(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))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACK(
x1,
x2) =
x1
s(
x1) =
s(
x1)
ack(
x1,
x2) =
ack(
x1,
x2)
0 =
0
Lexicographic path order with status [LPO].
Precedence:
ack2 > s1
Status:
ack2: [1,2]
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:
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:
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) 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) =
x2
s(
x1) =
s(
x1)
ack(
x1,
x2) =
ack(
x1,
x2)
0 =
0
Lexicographic path order with status [LPO].
Precedence:
ack2 > s1
Status:
ack2: [1,2]
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))
(8) 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.
(9) PisEmptyProof (EQUIVALENT transformation)
The TRS P is empty. Hence, there is no (P,Q,R) chain.
(10) TRUE