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__nats -> a__adx1(a__zeros)
a__zeros -> cons2(0, zeros)
a__incr1(cons2(X, Y)) -> cons2(s1(X), incr1(Y))
a__adx1(cons2(X, Y)) -> a__incr1(cons2(X, adx1(Y)))
a__hd1(cons2(X, Y)) -> mark1(X)
a__tl1(cons2(X, Y)) -> mark1(Y)
mark1(nats) -> a__nats
mark1(adx1(X)) -> a__adx1(mark1(X))
mark1(zeros) -> a__zeros
mark1(incr1(X)) -> a__incr1(mark1(X))
mark1(hd1(X)) -> a__hd1(mark1(X))
mark1(tl1(X)) -> a__tl1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(X1, X2)
mark1(0) -> 0
mark1(s1(X)) -> s1(X)
a__nats -> nats
a__adx1(X) -> adx1(X)
a__zeros -> zeros
a__incr1(X) -> incr1(X)
a__hd1(X) -> hd1(X)
a__tl1(X) -> tl1(X)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a__nats -> a__adx1(a__zeros)
a__zeros -> cons2(0, zeros)
a__incr1(cons2(X, Y)) -> cons2(s1(X), incr1(Y))
a__adx1(cons2(X, Y)) -> a__incr1(cons2(X, adx1(Y)))
a__hd1(cons2(X, Y)) -> mark1(X)
a__tl1(cons2(X, Y)) -> mark1(Y)
mark1(nats) -> a__nats
mark1(adx1(X)) -> a__adx1(mark1(X))
mark1(zeros) -> a__zeros
mark1(incr1(X)) -> a__incr1(mark1(X))
mark1(hd1(X)) -> a__hd1(mark1(X))
mark1(tl1(X)) -> a__tl1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(X1, X2)
mark1(0) -> 0
mark1(s1(X)) -> s1(X)
a__nats -> nats
a__adx1(X) -> adx1(X)
a__zeros -> zeros
a__incr1(X) -> incr1(X)
a__hd1(X) -> hd1(X)
a__tl1(X) -> tl1(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(tl1(X)) -> A__TL1(mark1(X))
MARK1(hd1(X)) -> A__HD1(mark1(X))
MARK1(tl1(X)) -> MARK1(X)
A__HD1(cons2(X, Y)) -> MARK1(X)
MARK1(adx1(X)) -> A__ADX1(mark1(X))
A__NATS -> A__ADX1(a__zeros)
MARK1(incr1(X)) -> A__INCR1(mark1(X))
MARK1(zeros) -> A__ZEROS
A__TL1(cons2(X, Y)) -> MARK1(Y)
MARK1(hd1(X)) -> MARK1(X)
MARK1(nats) -> A__NATS
A__NATS -> A__ZEROS
A__ADX1(cons2(X, Y)) -> A__INCR1(cons2(X, adx1(Y)))
MARK1(incr1(X)) -> MARK1(X)
MARK1(adx1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__nats -> a__adx1(a__zeros)
a__zeros -> cons2(0, zeros)
a__incr1(cons2(X, Y)) -> cons2(s1(X), incr1(Y))
a__adx1(cons2(X, Y)) -> a__incr1(cons2(X, adx1(Y)))
a__hd1(cons2(X, Y)) -> mark1(X)
a__tl1(cons2(X, Y)) -> mark1(Y)
mark1(nats) -> a__nats
mark1(adx1(X)) -> a__adx1(mark1(X))
mark1(zeros) -> a__zeros
mark1(incr1(X)) -> a__incr1(mark1(X))
mark1(hd1(X)) -> a__hd1(mark1(X))
mark1(tl1(X)) -> a__tl1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(X1, X2)
mark1(0) -> 0
mark1(s1(X)) -> s1(X)
a__nats -> nats
a__adx1(X) -> adx1(X)
a__zeros -> zeros
a__incr1(X) -> incr1(X)
a__hd1(X) -> hd1(X)
a__tl1(X) -> tl1(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(tl1(X)) -> A__TL1(mark1(X))
MARK1(hd1(X)) -> A__HD1(mark1(X))
MARK1(tl1(X)) -> MARK1(X)
A__HD1(cons2(X, Y)) -> MARK1(X)
MARK1(adx1(X)) -> A__ADX1(mark1(X))
A__NATS -> A__ADX1(a__zeros)
MARK1(incr1(X)) -> A__INCR1(mark1(X))
MARK1(zeros) -> A__ZEROS
A__TL1(cons2(X, Y)) -> MARK1(Y)
MARK1(hd1(X)) -> MARK1(X)
MARK1(nats) -> A__NATS
A__NATS -> A__ZEROS
A__ADX1(cons2(X, Y)) -> A__INCR1(cons2(X, adx1(Y)))
MARK1(incr1(X)) -> MARK1(X)
MARK1(adx1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__nats -> a__adx1(a__zeros)
a__zeros -> cons2(0, zeros)
a__incr1(cons2(X, Y)) -> cons2(s1(X), incr1(Y))
a__adx1(cons2(X, Y)) -> a__incr1(cons2(X, adx1(Y)))
a__hd1(cons2(X, Y)) -> mark1(X)
a__tl1(cons2(X, Y)) -> mark1(Y)
mark1(nats) -> a__nats
mark1(adx1(X)) -> a__adx1(mark1(X))
mark1(zeros) -> a__zeros
mark1(incr1(X)) -> a__incr1(mark1(X))
mark1(hd1(X)) -> a__hd1(mark1(X))
mark1(tl1(X)) -> a__tl1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(X1, X2)
mark1(0) -> 0
mark1(s1(X)) -> s1(X)
a__nats -> nats
a__adx1(X) -> adx1(X)
a__zeros -> zeros
a__incr1(X) -> incr1(X)
a__hd1(X) -> hd1(X)
a__tl1(X) -> tl1(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 7 less nodes.

