Termination w.r.t. Q of the given QTRS could not be shown:

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

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
ACK(x1, x2)  =  ACK(x1)
s(x1)  =  s(x1)
ack(x1, x2)  =  ack(x1)
0  =  0

Recursive Path Order [RPO].
Precedence:

s₁ > ACK₁ > ack₁
s₁ > 0 > ack₁

The following usable rules [FROCOS05] were oriented: none

(9) 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:

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.

(10) 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: Recursive Path Order [RPO].
Precedence:

s₁ > ACK₂

The following usable rules [FROCOS05] were oriented: none

(11) Obligation:

Q DP problem:
P is empty.
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.

(12) PisEmptyProof (EQUIVALENT transformation)

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

(13) TRUE

(14) 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.

(15) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

The following pairs can be oriented strictly and are deleted.

LE(s(x), s(y)) → LE(x, y)

The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LE(x1, x2)  =  x2
s(x1)  =  s(x1)

Recursive Path Order [RPO].
Precedence:

trivial

The following usable rules [FROCOS05] were oriented: none

(16) Obligation:

Q DP problem:
P is empty.
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.

(17) PisEmptyProof (EQUIVALENT transformation)

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

(18) TRUE

(19) 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.