(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
numbers → d(0)
d(x) → if(le(x, nr), x)
if(true, x) → cons(x, d(s(x)))
if(false, x) → nil
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
nr → ack(s(s(s(s(s(s(0)))))), 0)
ack(0, x) → s(x)
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:
numbers → d(0)
d(x) → if(le(x, nr), x)
if(true, x) → cons(x, d(s(x)))
if(false, x) → nil
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
nr → ack(s(s(s(s(s(s(0)))))), 0)
ack(0, x) → s(x)
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:
numbers
d(x0)
if(true, x0)
if(false, x0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
nr
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:
NUMBERS → D(0)
D(x) → IF(le(x, nr), x)
D(x) → LE(x, nr)
D(x) → NR
IF(true, x) → D(s(x))
LE(s(x), s(y)) → LE(x, y)
NR → ACK(s(s(s(s(s(s(0)))))), 0)
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:
numbers → d(0)
d(x) → if(le(x, nr), x)
if(true, x) → cons(x, d(s(x)))
if(false, x) → nil
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
nr → ack(s(s(s(s(s(s(0)))))), 0)
ack(0, x) → s(x)
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:
numbers
d(x0)
if(true, x0)
if(false, x0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
nr
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(5) DependencyGraphProof (EQUIVALENT transformation)
The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 3 SCCs with 4 less nodes.
(6) Complex Obligation (AND)
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
ACK(s(x), s(y)) → ACK(s(x), y)
ACK(s(x), 0) → ACK(x, s(0))
The TRS R consists of the following rules:
numbers → d(0)
d(x) → if(le(x, nr), x)
if(true, x) → cons(x, d(s(x)))
if(false, x) → nil
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
nr → ack(s(s(s(s(s(s(0)))))), 0)
ack(0, x) → s(x)
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:
numbers
d(x0)
if(true, x0)
if(false, x0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
nr
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(8) Obligation:
Q DP problem:
The TRS P consists of the following rules:
LE(s(x), s(y)) → LE(x, y)
The TRS R consists of the following rules:
numbers → d(0)
d(x) → if(le(x, nr), x)
if(true, x) → cons(x, d(s(x)))
if(false, x) → nil
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
nr → ack(s(s(s(s(s(s(0)))))), 0)
ack(0, x) → s(x)
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:
numbers
d(x0)
if(true, x0)
if(false, x0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
nr
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.
(9) Obligation:
Q DP problem:
The TRS P consists of the following rules:
IF(true, x) → D(s(x))
D(x) → IF(le(x, nr), x)
The TRS R consists of the following rules:
numbers → d(0)
d(x) → if(le(x, nr), x)
if(true, x) → cons(x, d(s(x)))
if(false, x) → nil
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
nr → ack(s(s(s(s(s(s(0)))))), 0)
ack(0, x) → s(x)
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:
numbers
d(x0)
if(true, x0)
if(false, x0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
nr
ack(0, x0)
ack(s(x0), 0)
ack(s(x0), s(x1))
We have to consider all minimal (P,Q,R)-chains.