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

0 QTRS
↳1 Overlay + Local Confluence (⇔)
↳2 QTRS
↳3 DependencyPairsProof (⇔)
↳4 QDP
↳5 QDPOrderProof (⇔)
↳6 QDP
↳7 PisEmptyProof (⇔)
↳8 TRUE

(0) Obligation:

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

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

Q is empty.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

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

The set Q consists of the following terms:

ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))

(3) DependencyPairsProof (EQUIVALENT transformation)

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

(4) Obligation:

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

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

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))

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

(5) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].


The following pairs can be oriented strictly and are deleted.


ACK(s(x), 0) → ACK(x, s(0))
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
ACK(s(x), s(y)) → ACK(s(x), y)
The remaining pairs can at least be oriented weakly.
Used ordering: Lexicographic Path Order [LPO].
Precedence:
ACK2 > ack2 > s1
ACK2 > ack2 > 0

The following usable rules [FROCOS05] were oriented:

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

(6) Obligation:

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

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

The set Q consists of the following terms:

ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))

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

(7) PisEmptyProof (EQUIVALENT transformation)

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

(8) TRUE