Termination w.r.t. Q proof of TRS/TRCSREx5_Zan97

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:

a__f₁(X) -> a__if₃(mark₁(X), c, f₁(true))
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), mark₁(X2), X3)
mark₁(c) -> c
mark₁(true) -> true
mark₁(false) -> false
a__f₁(X) -> f₁(X)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__f₁(X) -> a__if₃(mark₁(X), c, f₁(true))
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), mark₁(X2), X3)
mark₁(c) -> c
mark₁(true) -> true
mark₁(false) -> false
a__f₁(X) -> f₁(X)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)

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:

MARK₁(if₃(X1, X2, X3)) -> MARK₁(X1)
A__F₁(X) -> A__IF₃(mark₁(X), c, f₁(true))
MARK₁(f₁(X)) -> A__F₁(mark₁(X))
A__IF₃(false, X, Y) -> MARK₁(Y)
A__F₁(X) -> MARK₁(X)
A__IF₃(true, X, Y) -> MARK₁(X)
MARK₁(if₃(X1, X2, X3)) -> MARK₁(X2)
MARK₁(if₃(X1, X2, X3)) -> A__IF₃(mark₁(X1), mark₁(X2), X3)
MARK₁(f₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__f₁(X) -> a__if₃(mark₁(X), c, f₁(true))
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), mark₁(X2), X3)
mark₁(c) -> c
mark₁(true) -> true
mark₁(false) -> false
a__f₁(X) -> f₁(X)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

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

MARK₁(if₃(X1, X2, X3)) -> MARK₁(X1)
A__F₁(X) -> A__IF₃(mark₁(X), c, f₁(true))
MARK₁(f₁(X)) -> A__F₁(mark₁(X))
A__IF₃(false, X, Y) -> MARK₁(Y)
A__F₁(X) -> MARK₁(X)
A__IF₃(true, X, Y) -> MARK₁(X)
MARK₁(if₃(X1, X2, X3)) -> MARK₁(X2)
MARK₁(if₃(X1, X2, X3)) -> A__IF₃(mark₁(X1), mark₁(X2), X3)
MARK₁(f₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__f₁(X) -> a__if₃(mark₁(X), c, f₁(true))
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), mark₁(X2), X3)
mark₁(c) -> c
mark₁(true) -> true
mark₁(false) -> false
a__f₁(X) -> f₁(X)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)

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.

A__IF₃(false, X, Y) -> MARK₁(Y)

The remaining pairs can at least be oriented weakly.

MARK₁(if₃(X1, X2, X3)) -> MARK₁(X1)
A__F₁(X) -> A__IF₃(mark₁(X), c, f₁(true))
MARK₁(f₁(X)) -> A__F₁(mark₁(X))
A__F₁(X) -> MARK₁(X)
A__IF₃(true, X, Y) -> MARK₁(X)
MARK₁(if₃(X1, X2, X3)) -> MARK₁(X2)
MARK₁(if₃(X1, X2, X3)) -> A__IF₃(mark₁(X1), mark₁(X2), X3)
MARK₁(f₁(X)) -> MARK₁(X)

Used ordering: Polynomial interpretation [21]:

POL(A__F₁(x₁)) = 2 + 3·x₁
POL(A__IF₃(x₁, x₂, x₃)) = 2 + 2·x₁ + 3·x₁·x₂ + 2·x₂ + 2·x₂·x₃ + 2·x₃
POL(MARK₁(x₁)) = 2 + 2·x₁
POL(a__f₁(x₁)) = 2·x₁
POL(a__if₃(x₁, x₂, x₃)) = x₁ + 3·x₁·x₂ + 2·x₁·x₃ + 3·x₂ + x₂·x₃ + 3·x₃
POL(c) = 0
POL(f₁(x₁)) = 2·x₁
POL(false) = 1
POL(if₃(x₁, x₂, x₃)) = x₁ + 3·x₁·x₂ + 2·x₁·x₃ + 3·x₂ + x₂·x₃ + 3·x₃
POL(mark₁(x₁)) = x₁
POL(true) = 0

The following usable rules [14] were oriented:

mark₁(false) -> false
mark₁(true) -> true
a__if₃(true, X, Y) -> mark₁(X)
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), mark₁(X2), X3)
mark₁(f₁(X)) -> a__f₁(mark₁(X))
a__f₁(X) -> a__if₃(mark₁(X), c, f₁(true))
a__if₃(false, X, Y) -> mark₁(Y)
mark₁(c) -> c
a__f₁(X) -> f₁(X)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

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

MARK₁(if₃(X1, X2, X3)) -> MARK₁(X1)
A__F₁(X) -> A__IF₃(mark₁(X), c, f₁(true))
MARK₁(f₁(X)) -> A__F₁(mark₁(X))
A__F₁(X) -> MARK₁(X)
A__IF₃(true, X, Y) -> MARK₁(X)
MARK₁(if₃(X1, X2, X3)) -> MARK₁(X2)
MARK₁(if₃(X1, X2, X3)) -> A__IF₃(mark₁(X1), mark₁(X2), X3)
MARK₁(f₁(X)) -> MARK₁(X)

The TRS R consists of the following rules:

a__f₁(X) -> a__if₃(mark₁(X), c, f₁(true))
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), mark₁(X2), X3)
mark₁(c) -> c
mark₁(true) -> true
mark₁(false) -> false
a__f₁(X) -> f₁(X)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)

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.

