Termination w.r.t. Q of the given QTRS could be proven:

0 QTRS
↳1 DependencyPairsProof (⇔)
↳2 QDP
↳3 DependencyGraphProof (⇔)
↳4 AND
    ↳5 QDP
        ↳6 QDPOrderProof (⇔)
        ↳7 QDP
        ↳8 PisEmptyProof (⇔)
        ↳9 TRUE
    ↳10 QDP
        ↳11 QDPOrderProof (⇔)
        ↳12 QDP
        ↳13 QDPOrderProof (⇔)
        ↳14 QDP
        ↳15 PisEmptyProof (⇔)
        ↳16 TRUE

(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) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


PLUS(x, s(s(y))) → PLUS(s(x), y)
PLUS(s(s(x)), y) → PLUS(x, s(y))
The remaining pairs can at least be oriented weakly.
Used ordering: Polynomial interpretation [POLO]:

POL(PLUS(x1, x2)) = x1 + x2   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(7) 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.

(8) PisEmptyProof (EQUIVALENT transformation)

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

(9) TRUE

(10) 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.

(11) 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: Polynomial interpretation [POLO]:

POL(0) = 0   
POL(ACK(x1, x2)) = x1   
POL(ack(x1, x2)) = 0   
POL(plus(x1, x2)) = 0   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(12) 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.

(13) 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: Polynomial interpretation [POLO]:

POL(ACK(x1, x2)) = x2   
POL(s(x1)) = 1 + x1   

The following usable rules [FROCOS05] were oriented: none

(14) 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.

(15) PisEmptyProof (EQUIVALENT transformation)

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

(16) TRUE