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:

plus2(s1(s1(x)), y) -> s1(plus2(x, s1(y)))
plus2(x, s1(s1(y))) -> s1(plus2(s1(x), y))
plus2(s1(0), y) -> s1(y)
plus2(0, y) -> y
ack2(0, y) -> s1(y)
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, plus2(y, ack2(s1(x), y)))

Q is empty.


QTRS
  ↳ DependencyPairsProof

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

plus2(s1(s1(x)), y) -> s1(plus2(x, s1(y)))
plus2(x, s1(s1(y))) -> s1(plus2(s1(x), y))
plus2(s1(0), y) -> s1(y)
plus2(0, y) -> y
ack2(0, y) -> s1(y)
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, plus2(y, ack2(s1(x), y)))

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:

ACK2(s1(x), s1(y)) -> ACK2(s1(x), y)
ACK2(s1(x), s1(y)) -> ACK2(x, plus2(y, ack2(s1(x), y)))
PLUS2(x, s1(s1(y))) -> PLUS2(s1(x), y)
ACK2(s1(x), s1(y)) -> PLUS2(y, ack2(s1(x), y))
PLUS2(s1(s1(x)), y) -> PLUS2(x, s1(y))
ACK2(s1(x), 0) -> ACK2(x, s1(0))

The TRS R consists of the following rules:

plus2(s1(s1(x)), y) -> s1(plus2(x, s1(y)))
plus2(x, s1(s1(y))) -> s1(plus2(s1(x), y))
plus2(s1(0), y) -> s1(y)
plus2(0, y) -> y
ack2(0, y) -> s1(y)
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, plus2(y, ack2(s1(x), y)))

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:

ACK2(s1(x), s1(y)) -> ACK2(s1(x), y)
ACK2(s1(x), s1(y)) -> ACK2(x, plus2(y, ack2(s1(x), y)))
PLUS2(x, s1(s1(y))) -> PLUS2(s1(x), y)
ACK2(s1(x), s1(y)) -> PLUS2(y, ack2(s1(x), y))
PLUS2(s1(s1(x)), y) -> PLUS2(x, s1(y))
ACK2(s1(x), 0) -> ACK2(x, s1(0))

The TRS R consists of the following rules:

plus2(s1(s1(x)), y) -> s1(plus2(x, s1(y)))
plus2(x, s1(s1(y))) -> s1(plus2(s1(x), y))
plus2(s1(0), y) -> s1(y)
plus2(0, y) -> y
ack2(0, y) -> s1(y)
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, plus2(y, ack2(s1(x), y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 1 less node.

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

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

PLUS2(x, s1(s1(y))) -> PLUS2(s1(x), y)
PLUS2(s1(s1(x)), y) -> PLUS2(x, s1(y))

The TRS R consists of the following rules:

plus2(s1(s1(x)), y) -> s1(plus2(x, s1(y)))
plus2(x, s1(s1(y))) -> s1(plus2(s1(x), y))
plus2(s1(0), y) -> s1(y)
plus2(0, y) -> y
ack2(0, y) -> s1(y)
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, plus2(y, ack2(s1(x), y)))

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.


PLUS2(x, s1(s1(y))) -> PLUS2(s1(x), y)
PLUS2(s1(s1(x)), y) -> PLUS2(x, s1(y))
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(PLUS2(x1, x2)) = 2·x1 + 2·x2   
POL(s1(x1)) = 2 + x1   

The following usable rules [14] were oriented: none



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

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

plus2(s1(s1(x)), y) -> s1(plus2(x, s1(y)))
plus2(x, s1(s1(y))) -> s1(plus2(s1(x), y))
plus2(s1(0), y) -> s1(y)
plus2(0, y) -> y
ack2(0, y) -> s1(y)
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, plus2(y, ack2(s1(x), y)))

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.

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

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

ACK2(s1(x), s1(y)) -> ACK2(s1(x), y)
ACK2(s1(x), s1(y)) -> ACK2(x, plus2(y, ack2(s1(x), y)))
ACK2(s1(x), 0) -> ACK2(x, s1(0))

The TRS R consists of the following rules:

plus2(s1(s1(x)), y) -> s1(plus2(x, s1(y)))
plus2(x, s1(s1(y))) -> s1(plus2(s1(x), y))
plus2(s1(0), y) -> s1(y)
plus2(0, y) -> y
ack2(0, y) -> s1(y)
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, plus2(y, ack2(s1(x), y)))

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.


ACK2(s1(x), s1(y)) -> ACK2(x, plus2(y, ack2(s1(x), y)))
ACK2(s1(x), 0) -> ACK2(x, s1(0))
The remaining pairs can at least be oriented weakly.

ACK2(s1(x), s1(y)) -> ACK2(s1(x), y)
Used ordering: Polynomial interpretation [21]:

POL(0) = 2   
POL(ACK2(x1, x2)) = x1   
POL(ack2(x1, x2)) = 1 + x1 + x2   
POL(plus2(x1, x2)) = 1   
POL(s1(x1)) = 2 + x1   

The following usable rules [14] were oriented: none



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

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

ACK2(s1(x), s1(y)) -> ACK2(s1(x), y)

The TRS R consists of the following rules:

plus2(s1(s1(x)), y) -> s1(plus2(x, s1(y)))
plus2(x, s1(s1(y))) -> s1(plus2(s1(x), y))
plus2(s1(0), y) -> s1(y)
plus2(0, y) -> y
ack2(0, y) -> s1(y)
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, plus2(y, ack2(s1(x), y)))

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.


ACK2(s1(x), s1(y)) -> ACK2(s1(x), y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACK2(x1, x2)) = 2·x2   
POL(s1(x1)) = 1 + 2·x1   

The following usable rules [14] were oriented: none



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

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

plus2(s1(s1(x)), y) -> s1(plus2(x, s1(y)))
plus2(x, s1(s1(y))) -> s1(plus2(s1(x), y))
plus2(s1(0), y) -> s1(y)
plus2(0, y) -> y
ack2(0, y) -> s1(y)
ack2(s1(x), 0) -> ack2(x, s1(0))
ack2(s1(x), s1(y)) -> ack2(x, plus2(y, ack2(s1(x), y)))

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.