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(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(add2(X1, X2)) -> A__ADD2(mark1(X1), X2)
A__IF3(true, X, Y) -> MARK1(X)
The remaining pairs can at least be oriented weakly.

MARK1(first2(X1, X2)) -> MARK1(X1)
A__ADD2(0, X) -> MARK1(X)
MARK1(first2(X1, X2)) -> MARK1(X2)
A__AND2(true, X) -> MARK1(X)
MARK1(if3(X1, X2, X3)) -> A__IF3(mark1(X1), X2, X3)
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(A__ADD2(x1, x2)) = x2   
POL(A__AND2(x1, x2)) = x2   
POL(A__IF3(x1, x2, x3)) = 1 + x2 + x3   
POL(MARK1(x1)) = 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)) = 1 + x1 + x2   
POL(and2(x1, x2)) = 1 + x1 + x2   
POL(cons2(x1, x2)) = 0   
POL(false) = 0   
POL(first2(x1, x2)) = x1 + x2   
POL(from1(x1)) = 0   
POL(if3(x1, x2, x3)) = 1 + x1 + x2 + 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:

MARK1(first2(X1, X2)) -> MARK1(X1)
A__ADD2(0, X) -> MARK1(X)
MARK1(first2(X1, X2)) -> MARK1(X2)
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.
The approximation of the Dependency Graph [13,14,18] contains 1 SCC with 3 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof

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

MARK1(first2(X1, X2)) -> MARK1(X1)
MARK1(first2(X1, X2)) -> MARK1(X2)

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)
MARK1(first2(X1, X2)) -> MARK1(X2)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MARK1(x1)) = x1   
POL(first2(x1, x2)) = 1 + x1 + x2   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof
                    ↳ QDP
                      ↳ PisEmptyProof

Q DP problem:
P is empty.
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 TRS P is empty. Hence, there is no (P,Q,R) chain.