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 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.
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.

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

POL(minus(x₁, x₂)) = 2 + (2)x_1
POL(le(x₁, x₂)) = (4)x_2
POL(true) = 0
POL(pred(x₁)) = 5/4 + (1/4)x_1
POL(false) = 0
POL(s(x₁)) = 4 + (4)x_1
POL(GCD(x₁, x₂)) = 3/4 + (4)x_1 + (2)x_2
POL(IF(x₁, x₂, x₃)) = 3 + (2)x_2 + (2)x_3
POL(0) = 0

The value of delta used in the strict ordering is 9/4.
The following usable rules [17] were oriented:

minus(X, s(Y)) → pred(minus(X, Y))
minus(X, 0) → X
pred(s(X)) → X

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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))

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.