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__adx(a__zeros)
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
a__hd(cons(X, Y)) → mark(X)
a__tl(cons(X, Y)) → mark(Y)
mark(nats) → a__nats
mark(adx(X)) → a__adx(mark(X))
mark(zeros) → a__zeros
mark(incr(X)) → a__incr(mark(X))
mark(hd(X)) → a__hd(mark(X))
mark(tl(X)) → a__tl(mark(X))
mark(cons(X1, X2)) → cons(X1, X2)
mark(0) → 0
mark(s(X)) → s(X)
a__nats → nats
a__adx(X) → adx(X)
a__zeros → zeros
a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__tl(X) → tl(X)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
a__hd(cons(X, Y)) → mark(X)
a__tl(cons(X, Y)) → mark(Y)
mark(nats) → a__nats
mark(adx(X)) → a__adx(mark(X))
mark(zeros) → a__zeros
mark(incr(X)) → a__incr(mark(X))
mark(hd(X)) → a__hd(mark(X))
mark(tl(X)) → a__tl(mark(X))
mark(cons(X1, X2)) → cons(X1, X2)
mark(0) → 0
mark(s(X)) → s(X)
a__nats → nats
a__adx(X) → adx(X)
a__zeros → zeros
a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__tl(X) → tl(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(incr(X)) → A__INCR(mark(X))
A__TL(cons(X, Y)) → MARK(Y)
MARK(hd(X)) → A__HD(mark(X))
MARK(zeros) → A__ZEROS
A__NATS → A__ADX(a__zeros)
A__ADX(cons(X, Y)) → A__INCR(cons(X, adx(Y)))
MARK(incr(X)) → MARK(X)
A__NATS → A__ZEROS
A__HD(cons(X, Y)) → MARK(X)
MARK(adx(X)) → A__ADX(mark(X))
MARK(adx(X)) → MARK(X)
MARK(hd(X)) → MARK(X)
MARK(tl(X)) → MARK(X)
MARK(nats) → A__NATS
MARK(tl(X)) → A__TL(mark(X))

The TRS R consists of the following rules:

a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
a__hd(cons(X, Y)) → mark(X)
a__tl(cons(X, Y)) → mark(Y)
mark(nats) → a__nats
mark(adx(X)) → a__adx(mark(X))
mark(zeros) → a__zeros
mark(incr(X)) → a__incr(mark(X))
mark(hd(X)) → a__hd(mark(X))
mark(tl(X)) → a__tl(mark(X))
mark(cons(X1, X2)) → cons(X1, X2)
mark(0) → 0
mark(s(X)) → s(X)
a__nats → nats
a__adx(X) → adx(X)
a__zeros → zeros
a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__tl(X) → tl(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(incr(X)) → A__INCR(mark(X))
A__TL(cons(X, Y)) → MARK(Y)
MARK(hd(X)) → A__HD(mark(X))
MARK(zeros) → A__ZEROS
A__NATS → A__ADX(a__zeros)
A__ADX(cons(X, Y)) → A__INCR(cons(X, adx(Y)))
MARK(incr(X)) → MARK(X)
A__NATS → A__ZEROS
A__HD(cons(X, Y)) → MARK(X)
MARK(adx(X)) → A__ADX(mark(X))
MARK(adx(X)) → MARK(X)
MARK(hd(X)) → MARK(X)
MARK(tl(X)) → MARK(X)
MARK(nats) → A__NATS
MARK(tl(X)) → A__TL(mark(X))

The TRS R consists of the following rules:

a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
a__hd(cons(X, Y)) → mark(X)
a__tl(cons(X, Y)) → mark(Y)
mark(nats) → a__nats
mark(adx(X)) → a__adx(mark(X))
mark(zeros) → a__zeros
mark(incr(X)) → a__incr(mark(X))
mark(hd(X)) → a__hd(mark(X))
mark(tl(X)) → a__tl(mark(X))
mark(cons(X1, X2)) → cons(X1, X2)
mark(0) → 0
mark(s(X)) → s(X)
a__nats → nats
a__adx(X) → adx(X)
a__zeros → zeros
a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__tl(X) → tl(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 7 less nodes.

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

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

A__HD(cons(X, Y)) → MARK(X)
A__TL(cons(X, Y)) → MARK(Y)
MARK(hd(X)) → A__HD(mark(X))
MARK(adx(X)) → MARK(X)
MARK(hd(X)) → MARK(X)
MARK(tl(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(tl(X)) → A__TL(mark(X))

The TRS R consists of the following rules:

a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
a__hd(cons(X, Y)) → mark(X)
a__tl(cons(X, Y)) → mark(Y)
mark(nats) → a__nats
mark(adx(X)) → a__adx(mark(X))
mark(zeros) → a__zeros
mark(incr(X)) → a__incr(mark(X))
mark(hd(X)) → a__hd(mark(X))
mark(tl(X)) → a__tl(mark(X))
mark(cons(X1, X2)) → cons(X1, X2)
mark(0) → 0
mark(s(X)) → s(X)
a__nats → nats
a__adx(X) → adx(X)
a__zeros → zeros
a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__tl(X) → tl(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.


