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

minus₂(X, s₁(Y)) -> pred₁(minus₂(X, Y))
minus₂(X, 0) -> X
pred₁(s₁(X)) -> X
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
le₂(s₁(X), 0) -> false
le₂(0, Y) -> true
gcd₂(0, Y) -> 0
gcd₂(s₁(X), 0) -> s₁(X)
gcd₂(s₁(X), s₁(Y)) -> if₃(le₂(Y, X), s₁(X), s₁(Y))
if₃(true, s₁(X), s₁(Y)) -> gcd₂(minus₂(X, Y), s₁(Y))
if₃(false, s₁(X), s₁(Y)) -> gcd₂(minus₂(Y, X), s₁(X))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

minus₂(X, s₁(Y)) -> pred₁(minus₂(X, Y))
minus₂(X, 0) -> X
pred₁(s₁(X)) -> X
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
le₂(s₁(X), 0) -> false
le₂(0, Y) -> true
gcd₂(0, Y) -> 0
gcd₂(s₁(X), 0) -> s₁(X)
gcd₂(s₁(X), s₁(Y)) -> if₃(le₂(Y, X), s₁(X), s₁(Y))
if₃(true, s₁(X), s₁(Y)) -> gcd₂(minus₂(X, Y), s₁(Y))
if₃(false, s₁(X), s₁(Y)) -> gcd₂(minus₂(Y, X), s₁(X))

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:

IF₃(false, s₁(X), s₁(Y)) -> MINUS₂(Y, X)
IF₃(true, s₁(X), s₁(Y)) -> GCD₂(minus₂(X, Y), s₁(Y))
IF₃(false, s₁(X), s₁(Y)) -> GCD₂(minus₂(Y, X), s₁(X))
MINUS₂(X, s₁(Y)) -> PRED₁(minus₂(X, Y))
LE₂(s₁(X), s₁(Y)) -> LE₂(X, Y)
GCD₂(s₁(X), s₁(Y)) -> IF₃(le₂(Y, X), s₁(X), s₁(Y))
GCD₂(s₁(X), s₁(Y)) -> LE₂(Y, X)
IF₃(true, s₁(X), s₁(Y)) -> MINUS₂(X, Y)
MINUS₂(X, s₁(Y)) -> MINUS₂(X, Y)

The TRS R consists of the following rules:

minus₂(X, s₁(Y)) -> pred₁(minus₂(X, Y))
minus₂(X, 0) -> X
pred₁(s₁(X)) -> X
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
le₂(s₁(X), 0) -> false
le₂(0, Y) -> true
gcd₂(0, Y) -> 0
gcd₂(s₁(X), 0) -> s₁(X)
gcd₂(s₁(X), s₁(Y)) -> if₃(le₂(Y, X), s₁(X), s₁(Y))
if₃(true, s₁(X), s₁(Y)) -> gcd₂(minus₂(X, Y), s₁(Y))
if₃(false, s₁(X), s₁(Y)) -> gcd₂(minus₂(Y, X), s₁(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:

IF₃(false, s₁(X), s₁(Y)) -> MINUS₂(Y, X)
IF₃(true, s₁(X), s₁(Y)) -> GCD₂(minus₂(X, Y), s₁(Y))
IF₃(false, s₁(X), s₁(Y)) -> GCD₂(minus₂(Y, X), s₁(X))
MINUS₂(X, s₁(Y)) -> PRED₁(minus₂(X, Y))
LE₂(s₁(X), s₁(Y)) -> LE₂(X, Y)
GCD₂(s₁(X), s₁(Y)) -> IF₃(le₂(Y, X), s₁(X), s₁(Y))
GCD₂(s₁(X), s₁(Y)) -> LE₂(Y, X)
IF₃(true, s₁(X), s₁(Y)) -> MINUS₂(X, Y)
MINUS₂(X, s₁(Y)) -> MINUS₂(X, Y)

The TRS R consists of the following rules:

minus₂(X, s₁(Y)) -> pred₁(minus₂(X, Y))
minus₂(X, 0) -> X
pred₁(s₁(X)) -> X
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
le₂(s₁(X), 0) -> false
le₂(0, Y) -> true
gcd₂(0, Y) -> 0
gcd₂(s₁(X), 0) -> s₁(X)
gcd₂(s₁(X), s₁(Y)) -> if₃(le₂(Y, X), s₁(X), s₁(Y))
if₃(true, s₁(X), s₁(Y)) -> gcd₂(minus₂(X, Y), s₁(Y))
if₃(false, s₁(X), s₁(Y)) -> gcd₂(minus₂(Y, X), s₁(X))

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 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

LE₂(s₁(X), s₁(Y)) -> LE₂(X, Y)

The TRS R consists of the following rules:

minus₂(X, s₁(Y)) -> pred₁(minus₂(X, Y))
minus₂(X, 0) -> X
pred₁(s₁(X)) -> X
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
le₂(s₁(X), 0) -> false
le₂(0, Y) -> true
gcd₂(0, Y) -> 0
gcd₂(s₁(X), 0) -> s₁(X)
gcd₂(s₁(X), s₁(Y)) -> if₃(le₂(Y, X), s₁(X), s₁(Y))
if₃(true, s₁(X), s₁(Y)) -> gcd₂(minus₂(X, Y), s₁(Y))
if₃(false, s₁(X), s₁(Y)) -> gcd₂(minus₂(Y, X), s₁(X))

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.

