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))


Using Dependency Pairs [1,13] we result in the following initial DP problem:
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 [13,14,18] contains 1 SCC with 2 less nodes.

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

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


ACKIN2(s1(X), s1(Y)) -> U212(ackin2(s1(X), Y), X)
ACKIN2(s1(X), 0) -> ACKIN2(X, s1(0))
The remaining pairs can at least by weakly be oriented.

ACKIN2(s1(X), s1(Y)) -> ACKIN2(s1(X), Y)
U212(ackout1(X), Y) -> ACKIN2(Y, X)
Used ordering: Combined order from the following AFS and order.
ACKIN2(x1, x2)  =  x1
s1(x1)  =  s1(x1)
U212(x1, x2)  =  x2
ackout1(x1)  =  ackout
ackin2(x1, x2)  =  x1
0  =  0
u111(x1)  =  u11
u212(x1, x2)  =  u211(x2)
u221(x1)  =  u22

Lexicographic Path Order [19].
Precedence:
0 > s1 > u11 > ackout
0 > s1 > u211 > u22 > ackout

The following usable rules [14] were oriented:

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



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ QDP
              ↳ QDPOrderProof
                ↳ 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 [13,14,18] contains 1 SCC with 1 less node.

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

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.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


ACKIN2(s1(X), s1(Y)) -> ACKIN2(s1(X), Y)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
ACKIN2(x1, x2)  =  ACKIN1(x2)
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
s1 > ACKIN1

The following usable rules [14] were oriented: none



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