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:

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

Q is empty.


QTRS
  ↳ Non-Overlap Check

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

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

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
QTRS
      ↳ DependencyPairsProof

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

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

The set Q consists of the following terms:

ack2(0, x0)
ack2(s1(x0), 0)
ack2(s1(x0), s1(x1))


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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

ack2(0, x0)
ack2(s1(x0), 0)
ack2(s1(x0), s1(x1))

We have to consider all minimal (P,Q,R)-chains.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
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, ack2(s1(x), y))
ACK2(s1(x), 0) -> ACK2(x, s1(0))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

ack2(0, x0)
ack2(s1(x0), 0)
ack2(s1(x0), s1(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


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

ACK2(s1(x), s1(y)) -> ACK2(s1(x), y)
Used ordering: Combined order from the following AFS and order.
ACK2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
ack2(x1, x2)  =  ack2(x1, x2)
0  =  0

Lexicographic Path Order [19].
Precedence:
ack2 > s1
0 > s1


The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ 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:

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

The set Q consists of the following terms:

ack2(0, x0)
ack2(s1(x0), 0)
ack2(s1(x0), s1(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


ACK2(s1(x), s1(y)) -> ACK2(s1(x), y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ACK2(x1, x2)  =  ACK1(x2)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
[ACK1, s1]


The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ QDPOrderProof
QDP
                  ↳ PisEmptyProof

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

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

The set Q consists of the following terms:

ack2(0, x0)
ack2(s1(x0), 0)
ack2(s1(x0), s1(x1))

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.