Termination w.r.t. Q proof of /tmp/tmphV3wcd/Ex15_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__and(true, X) → mark(X)
a__and(false, Y) → false
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
a__from(X) → cons(X, from(s(X)))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(add(X1, X2)) → a__add(mark(X1), X2)
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(X)
mark(true) → true
mark(false) → false
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__add(X1, X2) → add(X1, X2)
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(X)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

a__and(true, X) → mark(X)
a__and(false, Y) → false
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
a__from(X) → cons(X, from(s(X)))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(add(X1, X2)) → a__add(mark(X1), X2)
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(X)
mark(true) → true
mark(false) → false
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__add(X1, X2) → add(X1, X2)
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(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:

MARK(first(X1, X2)) → MARK(X1)
A__ADD(0, X) → MARK(X)
MARK(add(X1, X2)) → A__ADD(mark(X1), X2)
MARK(first(X1, X2)) → MARK(X2)
A__IF(true, X, Y) → MARK(X)
A__AND(true, X) → MARK(X)
MARK(first(X1, X2)) → A__FIRST(mark(X1), mark(X2))
MARK(add(X1, X2)) → MARK(X1)
A__IF(false, X, Y) → MARK(Y)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(and(X1, X2)) → MARK(X1)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
MARK(from(X)) → A__FROM(X)

The TRS R consists of the following rules:

a__and(true, X) → mark(X)
a__and(false, Y) → false
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
a__from(X) → cons(X, from(s(X)))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(add(X1, X2)) → a__add(mark(X1), X2)
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(X)
mark(true) → true
mark(false) → false
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__add(X1, X2) → add(X1, X2)
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(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:

MARK(first(X1, X2)) → MARK(X1)
A__ADD(0, X) → MARK(X)
MARK(add(X1, X2)) → A__ADD(mark(X1), X2)
MARK(first(X1, X2)) → MARK(X2)
A__IF(true, X, Y) → MARK(X)
A__AND(true, X) → MARK(X)
MARK(first(X1, X2)) → A__FIRST(mark(X1), mark(X2))
MARK(add(X1, X2)) → MARK(X1)
A__IF(false, X, Y) → MARK(Y)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(and(X1, X2)) → MARK(X1)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
MARK(from(X)) → A__FROM(X)

The TRS R consists of the following rules:

a__and(true, X) → mark(X)
a__and(false, Y) → false
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
a__from(X) → cons(X, from(s(X)))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(add(X1, X2)) → a__add(mark(X1), X2)
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(X)
mark(true) → true
mark(false) → false
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__add(X1, X2) → add(X1, X2)
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 2 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

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

MARK(first(X1, X2)) → MARK(X1)
A__ADD(0, X) → MARK(X)
A__IF(false, X, Y) → MARK(Y)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
MARK(first(X1, X2)) → MARK(X2)
MARK(add(X1, X2)) → A__ADD(mark(X1), X2)
A__IF(true, X, Y) → MARK(X)
A__AND(true, X) → MARK(X)
MARK(add(X1, X2)) → MARK(X1)

The TRS R consists of the following rules:

a__and(true, X) → mark(X)
a__and(false, Y) → false
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
a__from(X) → cons(X, from(s(X)))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(add(X1, X2)) → a__add(mark(X1), X2)
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(X)
mark(true) → true
mark(false) → false
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__add(X1, X2) → add(X1, X2)
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(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.

MARK(first(X1, X2)) → MARK(X1)
A__IF(false, X, Y) → MARK(Y)
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(and(X1, X2)) → A__AND(mark(X1), X2)
MARK(and(X1, X2)) → MARK(X1)
MARK(first(X1, X2)) → MARK(X2)
MARK(add(X1, X2)) → A__ADD(mark(X1), X2)
A__IF(true, X, Y) → MARK(X)
MARK(add(X1, X2)) → MARK(X1)

The remaining pairs can at least be oriented weakly.

A__ADD(0, X) → MARK(X)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
A__AND(true, X) → MARK(X)

Used ordering: Polynomial interpretation [25,35]:

POL(A__IF(x₁, x₂, x₃)) = 2 + (2)x_2 + (2)x_3
POL(from(x₁)) = 1 + (2)x_1
POL(a__and(x₁, x₂)) = 1 + (3)x_2
POL(true) = 2
POL(a__add(x₁, x₂)) = 3 + (3)x_1 + (2)x_2
POL(mark(x₁)) = 3
POL(a__from(x₁)) = 4 + (3)x_1
POL(and(x₁, x₂)) = 3 + x_1 + (4)x_2
POL(first(x₁, x₂)) = 1 + (4)x_1 + x_2
POL(0) = 3
POL(a__first(x₁, x₂)) = 2 + x_1 + x_2
POL(cons(x₁, x₂)) = 2 + (4)x_1
POL(A__AND(x₁, x₂)) = 1 + (3)x_2
POL(add(x₁, x₂)) = 3 + x_1 + (4)x_2
POL(if(x₁, x₂, x₃)) = 1 + x_1 + (2)x_2 + (3)x_3
POL(MARK(x₁)) = 1 + x_1
POL(A__ADD(x₁, x₂)) = 1 + (2)x_2
POL(false) = 3
POL(s(x₁)) = 4 + x_1
POL(nil) = 3
POL(a__if(x₁, x₂, x₃)) = 2 + (3)x_2 + (4)x_3

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

          ↳ QDPOrderProof

            ↳ QDP

              ↳ DependencyGraphProof

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

A__ADD(0, X) → MARK(X)
MARK(if(X1, X2, X3)) → A__IF(mark(X1), X2, X3)
A__AND(true, X) → MARK(X)

The TRS R consists of the following rules:

a__and(true, X) → mark(X)
a__and(false, Y) → false
a__if(true, X, Y) → mark(X)
a__if(false, X, Y) → mark(Y)
a__add(0, X) → mark(X)
a__add(s(X), Y) → s(add(X, Y))
a__first(0, X) → nil
a__first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
a__from(X) → cons(X, from(s(X)))
mark(and(X1, X2)) → a__and(mark(X1), X2)
mark(if(X1, X2, X3)) → a__if(mark(X1), X2, X3)
mark(add(X1, X2)) → a__add(mark(X1), X2)
mark(first(X1, X2)) → a__first(mark(X1), mark(X2))
mark(from(X)) → a__from(X)
mark(true) → true
mark(false) → false
mark(0) → 0
mark(s(X)) → s(X)
mark(nil) → nil
mark(cons(X1, X2)) → cons(X1, X2)
a__and(X1, X2) → and(X1, X2)
a__if(X1, X2, X3) → if(X1, X2, X3)
a__add(X1, X2) → add(X1, X2)
a__first(X1, X2) → first(X1, X2)
a__from(X) → from(X)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 3 less nodes.