(0) Obligation:
Q restricted rewrite system:
The TRS R consists of the following rules:
double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false
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:
DOUBLE(x) → PERMUTE(x, x, a)
PERMUTE(x, y, a) → PERMUTE(isZero(x), x, b)
PERMUTE(x, y, a) → ISZERO(x)
PERMUTE(false, x, b) → PERMUTE(ack(x, x), p(x), c)
PERMUTE(false, x, b) → ACK(x, x)
PERMUTE(false, x, b) → P(x)
PERMUTE(y, x, c) → PERMUTE(x, y, a)
ACK(0, x) → PLUS(x, s(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)
PLUS(s(x), y) → PLUS(x, s(y))
PLUS(x, s(s(y))) → PLUS(s(x), y)
The TRS R consists of the following rules:
double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false
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 3 SCCs with 5 less nodes.
(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(x), y) → PLUS(x, s(y))
The TRS R consists of the following rules:
double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false
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, 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:
double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false
Q is empty.
We have to consider all minimal (P,Q,R)-chains.
(7) Obligation:
Q DP problem:
The TRS P consists of the following rules:
PERMUTE(false, x, b) → PERMUTE(ack(x, x), p(x), c)
PERMUTE(y, x, c) → PERMUTE(x, y, a)
PERMUTE(x, y, a) → PERMUTE(isZero(x), x, b)
The TRS R consists of the following rules:
double(x) → permute(x, x, a)
permute(x, y, a) → permute(isZero(x), x, b)
permute(false, x, b) → permute(ack(x, x), p(x), c)
permute(true, x, b) → 0
permute(y, x, c) → s(s(permute(x, y, a)))
p(0) → 0
p(s(x)) → x
ack(0, x) → plus(x, s(0))
ack(s(x), 0) → ack(x, s(0))
ack(s(x), s(y)) → ack(x, ack(s(x), y))
plus(0, y) → y
plus(s(x), y) → plus(x, s(y))
plus(x, s(s(y))) → s(plus(s(x), y))
plus(x, s(0)) → s(x)
plus(x, 0) → x
isZero(0) → true
isZero(s(x)) → false
Q is empty.
We have to consider all minimal (P,Q,R)-chains.