Termination w.r.t. Q proof of TRS/TRCSREx15_Luc98

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__and₂(true, X) -> mark₁(X)
a__and₂(false, Y) -> false
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
a__add₂(0, X) -> mark₁(X)
a__add₂(s₁(X), Y) -> s₁(add₂(X, Y))
a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
a__from₁(X) -> cons₂(X, from₁(s₁(X)))
mark₁(and₂(X1, X2)) -> a__and₂(mark₁(X1), X2)
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), X2, X3)
mark₁(add₂(X1, X2)) -> a__add₂(mark₁(X1), X2)
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(X)
mark₁(true) -> true
mark₁(false) -> false
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
a__and₂(X1, X2) -> and₂(X1, X2)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)
a__add₂(X1, X2) -> add₂(X1, X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__and₂(true, X) -> mark₁(X)
a__and₂(false, Y) -> false
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
a__add₂(0, X) -> mark₁(X)
a__add₂(s₁(X), Y) -> s₁(add₂(X, Y))
a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
a__from₁(X) -> cons₂(X, from₁(s₁(X)))
mark₁(and₂(X1, X2)) -> a__and₂(mark₁(X1), X2)
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), X2, X3)
mark₁(add₂(X1, X2)) -> a__add₂(mark₁(X1), X2)
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(X)
mark₁(true) -> true
mark₁(false) -> false
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
a__and₂(X1, X2) -> and₂(X1, X2)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)
a__add₂(X1, X2) -> add₂(X1, X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(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:

MARK₁(first₂(X1, X2)) -> A__FIRST₂(mark₁(X1), mark₁(X2))
MARK₁(and₂(X1, X2)) -> MARK₁(X1)
MARK₁(from₁(X)) -> A__FROM₁(X)
A__IF₃(false, X, Y) -> MARK₁(Y)
MARK₁(first₂(X1, X2)) -> MARK₁(X2)
A__AND₂(true, X) -> MARK₁(X)
MARK₁(first₂(X1, X2)) -> MARK₁(X1)
A__ADD₂(0, X) -> MARK₁(X)
MARK₁(if₃(X1, X2, X3)) -> MARK₁(X1)
MARK₁(add₂(X1, X2)) -> MARK₁(X1)
MARK₁(and₂(X1, X2)) -> A__AND₂(mark₁(X1), X2)
MARK₁(add₂(X1, X2)) -> A__ADD₂(mark₁(X1), X2)
A__IF₃(true, X, Y) -> MARK₁(X)
MARK₁(if₃(X1, X2, X3)) -> A__IF₃(mark₁(X1), X2, X3)

The TRS R consists of the following rules:

a__and₂(true, X) -> mark₁(X)
a__and₂(false, Y) -> false
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
a__add₂(0, X) -> mark₁(X)
a__add₂(s₁(X), Y) -> s₁(add₂(X, Y))
a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
a__from₁(X) -> cons₂(X, from₁(s₁(X)))
mark₁(and₂(X1, X2)) -> a__and₂(mark₁(X1), X2)
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), X2, X3)
mark₁(add₂(X1, X2)) -> a__add₂(mark₁(X1), X2)
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(X)
mark₁(true) -> true
mark₁(false) -> false
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
a__and₂(X1, X2) -> and₂(X1, X2)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)
a__add₂(X1, X2) -> add₂(X1, X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(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:

MARK₁(first₂(X1, X2)) -> A__FIRST₂(mark₁(X1), mark₁(X2))
MARK₁(and₂(X1, X2)) -> MARK₁(X1)
MARK₁(from₁(X)) -> A__FROM₁(X)
A__IF₃(false, X, Y) -> MARK₁(Y)
MARK₁(first₂(X1, X2)) -> MARK₁(X2)
A__AND₂(true, X) -> MARK₁(X)
MARK₁(first₂(X1, X2)) -> MARK₁(X1)
A__ADD₂(0, X) -> MARK₁(X)
MARK₁(if₃(X1, X2, X3)) -> MARK₁(X1)
MARK₁(add₂(X1, X2)) -> MARK₁(X1)
MARK₁(and₂(X1, X2)) -> A__AND₂(mark₁(X1), X2)
MARK₁(add₂(X1, X2)) -> A__ADD₂(mark₁(X1), X2)
A__IF₃(true, X, Y) -> MARK₁(X)
MARK₁(if₃(X1, X2, X3)) -> A__IF₃(mark₁(X1), X2, X3)

The TRS R consists of the following rules:

a__and₂(true, X) -> mark₁(X)
a__and₂(false, Y) -> false
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
a__add₂(0, X) -> mark₁(X)
a__add₂(s₁(X), Y) -> s₁(add₂(X, Y))
a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
a__from₁(X) -> cons₂(X, from₁(s₁(X)))
mark₁(and₂(X1, X2)) -> a__and₂(mark₁(X1), X2)
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), X2, X3)
mark₁(add₂(X1, X2)) -> a__add₂(mark₁(X1), X2)
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(X)
mark₁(true) -> true
mark₁(false) -> false
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
a__and₂(X1, X2) -> and₂(X1, X2)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)
a__add₂(X1, X2) -> add₂(X1, X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(X)

