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
  ↳ 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)

Q is empty.

Using Dependency Pairs [1,13] we result in the following initial DP problem:
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)

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:

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)

Q is empty.
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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof

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)

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


The following pairs can be oriented strictly and are deleted.


ACK_IN2(s1(m), 0) -> ACK_IN2(m, s1(0))
ACK_IN2(s1(m), s1(n)) -> U212(ack_in2(s1(m), n), m)
The remaining pairs can at least be oriented weakly.

ACK_IN2(s1(m), s1(n)) -> ACK_IN2(s1(m), n)
U212(ack_out1(n), m) -> ACK_IN2(m, n)
Used ordering: Polynomial interpretation [21]:

POL(0) = 3   
POL(ACK_IN2(x1, x2)) = 3·x1   
POL(U212(x1, x2)) = 3·x2   
POL(ack_in2(x1, x2)) = x1 + 3·x1·x2 + 2·x2   
POL(ack_out1(x1)) = 2   
POL(s1(x1)) = 3 + 3·x1 + 2·x12   
POL(u111(x1)) = 0   
POL(u212(x1, x2)) = 0   
POL(u221(x1)) = 0   

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ 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)

Q is empty.
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
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ QDP
          ↳ QDPOrderProof
            ↳ QDP
              ↳ DependencyGraphProof
                ↳ QDP
                  ↳ QDPOrderProof

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)

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


The following pairs can be oriented strictly and are deleted.


ACK_IN2(s1(m), s1(n)) -> ACK_IN2(s1(m), n)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(ACK_IN2(x1, x2)) = x1·x2   
POL(s1(x1)) = 2 + 3·x1 + 3·x12   

The following usable rules [14] were oriented: none



↳ 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:

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