Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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.

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 contains 2 SCCs with 1 less node.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
QDP
          ↳ 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.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
QDP
            ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACK2(s1(x), s1(y)) -> ACK2(x, plus2(y, ack2(s1(x), y)))
ACK2(s1(x), 0) -> ACK2(x, s1(0))
Used argument filtering: ACK2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
QDP
                ↳ QDPAfsSolverProof

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.
By using an argument filtering and a montonic ordering, at least one Dependency Pair of this SCC can be strictly oriented.

ACK2(s1(x), s1(y)) -> ACK2(s1(x), y)
Used argument filtering: ACK2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial


↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
          ↳ QDP
            ↳ QDPAfsSolverProof
              ↳ QDP
                ↳ QDPAfsSolverProof
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.