Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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

plus(s(s(x)), y) → s(plus(x, s(y)))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(s(0), y) → s(y)
plus(0, y) → y
ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, plus(y, ack(s(x), y)))

Q is empty.

(1) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(2) Obligation:

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

PLUS(s(s(x)), y) → PLUS(x, s(y))
PLUS(x, s(s(y))) → PLUS(s(x), y)
ACK(s(x), 0) → ACK(x, s(0))
ACK(s(x), s(y)) → ACK(x, plus(y, ack(s(x), y)))
ACK(s(x), s(y)) → PLUS(y, ack(s(x), y))
ACK(s(x), s(y)) → ACK(s(x), y)

The TRS R consists of the following rules:

plus(s(s(x)), y) → s(plus(x, s(y)))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(s(0), y) → s(y)
plus(0, y) → y
ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, plus(y, ack(s(x), y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(3) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 1 less node.

(4) Complex Obligation (AND)

(5) Obligation:

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

PLUS(x, s(s(y))) → PLUS(s(x), y)
PLUS(s(s(x)), y) → PLUS(x, s(y))

The TRS R consists of the following rules:

plus(s(s(x)), y) → s(plus(x, s(y)))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(s(0), y) → s(y)
plus(0, y) → y
ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, plus(y, ack(s(x), y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(6) Obligation:

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

ACK(s(x), s(y)) → ACK(x, plus(y, ack(s(x), y)))
ACK(s(x), 0) → ACK(x, s(0))
ACK(s(x), s(y)) → ACK(s(x), y)

The TRS R consists of the following rules:

plus(s(s(x)), y) → s(plus(x, s(y)))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(s(0), y) → s(y)
plus(0, y) → y
ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, plus(y, ack(s(x), y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(7) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACK(s(x), s(y)) → ACK(x, plus(y, ack(s(x), y)))
ACK(s(x), 0) → ACK(x, s(0))

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACK(x1, x2)  =  ACK(x1)
s(x1)  =  s(x1)
plus(x1, x2)  =  plus
ack(x1, x2)  =  ack(x1, x2)
0  =  0

Recursive Path Order [RPO].
Precedence:

s₁ > ACK₁
s₁ > plus
ack₂ > plus
0 > ACK₁

The following usable rules [FROCOS05] were oriented: none

(8) Obligation:

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

ACK(s(x), s(y)) → ACK(s(x), y)

The TRS R consists of the following rules:

plus(s(s(x)), y) → s(plus(x, s(y)))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(s(0), y) → s(y)
plus(0, y) → y
ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, plus(y, ack(s(x), y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(9) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

ACK(s(x), s(y)) → ACK(s(x), y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACK(x1, x2)  =  ACK(x2)
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:

s₁ > ACK₁

The following usable rules [FROCOS05] were oriented: none

(10) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

plus(s(s(x)), y) → s(plus(x, s(y)))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(s(0), y) → s(y)
plus(0, y) → y
ack(0, y) → s(y)
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, plus(y, ack(s(x), y)))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.

(11) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(12) TRUE