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:

ack_in2(0, n) -> ack_out1(s1(n))
ack_in2(s1(m), 0) -> u111(ack_in2(m, s1(0)))
u111(ack_out1(n)) -> ack_out1(n)
ack_in2(s1(m), s1(n)) -> u212(ack_in2(s1(m), n), m)
u212(ack_out1(n), m) -> u221(ack_in2(m, n))
u221(ack_out1(n)) -> ack_out1(n)

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

ack_in2(0, n) -> ack_out1(s1(n))
ack_in2(s1(m), 0) -> u111(ack_in2(m, s1(0)))
u111(ack_out1(n)) -> ack_out1(n)
ack_in2(s1(m), s1(n)) -> u212(ack_in2(s1(m), n), m)
u212(ack_out1(n), m) -> u221(ack_in2(m, n))
u221(ack_out1(n)) -> ack_out1(n)

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:

ack_in2(0, n) -> ack_out1(s1(n))
ack_in2(s1(m), 0) -> u111(ack_in2(m, s1(0)))
u111(ack_out1(n)) -> ack_out1(n)
ack_in2(s1(m), s1(n)) -> u212(ack_in2(s1(m), n), m)
u212(ack_out1(n), m) -> u221(ack_in2(m, n))
u221(ack_out1(n)) -> ack_out1(n)

The set Q consists of the following terms:

ack_in2(0, x0)
ack_in2(s1(x0), 0)
u111(ack_out1(x0))
ack_in2(s1(x0), s1(x1))
u212(ack_out1(x0), x1)
u221(ack_out1(x0))


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

ACK_IN2(s1(m), s1(n)) -> ACK_IN2(s1(m), n)
ACK_IN2(s1(m), 0) -> U111(ack_in2(m, s1(0)))
ACK_IN2(s1(m), 0) -> ACK_IN2(m, s1(0))
U212(ack_out1(n), m) -> ACK_IN2(m, n)
ACK_IN2(s1(m), s1(n)) -> U212(ack_in2(s1(m), n), m)
U212(ack_out1(n), m) -> U221(ack_in2(m, n))

The TRS R consists of the following rules:

ack_in2(0, n) -> ack_out1(s1(n))
ack_in2(s1(m), 0) -> u111(ack_in2(m, s1(0)))
u111(ack_out1(n)) -> ack_out1(n)
ack_in2(s1(m), s1(n)) -> u212(ack_in2(s1(m), n), m)
u212(ack_out1(n), m) -> u221(ack_in2(m, n))
u221(ack_out1(n)) -> ack_out1(n)

The set Q consists of the following terms:

ack_in2(0, x0)
ack_in2(s1(x0), 0)
u111(ack_out1(x0))
ack_in2(s1(x0), s1(x1))
u212(ack_out1(x0), x1)
u221(ack_out1(x0))

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

ACK_IN2(s1(m), s1(n)) -> ACK_IN2(s1(m), n)
ACK_IN2(s1(m), 0) -> U111(ack_in2(m, s1(0)))
ACK_IN2(s1(m), 0) -> ACK_IN2(m, s1(0))
U212(ack_out1(n), m) -> ACK_IN2(m, n)
ACK_IN2(s1(m), s1(n)) -> U212(ack_in2(s1(m), n), m)
U212(ack_out1(n), m) -> U221(ack_in2(m, n))

The TRS R consists of the following rules:

ack_in2(0, n) -> ack_out1(s1(n))
ack_in2(s1(m), 0) -> u111(ack_in2(m, s1(0)))
u111(ack_out1(n)) -> ack_out1(n)
ack_in2(s1(m), s1(n)) -> u212(ack_in2(s1(m), n), m)
u212(ack_out1(n), m) -> u221(ack_in2(m, n))
u221(ack_out1(n)) -> ack_out1(n)

The set Q consists of the following terms:

ack_in2(0, x0)
ack_in2(s1(x0), 0)
u111(ack_out1(x0))
ack_in2(s1(x0), s1(x1))
u212(ack_out1(x0), x1)
u221(ack_out1(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 2 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPAfsSolverProof

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

ACK_IN2(s1(m), s1(n)) -> ACK_IN2(s1(m), n)
ACK_IN2(s1(m), 0) -> ACK_IN2(m, s1(0))
U212(ack_out1(n), m) -> ACK_IN2(m, n)
ACK_IN2(s1(m), s1(n)) -> U212(ack_in2(s1(m), n), m)