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

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

MARK₁(first₂(X1, X2)) -> MARK₁(X1)
A__ADD₂(0, X) -> MARK₁(X)
MARK₁(if₃(X1, X2, X3)) -> MARK₁(X1)
MARK₁(add₂(X1, X2)) -> MARK₁(X1)
MARK₁(and₂(X1, X2)) -> MARK₁(X1)
MARK₁(and₂(X1, X2)) -> A__AND₂(mark₁(X1), X2)
A__IF₃(false, X, Y) -> MARK₁(Y)
MARK₁(first₂(X1, X2)) -> MARK₁(X2)
MARK₁(add₂(X1, X2)) -> A__ADD₂(mark₁(X1), X2)
A__IF₃(true, X, Y) -> MARK₁(X)
A__AND₂(true, X) -> MARK₁(X)
MARK₁(if₃(X1, X2, X3)) -> A__IF₃(mark₁(X1), X2, X3)

The TRS R consists of the following rules:

a__and₂(true, X) -> mark₁(X)
a__and₂(false, Y) -> false
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
a__add₂(0, X) -> mark₁(X)
a__add₂(s₁(X), Y) -> s₁(add₂(X, Y))
a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
a__from₁(X) -> cons₂(X, from₁(s₁(X)))
mark₁(and₂(X1, X2)) -> a__and₂(mark₁(X1), X2)
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), X2, X3)
mark₁(add₂(X1, X2)) -> a__add₂(mark₁(X1), X2)
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(X)
mark₁(true) -> true
mark₁(false) -> false
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
a__and₂(X1, X2) -> and₂(X1, X2)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)
a__add₂(X1, X2) -> add₂(X1, X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(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.

MARK₁(first₂(X1, X2)) -> MARK₁(X1)
A__ADD₂(0, X) -> MARK₁(X)
MARK₁(if₃(X1, X2, X3)) -> MARK₁(X1)
MARK₁(add₂(X1, X2)) -> MARK₁(X1)
MARK₁(and₂(X1, X2)) -> MARK₁(X1)
MARK₁(and₂(X1, X2)) -> A__AND₂(mark₁(X1), X2)
A__IF₃(false, X, Y) -> MARK₁(Y)
MARK₁(first₂(X1, X2)) -> MARK₁(X2)
MARK₁(add₂(X1, X2)) -> A__ADD₂(mark₁(X1), X2)
A__IF₃(true, X, Y) -> MARK₁(X)
MARK₁(if₃(X1, X2, X3)) -> A__IF₃(mark₁(X1), X2, X3)

The remaining pairs can at least be oriented weakly.

A__AND₂(true, X) -> MARK₁(X)

Used ordering: Polynomial interpretation [21]:

POL(0) = 0
POL(A__ADD₂(x₁, x₂)) = 2 + 2·x₂
POL(A__AND₂(x₁, x₂)) = 3·x₂
POL(A__IF₃(x₁, x₂, x₃)) = 1 + 3·x₂ + 3·x₃
POL(MARK₁(x₁)) = 2·x₁
POL(a__add₂(x₁, x₂)) = 0
POL(a__and₂(x₁, x₂)) = 0
POL(a__first₂(x₁, x₂)) = 0
POL(a__from₁(x₁)) = 0
POL(a__if₃(x₁, x₂, x₃)) = 0
POL(add₂(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(and₂(x₁, x₂)) = 1 + 2·x₁ + 2·x₂
POL(cons₂(x₁, x₂)) = 0
POL(false) = 0
POL(first₂(x₁, x₂)) = 3 + x₁ + 2·x₂
POL(from₁(x₁)) = 0
POL(if₃(x₁, x₂, x₃)) = 1 + 2·x₁ + 2·x₂ + 2·x₃
POL(mark₁(x₁)) = 0
POL(nil) = 0
POL(s₁(x₁)) = 0
POL(true) = 0

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

A__AND₂(true, X) -> MARK₁(X)

The TRS R consists of the following rules:

a__and₂(true, X) -> mark₁(X)
a__and₂(false, Y) -> false
a__if₃(true, X, Y) -> mark₁(X)
a__if₃(false, X, Y) -> mark₁(Y)
a__add₂(0, X) -> mark₁(X)
a__add₂(s₁(X), Y) -> s₁(add₂(X, Y))
a__first₂(0, X) -> nil
a__first₂(s₁(X), cons₂(Y, Z)) -> cons₂(Y, first₂(X, Z))
a__from₁(X) -> cons₂(X, from₁(s₁(X)))
mark₁(and₂(X1, X2)) -> a__and₂(mark₁(X1), X2)
mark₁(if₃(X1, X2, X3)) -> a__if₃(mark₁(X1), X2, X3)
mark₁(add₂(X1, X2)) -> a__add₂(mark₁(X1), X2)
mark₁(first₂(X1, X2)) -> a__first₂(mark₁(X1), mark₁(X2))
mark₁(from₁(X)) -> a__from₁(X)
mark₁(true) -> true
mark₁(false) -> false
mark₁(0) -> 0
mark₁(s₁(X)) -> s₁(X)
mark₁(nil) -> nil
mark₁(cons₂(X1, X2)) -> cons₂(X1, X2)
a__and₂(X1, X2) -> and₂(X1, X2)
a__if₃(X1, X2, X3) -> if₃(X1, X2, X3)
a__add₂(X1, X2) -> add₂(X1, X2)
a__first₂(X1, X2) -> first₂(X1, X2)
a__from₁(X) -> from₁(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.