A__HD(cons(X, Y)) → MARK(X)
MARK(hd(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.

A__TL(cons(X, Y)) → MARK(Y)
MARK(hd(X)) → A__HD(mark(X))
MARK(adx(X)) → MARK(X)
MARK(tl(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(tl(X)) → A__TL(mark(X))
Used ordering: Polynomial interpretation [25,35]:

POL(a__hd(x1)) = 1 + (4)x_1   
POL(a__adx(x1)) = x_1   
POL(a__tl(x1)) = (4)x_1   
POL(a__zeros) = 0   
POL(A__TL(x1)) = (4)x_1   
POL(mark(x1)) = (2)x_1   
POL(0) = 0   
POL(hd(x1)) = 1 + (4)x_1   
POL(cons(x1, x2)) = (4)x_1 + (2)x_2   
POL(MARK(x1)) = (4)x_1   
POL(a__nats) = 0   
POL(adx(x1)) = x_1   
POL(a__incr(x1)) = x_1   
POL(incr(x1)) = x_1   
POL(tl(x1)) = (4)x_1   
POL(zeros) = 0   
POL(A__HD(x1)) = 4 + (4)x_1   
POL(s(x1)) = x_1   
POL(nats) = 0   
The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented:

a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__adx(X) → adx(X)
a__zeros → zeros
a__tl(X) → tl(X)
a__zeros → cons(0, zeros)
a__nats → a__adx(a__zeros)
a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(incr(X)) → a__incr(mark(X))
mark(zeros) → a__zeros
a__hd(cons(X, Y)) → mark(X)
mark(hd(X)) → a__hd(mark(X))
mark(tl(X)) → a__tl(mark(X))
a__tl(cons(X, Y)) → mark(Y)
mark(0) → 0
mark(cons(X1, X2)) → cons(X1, X2)
a__nats → nats
mark(s(X)) → s(X)



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

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

MARK(adx(X)) → MARK(X)
MARK(hd(X)) → A__HD(mark(X))
A__TL(cons(X, Y)) → MARK(Y)
MARK(tl(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(tl(X)) → A__TL(mark(X))

The TRS R consists of the following rules:

a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
a__hd(cons(X, Y)) → mark(X)
a__tl(cons(X, Y)) → mark(Y)
mark(nats) → a__nats
mark(adx(X)) → a__adx(mark(X))
mark(zeros) → a__zeros
mark(incr(X)) → a__incr(mark(X))
mark(hd(X)) → a__hd(mark(X))
mark(tl(X)) → a__tl(mark(X))
mark(cons(X1, X2)) → cons(X1, X2)
mark(0) → 0
mark(s(X)) → s(X)
a__nats → nats
a__adx(X) → adx(X)
a__zeros → zeros
a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__tl(X) → tl(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 1 less node.

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

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

A__TL(cons(X, Y)) → MARK(Y)
MARK(adx(X)) → MARK(X)
MARK(tl(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
MARK(tl(X)) → A__TL(mark(X))

The TRS R consists of the following rules:

a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
a__hd(cons(X, Y)) → mark(X)
a__tl(cons(X, Y)) → mark(Y)
mark(nats) → a__nats
mark(adx(X)) → a__adx(mark(X))
mark(zeros) → a__zeros
mark(incr(X)) → a__incr(mark(X))
mark(hd(X)) → a__hd(mark(X))
mark(tl(X)) → a__tl(mark(X))
mark(cons(X1, X2)) → cons(X1, X2)
mark(0) → 0
mark(s(X)) → s(X)
a__nats → nats
a__adx(X) → adx(X)
a__zeros → zeros
a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__tl(X) → tl(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(adx(X)) → MARK(X)
MARK(incr(X)) → MARK(X)
The remaining pairs can at least be oriented weakly.