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

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

MARK1(tl1(X)) -> A__TL1(mark1(X))
MARK1(hd1(X)) -> A__HD1(mark1(X))
MARK1(tl1(X)) -> MARK1(X)
A__HD1(cons2(X, Y)) -> MARK1(X)
MARK1(incr1(X)) -> MARK1(X)
A__TL1(cons2(X, Y)) -> MARK1(Y)
MARK1(hd1(X)) -> MARK1(X)
MARK1(adx1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__nats -> a__adx1(a__zeros)
a__zeros -> cons2(0, zeros)
a__incr1(cons2(X, Y)) -> cons2(s1(X), incr1(Y))
a__adx1(cons2(X, Y)) -> a__incr1(cons2(X, adx1(Y)))
a__hd1(cons2(X, Y)) -> mark1(X)
a__tl1(cons2(X, Y)) -> mark1(Y)
mark1(nats) -> a__nats
mark1(adx1(X)) -> a__adx1(mark1(X))
mark1(zeros) -> a__zeros
mark1(incr1(X)) -> a__incr1(mark1(X))
mark1(hd1(X)) -> a__hd1(mark1(X))
mark1(tl1(X)) -> a__tl1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(X1, X2)
mark1(0) -> 0
mark1(s1(X)) -> s1(X)
a__nats -> nats
a__adx1(X) -> adx1(X)
a__zeros -> zeros
a__incr1(X) -> incr1(X)
a__hd1(X) -> hd1(X)
a__tl1(X) -> tl1(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(tl1(X)) -> MARK1(X)
A__HD1(cons2(X, Y)) -> MARK1(X)
A__TL1(cons2(X, Y)) -> MARK1(Y)
MARK1(hd1(X)) -> MARK1(X)
MARK1(adx1(X)) -> MARK1(X)
The remaining pairs can at least be oriented weakly.

MARK1(tl1(X)) -> A__TL1(mark1(X))
MARK1(hd1(X)) -> A__HD1(mark1(X))
MARK1(incr1(X)) -> MARK1(X)
Used ordering: Polynomial interpretation [21]:

POL(0) = 0   
POL(A__HD1(x1)) = 1 + x1   
POL(A__TL1(x1)) = 1 + x1   
POL(MARK1(x1)) = x1   
POL(a__adx1(x1)) = 1 + x1   
POL(a__hd1(x1)) = 1 + x1   
POL(a__incr1(x1)) = x1   
POL(a__nats) = 1   
POL(a__tl1(x1)) = 1 + x1   
POL(a__zeros) = 0   
POL(adx1(x1)) = 1 + x1   
POL(cons2(x1, x2)) = x1 + x2   
POL(hd1(x1)) = 1 + x1   
POL(incr1(x1)) = x1   
POL(mark1(x1)) = x1   
POL(nats) = 1   
POL(s1(x1)) = 0   
POL(tl1(x1)) = 1 + x1   
POL(zeros) = 0   