LE₂(s₁(X), s₁(Y)) -> LE₂(X, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( LE₂(x₁, x₂) ) = 3x₂ + 3

POL( s₁(x₁) ) = x₁ + 3

The following usable rules [14] 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:

minus₂(X, s₁(Y)) -> pred₁(minus₂(X, Y))
minus₂(X, 0) -> X
pred₁(s₁(X)) -> X
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
le₂(s₁(X), 0) -> false
le₂(0, Y) -> true
gcd₂(0, Y) -> 0
gcd₂(s₁(X), 0) -> s₁(X)
gcd₂(s₁(X), s₁(Y)) -> if₃(le₂(Y, X), s₁(X), s₁(Y))
if₃(true, s₁(X), s₁(Y)) -> gcd₂(minus₂(X, Y), s₁(Y))
if₃(false, s₁(X), s₁(Y)) -> gcd₂(minus₂(Y, X), s₁(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:

MINUS₂(X, s₁(Y)) -> MINUS₂(X, Y)

The TRS R consists of the following rules:

minus₂(X, s₁(Y)) -> pred₁(minus₂(X, Y))
minus₂(X, 0) -> X
pred₁(s₁(X)) -> X
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
le₂(s₁(X), 0) -> false
le₂(0, Y) -> true
gcd₂(0, Y) -> 0
gcd₂(s₁(X), 0) -> s₁(X)
gcd₂(s₁(X), s₁(Y)) -> if₃(le₂(Y, X), s₁(X), s₁(Y))
if₃(true, s₁(X), s₁(Y)) -> gcd₂(minus₂(X, Y), s₁(Y))
if₃(false, s₁(X), s₁(Y)) -> gcd₂(minus₂(Y, X), s₁(X))

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.

MINUS₂(X, s₁(Y)) -> MINUS₂(X, Y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MINUS₂(x₁, x₂) ) = 2x₁ + 2x₂ + 2

POL( s₁(x₁) ) = 3x₁ + 1

The following usable rules [14] 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:

minus₂(X, s₁(Y)) -> pred₁(minus₂(X, Y))
minus₂(X, 0) -> X
pred₁(s₁(X)) -> X
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
le₂(s₁(X), 0) -> false
le₂(0, Y) -> true
gcd₂(0, Y) -> 0
gcd₂(s₁(X), 0) -> s₁(X)
gcd₂(s₁(X), s₁(Y)) -> if₃(le₂(Y, X), s₁(X), s₁(Y))
if₃(true, s₁(X), s₁(Y)) -> gcd₂(minus₂(X, Y), s₁(Y))
if₃(false, s₁(X), s₁(Y)) -> gcd₂(minus₂(Y, X), s₁(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:

IF₃(true, s₁(X), s₁(Y)) -> GCD₂(minus₂(X, Y), s₁(Y))
IF₃(false, s₁(X), s₁(Y)) -> GCD₂(minus₂(Y, X), s₁(X))
GCD₂(s₁(X), s₁(Y)) -> IF₃(le₂(Y, X), s₁(X), s₁(Y))

The TRS R consists of the following rules:

minus₂(X, s₁(Y)) -> pred₁(minus₂(X, Y))
minus₂(X, 0) -> X
pred₁(s₁(X)) -> X
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
le₂(s₁(X), 0) -> false
le₂(0, Y) -> true
gcd₂(0, Y) -> 0
gcd₂(s₁(X), 0) -> s₁(X)
gcd₂(s₁(X), s₁(Y)) -> if₃(le₂(Y, X), s₁(X), s₁(Y))
if₃(true, s₁(X), s₁(Y)) -> gcd₂(minus₂(X, Y), s₁(Y))
if₃(false, s₁(X), s₁(Y)) -> gcd₂(minus₂(Y, X), s₁(X))

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.

IF₃(true, s₁(X), s₁(Y)) -> GCD₂(minus₂(X, Y), s₁(Y))
IF₃(false, s₁(X), s₁(Y)) -> GCD₂(minus₂(Y, X), s₁(X))

The remaining pairs can at least be oriented weakly.

GCD₂(s₁(X), s₁(Y)) -> IF₃(le₂(Y, X), s₁(X), s₁(Y))

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( minus₂(x₁, x₂) ) = x₁

POL( true ) = 1

POL( GCD₂(x₁, x₂) ) = max{0, 3x₁ + 3x₂ - 2}

POL( false ) = 1

POL( IF₃(x₁, ..., x₃) ) = max{0, 3x₂ + 3x₃ - 2}

POL( 0 ) = max{0, -2}

POL( s₁(x₁) ) = x₁ + 3

POL( le₂(x₁, x₂) ) = max{0, -3}

POL( pred₁(x₁) ) = max{0, x₁ - 2}

The following usable rules [14] were oriented:

minus₂(X, 0) -> X
pred₁(s₁(X)) -> X
minus₂(X, s₁(Y)) -> pred₁(minus₂(X, Y))

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

GCD₂(s₁(X), s₁(Y)) -> IF₃(le₂(Y, X), s₁(X), s₁(Y))

The TRS R consists of the following rules:

minus₂(X, s₁(Y)) -> pred₁(minus₂(X, Y))
minus₂(X, 0) -> X
pred₁(s₁(X)) -> X
le₂(s₁(X), s₁(Y)) -> le₂(X, Y)
le₂(s₁(X), 0) -> false
le₂(0, Y) -> true
gcd₂(0, Y) -> 0
gcd₂(s₁(X), 0) -> s₁(X)
gcd₂(s₁(X), s₁(Y)) -> if₃(le₂(Y, X), s₁(X), s₁(Y))
if₃(true, s₁(X), s₁(Y)) -> gcd₂(minus₂(X, Y), s₁(Y))
if₃(false, s₁(X), s₁(Y)) -> gcd₂(minus₂(Y, X), s₁(X))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.