The TRS R consists of the following rules:

ack_in2(0, n) -> ack_out1(s1(n))
ack_in2(s1(m), 0) -> u111(ack_in2(m, s1(0)))
u111(ack_out1(n)) -> ack_out1(n)
ack_in2(s1(m), s1(n)) -> u212(ack_in2(s1(m), n), m)
u212(ack_out1(n), m) -> u221(ack_in2(m, n))
u221(ack_out1(n)) -> ack_out1(n)

The set Q consists of the following terms:

ack_in2(0, x0)
ack_in2(s1(x0), 0)
u111(ack_out1(x0))
ack_in2(s1(x0), s1(x1))
u212(ack_out1(x0), x1)
u221(ack_out1(x0))

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.

ACK_IN2(s1(m), 0) -> ACK_IN2(m, s1(0))
ACK_IN2(s1(m), s1(n)) -> U212(ack_in2(s1(m), n), m)
Used argument filtering: ACK_IN2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
U212(x1, x2)  =  x2
ack_in2(x1, x2)  =  ack_in
0  =  0
u111(x1)  =  u11
u212(x1, x2)  =  u21
ack_out1(x1)  =  ack_out
u221(x1)  =  u22
Used ordering: Quasi Precedence: [ack_in, u21] > u11 > ack_out [ack_in, u21] > u22 > ack_out 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPAfsSolverProof
                ↳ QDP
                  ↳ DependencyGraphProof

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

ACK_IN2(s1(m), s1(n)) -> ACK_IN2(s1(m), n)
U212(ack_out1(n), m) -> ACK_IN2(m, n)

The TRS R consists of the following rules:

ack_in2(0, n) -> ack_out1(s1(n))
ack_in2(s1(m), 0) -> u111(ack_in2(m, s1(0)))
u111(ack_out1(n)) -> ack_out1(n)
ack_in2(s1(m), s1(n)) -> u212(ack_in2(s1(m), n), m)
u212(ack_out1(n), m) -> u221(ack_in2(m, n))
u221(ack_out1(n)) -> ack_out1(n)

The set Q consists of the following terms:

ack_in2(0, x0)
ack_in2(s1(x0), 0)
u111(ack_out1(x0))
ack_in2(s1(x0), s1(x1))
u212(ack_out1(x0), x1)
u221(ack_out1(x0))

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph contains 1 SCC with 1 less node.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPAfsSolverProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ QDPAfsSolverProof

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

ACK_IN2(s1(m), s1(n)) -> ACK_IN2(s1(m), n)

The TRS R consists of the following rules:

ack_in2(0, n) -> ack_out1(s1(n))
ack_in2(s1(m), 0) -> u111(ack_in2(m, s1(0)))
u111(ack_out1(n)) -> ack_out1(n)
ack_in2(s1(m), s1(n)) -> u212(ack_in2(s1(m), n), m)
u212(ack_out1(n), m) -> u221(ack_in2(m, n))
u221(ack_out1(n)) -> ack_out1(n)

The set Q consists of the following terms:

ack_in2(0, x0)
ack_in2(s1(x0), 0)
u111(ack_out1(x0))
ack_in2(s1(x0), s1(x1))
u212(ack_out1(x0), x1)
u221(ack_out1(x0))

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.

ACK_IN2(s1(m), s1(n)) -> ACK_IN2(s1(m), n)
Used argument filtering: ACK_IN2(x1, x2)  =  x2
s1(x1)  =  s1(x1)
Used ordering: Quasi Precedence: trivial 


↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPAfsSolverProof
                ↳ QDP
                  ↳ DependencyGraphProof
                    ↳ QDP
                      ↳ QDPAfsSolverProof
                        ↳ QDP
                          ↳ PisEmptyProof

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

ack_in2(0, n) -> ack_out1(s1(n))
ack_in2(s1(m), 0) -> u111(ack_in2(m, s1(0)))
u111(ack_out1(n)) -> ack_out1(n)
ack_in2(s1(m), s1(n)) -> u212(ack_in2(s1(m), n), m)
u212(ack_out1(n), m) -> u221(ack_in2(m, n))
u221(ack_out1(n)) -> ack_out1(n)

The set Q consists of the following terms:

ack_in2(0, x0)
ack_in2(s1(x0), 0)
u111(ack_out1(x0))
ack_in2(s1(x0), s1(x1))
u212(ack_out1(x0), x1)
u221(ack_out1(x0))

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