The following usable rules [14] were oriented:

a__adx1(cons2(X, Y)) -> a__incr1(cons2(X, adx1(Y)))
a__zeros -> cons2(0, zeros)
a__adx1(X) -> adx1(X)
mark1(incr1(X)) -> a__incr1(mark1(X))
a__hd1(X) -> hd1(X)
mark1(nats) -> a__nats
a__nats -> nats
a__nats -> a__adx1(a__zeros)
mark1(hd1(X)) -> a__hd1(mark1(X))
a__hd1(cons2(X, Y)) -> mark1(X)
mark1(tl1(X)) -> a__tl1(mark1(X))
a__tl1(cons2(X, Y)) -> mark1(Y)
mark1(adx1(X)) -> a__adx1(mark1(X))
a__incr1(X) -> incr1(X)
mark1(s1(X)) -> s1(X)
a__zeros -> zeros
a__incr1(cons2(X, Y)) -> cons2(s1(X), incr1(Y))
mark1(zeros) -> a__zeros
mark1(0) -> 0
mark1(cons2(X1, X2)) -> cons2(X1, X2)
a__tl1(X) -> tl1(X)



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

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

MARK1(hd1(X)) -> A__HD1(mark1(X))
MARK1(tl1(X)) -> A__TL1(mark1(X))
MARK1(incr1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__nats -> a__adx1(a__zeros)
a__zeros -> cons2(0, zeros)
a__incr1(cons2(X, Y)) -> cons2(s1(X), incr1(Y))
a__adx1(cons2(X, Y)) -> a__incr1(cons2(X, adx1(Y)))
a__hd1(cons2(X, Y)) -> mark1(X)
a__tl1(cons2(X, Y)) -> mark1(Y)
mark1(nats) -> a__nats
mark1(adx1(X)) -> a__adx1(mark1(X))
mark1(zeros) -> a__zeros
mark1(incr1(X)) -> a__incr1(mark1(X))
mark1(hd1(X)) -> a__hd1(mark1(X))
mark1(tl1(X)) -> a__tl1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(X1, X2)
mark1(0) -> 0
mark1(s1(X)) -> s1(X)
a__nats -> nats
a__adx1(X) -> adx1(X)
a__zeros -> zeros
a__incr1(X) -> incr1(X)
a__hd1(X) -> hd1(X)
a__tl1(X) -> tl1(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
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof

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

MARK1(incr1(X)) -> MARK1(X)

The TRS R consists of the following rules:

a__nats -> a__adx1(a__zeros)
a__zeros -> cons2(0, zeros)
a__incr1(cons2(X, Y)) -> cons2(s1(X), incr1(Y))
a__adx1(cons2(X, Y)) -> a__incr1(cons2(X, adx1(Y)))
a__hd1(cons2(X, Y)) -> mark1(X)
a__tl1(cons2(X, Y)) -> mark1(Y)
mark1(nats) -> a__nats
mark1(adx1(X)) -> a__adx1(mark1(X))
mark1(zeros) -> a__zeros
mark1(incr1(X)) -> a__incr1(mark1(X))
mark1(hd1(X)) -> a__hd1(mark1(X))
mark1(tl1(X)) -> a__tl1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(X1, X2)
mark1(0) -> 0
mark1(s1(X)) -> s1(X)
a__nats -> nats
a__adx1(X) -> adx1(X)
a__zeros -> zeros
a__incr1(X) -> incr1(X)
a__hd1(X) -> hd1(X)
a__tl1(X) -> tl1(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(incr1(X)) -> MARK1(X)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MARK1(x1)) = x1   
POL(incr1(x1)) = 1 + x1   

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__nats -> a__adx1(a__zeros)
a__zeros -> cons2(0, zeros)
a__incr1(cons2(X, Y)) -> cons2(s1(X), incr1(Y))
a__adx1(cons2(X, Y)) -> a__incr1(cons2(X, adx1(Y)))
a__hd1(cons2(X, Y)) -> mark1(X)
a__tl1(cons2(X, Y)) -> mark1(Y)
mark1(nats) -> a__nats
mark1(adx1(X)) -> a__adx1(mark1(X))
mark1(zeros) -> a__zeros
mark1(incr1(X)) -> a__incr1(mark1(X))
mark1(hd1(X)) -> a__hd1(mark1(X))
mark1(tl1(X)) -> a__tl1(mark1(X))
mark1(cons2(X1, X2)) -> cons2(X1, X2)
mark1(0) -> 0
mark1(s1(X)) -> s1(X)
a__nats -> nats
a__adx1(X) -> adx1(X)
a__zeros -> zeros
a__incr1(X) -> incr1(X)
a__hd1(X) -> hd1(X)
a__tl1(X) -> tl1(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.