MARK₁(f₁(X)) -> A__F₁(mark₁(X))
MARK₁(f₁(X)) -> MARK₁(X)

The remaining pairs can at least be oriented weakly.

MARK₁(if₃(X1, X2, X3)) -> MARK₁(X1)
A__F₁(X) -> A__IF₃(mark₁(X), c, f₁(true))
A__F₁(X) -> MARK₁(X)
A__IF₃(true, X, Y) -> MARK₁(X)
MARK₁(if₃(X1, X2, X3)) -> MARK₁(X2)
MARK₁(if₃(X1, X2, X3)) -> A__IF₃(mark₁(X1), mark₁(X2), X3)

Used ordering: Polynomial interpretation [21]:

POL(A__F₁(x₁)) = 1 + 3·x₁
POL(A__IF₃(x₁, x₂, x₃)) = 1 + x₁ + 3·x₂
POL(MARK₁(x₁)) = 1 + 3·x₁
POL(a__f₁(x₁)) = 2 + 3·x₁
POL(a__if₃(x₁, x₂, x₃)) = x₁ + x₁·x₃ + 3·x₂
POL(c) = 0
POL(f₁(x₁)) = 2 + 3·x₁
POL(false) = 2
POL(if₃(x₁, x₂, x₃)) = x₁ + x₁·x₃ + 3·x₂
POL(mark₁(x₁)) = x₁
POL(true) = 0

The following usable rules [14] were oriented:

mark₁(false) -> false
mark₁(true) -> true
a__if₃(true, X, Y) -> mark₁(X)
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), mark₁(X2), X3)
mark₁(f₁(X)) -> a__f₁(mark₁(X))
a__f₁(X) -> a__if₃(mark₁(X), c, f₁(true))
a__if₃(false, X, Y) -> mark₁(Y)
mark₁(c) -> c
a__f₁(X) -> f₁(X)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

MARK₁(if₃(X1, X2, X3)) -> MARK₁(X1)
A__F₁(X) -> A__IF₃(mark₁(X), c, f₁(true))
A__F₁(X) -> MARK₁(X)
A__IF₃(true, X, Y) -> MARK₁(X)
MARK₁(if₃(X1, X2, X3)) -> MARK₁(X2)
MARK₁(if₃(X1, X2, X3)) -> A__IF₃(mark₁(X1), mark₁(X2), X3)

The TRS R consists of the following rules:

a__f₁(X) -> a__if₃(mark₁(X), c, f₁(true))
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), mark₁(X2), X3)
mark₁(c) -> c
mark₁(true) -> true
mark₁(false) -> false
a__f₁(X) -> f₁(X)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

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

MARK₁(if₃(X1, X2, X3)) -> MARK₁(X1)
A__IF₃(true, X, Y) -> MARK₁(X)
MARK₁(if₃(X1, X2, X3)) -> MARK₁(X2)
MARK₁(if₃(X1, X2, X3)) -> A__IF₃(mark₁(X1), mark₁(X2), X3)

The TRS R consists of the following rules:

a__f₁(X) -> a__if₃(mark₁(X), c, f₁(true))
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), mark₁(X2), X3)
mark₁(c) -> c
mark₁(true) -> true
mark₁(false) -> false
a__f₁(X) -> f₁(X)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)

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.

MARK₁(if₃(X1, X2, X3)) -> MARK₁(X1)
MARK₁(if₃(X1, X2, X3)) -> MARK₁(X2)
MARK₁(if₃(X1, X2, X3)) -> A__IF₃(mark₁(X1), mark₁(X2), X3)

The remaining pairs can at least be oriented weakly.

A__IF₃(true, X, Y) -> MARK₁(X)

Used ordering: Polynomial interpretation [21]:

POL(A__IF₃(x₁, x₂, x₃)) = 2·x₂
POL(MARK₁(x₁)) = x₁
POL(a__f₁(x₁)) = 1 + 2·x₁
POL(a__if₃(x₁, x₂, x₃)) = 1 + x₁ + x₁·x₂ + x₁·x₃ + 2·x₂
POL(c) = 0
POL(f₁(x₁)) = 1 + 2·x₁
POL(false) = 2
POL(if₃(x₁, x₂, x₃)) = 1 + x₁ + x₁·x₂ + x₁·x₃ + 2·x₂
POL(mark₁(x₁)) = x₁
POL(true) = 0

The following usable rules [14] were oriented:

a__if₃(true, X, Y) -> mark₁(X)
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), mark₁(X2), X3)
mark₁(f₁(X)) -> a__f₁(mark₁(X))
a__f₁(X) -> a__if₃(mark₁(X), c, f₁(true))
a__if₃(false, X, Y) -> mark₁(Y)
mark₁(c) -> c
mark₁(false) -> false
a__f₁(X) -> f₁(X)
mark₁(true) -> true
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ QDPOrderProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ QDP

                  ↳ QDPOrderProof

                    ↳ QDP

                      ↳ DependencyGraphProof

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

A__IF₃(true, X, Y) -> MARK₁(X)

The TRS R consists of the following rules:

a__f₁(X) -> a__if₃(mark₁(X), c, f₁(true))
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
mark₁(f₁(X)) -> a__f₁(mark₁(X))
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), mark₁(X2), X3)
mark₁(c) -> c
mark₁(true) -> true
mark₁(false) -> false
a__f₁(X) -> f₁(X)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)

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.