Termination w.r.t. Q proof of TRS/Cimeack

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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

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.