A__TL(cons(X, Y)) → MARK(Y)
MARK(tl(X)) → MARK(X)
MARK(tl(X)) → A__TL(mark(X))
Used ordering: Polynomial interpretation [25,35]:

POL(a__hd(x1)) = 4 + (4)x_1   
POL(a__adx(x1)) = 2 + (2)x_1   
POL(a__tl(x1)) = (2)x_1   
POL(a__zeros) = 0   
POL(A__TL(x1)) = (4)x_1   
POL(mark(x1)) = (2)x_1   
POL(0) = 0   
POL(hd(x1)) = 2 + (4)x_1   
POL(cons(x1, x2)) = (4)x_1 + x_2   
POL(MARK(x1)) = (4)x_1   
POL(a__nats) = 4   
POL(adx(x1)) = 1 + (2)x_1   
POL(a__incr(x1)) = 1 + x_1   
POL(incr(x1)) = 1 + x_1   
POL(zeros) = 0   
POL(tl(x1)) = (2)x_1   
POL(s(x1)) = 0   
POL(nats) = 4   
The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented:

a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__adx(X) → adx(X)
a__zeros → zeros
a__tl(X) → tl(X)
a__zeros → cons(0, zeros)
a__nats → a__adx(a__zeros)
a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(incr(X)) → a__incr(mark(X))
mark(zeros) → a__zeros
a__hd(cons(X, Y)) → mark(X)
mark(hd(X)) → a__hd(mark(X))
mark(tl(X)) → a__tl(mark(X))
a__tl(cons(X, Y)) → mark(Y)
mark(0) → 0
mark(cons(X1, X2)) → cons(X1, X2)
a__nats → nats
mark(s(X)) → s(X)



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

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

A__TL(cons(X, Y)) → MARK(Y)
MARK(tl(X)) → MARK(X)
MARK(tl(X)) → A__TL(mark(X))

The TRS R consists of the following rules:

a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
a__hd(cons(X, Y)) → mark(X)
a__tl(cons(X, Y)) → mark(Y)
mark(nats) → a__nats
mark(adx(X)) → a__adx(mark(X))
mark(zeros) → a__zeros
mark(incr(X)) → a__incr(mark(X))
mark(hd(X)) → a__hd(mark(X))
mark(tl(X)) → a__tl(mark(X))
mark(cons(X1, X2)) → cons(X1, X2)
mark(0) → 0
mark(s(X)) → s(X)
a__nats → nats
a__adx(X) → adx(X)
a__zeros → zeros
a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__tl(X) → tl(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.


A__TL(cons(X, Y)) → MARK(Y)
MARK(tl(X)) → MARK(X)
MARK(tl(X)) → A__TL(mark(X))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(a__hd(x1)) = 4 + (3)x_1   
POL(a__adx(x1)) = 4 + x_1   
POL(a__tl(x1)) = 4 + x_1   
POL(a__zeros) = 0   
POL(A__TL(x1)) = 2 + x_1   
POL(mark(x1)) = (4)x_1   
POL(0) = 0   
POL(hd(x1)) = 2 + (3)x_1   
POL(cons(x1, x2)) = (2)x_1 + (4)x_2   
POL(MARK(x1)) = (4)x_1   
POL(a__nats) = 4   
POL(adx(x1)) = 4 + x_1   
POL(incr(x1)) = 0   
POL(a__incr(x1)) = 0   
POL(zeros) = 0   
POL(tl(x1)) = 4 + x_1   
POL(s(x1)) = 0   
POL(nats) = 1   
The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented:

a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__adx(X) → adx(X)
a__zeros → zeros
a__tl(X) → tl(X)
a__zeros → cons(0, zeros)
a__nats → a__adx(a__zeros)
a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
mark(adx(X)) → a__adx(mark(X))
mark(nats) → a__nats
mark(incr(X)) → a__incr(mark(X))
mark(zeros) → a__zeros
a__hd(cons(X, Y)) → mark(X)
mark(hd(X)) → a__hd(mark(X))
mark(tl(X)) → a__tl(mark(X))
a__tl(cons(X, Y)) → mark(Y)
mark(0) → 0
mark(cons(X1, X2)) → cons(X1, X2)
a__nats → nats
mark(s(X)) → s(X)



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

Q DP problem:
P is empty.
The TRS R consists of the following rules:

a__nats → a__adx(a__zeros)
a__zeros → cons(0, zeros)
a__incr(cons(X, Y)) → cons(s(X), incr(Y))
a__adx(cons(X, Y)) → a__incr(cons(X, adx(Y)))
a__hd(cons(X, Y)) → mark(X)
a__tl(cons(X, Y)) → mark(Y)
mark(nats) → a__nats
mark(adx(X)) → a__adx(mark(X))
mark(zeros) → a__zeros
mark(incr(X)) → a__incr(mark(X))
mark(hd(X)) → a__hd(mark(X))
mark(tl(X)) → a__tl(mark(X))
mark(cons(X1, X2)) → cons(X1, X2)
mark(0) → 0
mark(s(X)) → s(X)
a__nats → nats
a__adx(X) → adx(X)
a__zeros → zeros
a__incr(X) → incr(X)
a__hd(X) → hd(X)
a__tl(X) → tl(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.