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 Order [17,21] with Interpretation:

POL( U212(x1, x2) ) = 3x2 + 3


POL( ack_in2(x1, x2) ) = 3x2 + 1


POL( u221(x1) ) = max{0, -3}


POL( u212(x1, x2) ) = 2x1 + 3x2 + 1


POL( s1(x1) ) = 2x1 + 3


POL( 0 ) = max{0, -3}


POL( ACK_IN2(x1, x2) ) = 3x1 + 3


POL( u111(x1) ) = max{0, -3}


POL( ack_out1(x1) ) = 3x1 + 3



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 Order [17,21] with Interpretation:

POL( s1(x1) ) = x1 + 3


POL( ACK_IN2(x1, x2) ) = max{0, 3x1 + x2 - 1}



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.