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:

ackin2(0, X) -> ackout1(s1(X))
ackin2(s1(X), 0) -> u111(ackin2(X, s1(0)))
u111(ackout1(X)) -> ackout1(X)
ackin2(s1(X), s1(Y)) -> u212(ackin2(s1(X), Y), X)
u212(ackout1(X), Y) -> u221(ackin2(Y, X))
u221(ackout1(X)) -> ackout1(X)

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

ackin2(0, X) -> ackout1(s1(X))
ackin2(s1(X), 0) -> u111(ackin2(X, s1(0)))
u111(ackout1(X)) -> ackout1(X)
ackin2(s1(X), s1(Y)) -> u212(ackin2(s1(X), Y), X)
u212(ackout1(X), Y) -> u221(ackin2(Y, X))
u221(ackout1(X)) -> ackout1(X)

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:

ackin2(0, X) -> ackout1(s1(X))
ackin2(s1(X), 0) -> u111(ackin2(X, s1(0)))
u111(ackout1(X)) -> ackout1(X)
ackin2(s1(X), s1(Y)) -> u212(ackin2(s1(X), Y), X)
u212(ackout1(X), Y) -> u221(ackin2(Y, X))
u221(ackout1(X)) -> ackout1(X)

The set Q consists of the following terms:

ackin2(0, x0)
ackin2(s1(x0), 0)
u111(ackout1(x0))
ackin2(s1(x0), s1(x1))
u212(ackout1(x0), x1)
u221(ackout1(x0))


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

U212(ackout1(X), Y) -> ACKIN2(Y, X)
ACKIN2(s1(X), s1(Y)) -> ACKIN2(s1(X), Y)
ACKIN2(s1(X), 0) -> U111(ackin2(X, s1(0)))
ACKIN2(s1(X), s1(Y)) -> U212(ackin2(s1(X), Y), X)
ACKIN2(s1(X), 0) -> ACKIN2(X, s1(0))
U212(ackout1(X), Y) -> U221(ackin2(Y, X))

The TRS R consists of the following rules:

ackin2(0, X) -> ackout1(s1(X))
ackin2(s1(X), 0) -> u111(ackin2(X, s1(0)))
u111(ackout1(X)) -> ackout1(X)
ackin2(s1(X), s1(Y)) -> u212(ackin2(s1(X), Y), X)
u212(ackout1(X), Y) -> u221(ackin2(Y, X))
u221(ackout1(X)) -> ackout1(X)

The set Q consists of the following terms:

ackin2(0, x0)
ackin2(s1(x0), 0)
u111(ackout1(x0))
ackin2(s1(x0), s1(x1))
u212(ackout1(x0), x1)
u221(ackout1(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:

U212(ackout1(X), Y) -> ACKIN2(Y, X)
ACKIN2(s1(X), s1(Y)) -> ACKIN2(s1(X), Y)
ACKIN2(s1(X), 0) -> U111(ackin2(X, s1(0)))
ACKIN2(s1(X), s1(Y)) -> U212(ackin2(s1(X), Y), X)
ACKIN2(s1(X), 0) -> ACKIN2(X, s1(0))
U212(ackout1(X), Y) -> U221(ackin2(Y, X))

The TRS R consists of the following rules:

ackin2(0, X) -> ackout1(s1(X))
ackin2(s1(X), 0) -> u111(ackin2(X, s1(0)))
u111(ackout1(X)) -> ackout1(X)
ackin2(s1(X), s1(Y)) -> u212(ackin2(s1(X), Y), X)
u212(ackout1(X), Y) -> u221(ackin2(Y, X))
u221(ackout1(X)) -> ackout1(X)

The set Q consists of the following terms:

ackin2(0, x0)
ackin2(s1(x0), 0)
u111(ackout1(x0))
ackin2(s1(x0), s1(x1))
u212(ackout1(x0), x1)
u221(ackout1(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:

ACKIN2(s1(X), s1(Y)) -> ACKIN2(s1(X), Y)
U212(ackout1(X), Y) -> ACKIN2(Y, X)
ACKIN2(s1(X), s1(Y)) -> U212(ackin2(s1(X), Y), X)
ACKIN2(s1(X), 0) -> ACKIN2(X, s1(0))

The TRS R consists of the following rules:

ackin2(0, X) -> ackout1(s1(X))
ackin2(s1(X), 0) -> u111(ackin2(X, s1(0)))
u111(ackout1(X)) -> ackout1(X)
ackin2(s1(X), s1(Y)) -> u212(ackin2(s1(X), Y), X)
u212(ackout1(X), Y) -> u221(ackin2(Y, X))
u221(ackout1(X)) -> ackout1(X)

The set Q consists of the following terms:

ackin2(0, x0)
ackin2(s1(x0), 0)
u111(ackout1(x0))
ackin2(s1(x0), s1(x1))
u212(ackout1(x0), x1)
u221(ackout1(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.

ACKIN2(s1(X), s1(Y)) -> U212(ackin2(s1(X), Y), X)
ACKIN2(s1(X), 0) -> ACKIN2(X, s1(0))
Used argument filtering: ACKIN2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
U212(x1, x2)  =  x2
Used ordering: Quasi Precedence: trivial 


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

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

U212(ackout1(X), Y) -> ACKIN2(Y, X)
ACKIN2(s1(X), s1(Y)) -> ACKIN2(s1(X), Y)

The TRS R consists of the following rules:

ackin2(0, X) -> ackout1(s1(X))
ackin2(s1(X), 0) -> u111(ackin2(X, s1(0)))
u111(ackout1(X)) -> ackout1(X)
ackin2(s1(X), s1(Y)) -> u212(ackin2(s1(X), Y), X)
u212(ackout1(X), Y) -> u221(ackin2(Y, X))
u221(ackout1(X)) -> ackout1(X)

The set Q consists of the following terms:

ackin2(0, x0)
ackin2(s1(x0), 0)
u111(ackout1(x0))
ackin2(s1(x0), s1(x1))
u212(ackout1(x0), x1)
u221(ackout1(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:

ACKIN2(s1(X), s1(Y)) -> ACKIN2(s1(X), Y)

The TRS R consists of the following rules:

ackin2(0, X) -> ackout1(s1(X))
ackin2(s1(X), 0) -> u111(ackin2(X, s1(0)))
u111(ackout1(X)) -> ackout1(X)
ackin2(s1(X), s1(Y)) -> u212(ackin2(s1(X), Y), X)
u212(ackout1(X), Y) -> u221(ackin2(Y, X))
u221(ackout1(X)) -> ackout1(X)

The set Q consists of the following terms:

ackin2(0, x0)
ackin2(s1(x0), 0)
u111(ackout1(x0))
ackin2(s1(x0), s1(x1))
u212(ackout1(x0), x1)
u221(ackout1(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.

ACKIN2(s1(X), s1(Y)) -> ACKIN2(s1(X), Y)
Used argument filtering: ACKIN2(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:

ackin2(0, X) -> ackout1(s1(X))
ackin2(s1(X), 0) -> u111(ackin2(X, s1(0)))
u111(ackout1(X)) -> ackout1(X)
ackin2(s1(X), s1(Y)) -> u212(ackin2(s1(X), Y), X)
u212(ackout1(X), Y) -> u221(ackin2(Y, X))
u221(ackout1(X)) -> ackout1(X)

The set Q consists of the following terms:

ackin2(0, x0)
ackin2(s1(x0), 0)
u111(ackout1(x0))
ackin2(s1(x0), s1(x1))
u212(ackout1(x0), x1)
u221(ackout1(x0))

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