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

(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) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(9) 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:

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

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) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

numbers
d(x0)
if(true, x0)
if(false, x0)
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
nr

(11) 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:

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

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.

(12) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• ACK(s(x), 0) → ACK(x, s(0))
  The graph contains the following edges 1 > 1

• ACK(s(x), s(y)) → ACK(s(x), y)
  The graph contains the following edges 1 >= 1, 2 > 2

• ACK(s(x), s(y)) → ACK(x, ack(s(x), y))
  The graph contains the following edges 1 > 1

(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) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(16) Obligation:

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

LE(s(x), s(y)) → LE(x, y)

R is empty.
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) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

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))

(18) Obligation:

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

LE(s(x), s(y)) → LE(x, y)

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

(19) QDPSizeChangeProof (EQUIVALENT transformation)

By using the subterm criterion [SUBTERM_CRITERION] together with the size-change analysis [AAECC05] we have proven that there are no infinite chains for this DP problem.

From the DPs we obtained the following set of size-change graphs:

• LE(s(x), s(y)) → LE(x, y)
  The graph contains the following edges 1 > 1, 2 > 2

(20) TRUE

(21) 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.

(22) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(23) 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:

nr → ack(s(s(s(s(s(s(0)))))), 0)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
ack(0, x) → s(x)

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.

(24) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

numbers
d(x0)
if(true, x0)
if(false, x0)

(25) 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:

nr → ack(s(s(s(s(s(s(0)))))), 0)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
ack(0, x) → s(x)

The set Q consists of the following terms:

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.

(26) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule D(x) → IF(le(x, nr), x) we obtained the following new rules [LPAR04]:

D(s(z0)) → IF(le(s(z0), nr), s(z0))

(27) Obligation:

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

IF(true, x) → D(s(x))
D(s(z0)) → IF(le(s(z0), nr), s(z0))

The TRS R consists of the following rules:

nr → ack(s(s(s(s(s(s(0)))))), 0)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
ack(0, x) → s(x)

The set Q consists of the following terms:

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.

(28) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, nr), x) at position [0,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(s(s(0)))))), 0)), x)

(29) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(s(s(0)))))), 0)), x)

The TRS R consists of the following rules:

nr → ack(s(s(s(s(s(s(0)))))), 0)
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
ack(0, x) → s(x)

The set Q consists of the following terms:

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.

(30) UsableRulesProof (EQUIVALENT transformation)

As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [LPAR04] we can delete all non-usable rules [FROCOS05] from R.

(31) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(s(s(0)))))), 0)), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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.

(32) QReductionProof (EQUIVALENT transformation)

We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.[THIEMANN].

nr

(33) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(s(s(0)))))), 0)), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(34) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(s(s(0)))))), 0)), x) at position [0,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(s(0))))), s(0))), x)

(35) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(s(0))))), s(0))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(36) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(s(0))))), s(0))), x) at position [0,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(s(s(0))))), 0))), x)

(37) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(s(s(0))))), 0))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(38) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(s(s(0))))), 0))), x) at position [0,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(s(0)))), s(0)))), x)

(39) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(s(0)))), s(0)))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(40) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(s(0)))), s(0)))), x) at position [0,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(s(s(0)))), 0)))), x)

(41) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(s(s(0)))), 0)))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(42) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(s(s(0)))), 0)))), x) at position [0,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(s(0))), s(0))))), x)

(43) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(s(0))), s(0))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(44) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(s(0))), s(0))))), x) at position [0,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(s(s(0))), 0))))), x)

(45) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(s(s(0))), 0))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(46) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(s(s(0))), 0))))), x) at position [0,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(s(0)), s(0)))))), x)

(47) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(s(0)), s(0)))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(48) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(s(0)), s(0)))))), x) at position [0,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(s(s(0)), 0)))))), x)

(49) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(s(s(0)), 0)))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(50) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(s(s(0)), 0)))))), x) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(s(0), s(0))))))), x)

(51) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(s(0), s(0))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(52) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(s(0), s(0))))))), x) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(0, ack(s(0), 0))))))), x)

(53) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(0, ack(s(0), 0))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(54) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), ack(0, ack(s(0), 0))))))), x) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), s(ack(s(0), 0))))))), x)

(55) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), s(ack(s(0), 0))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(56) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(s(0), s(ack(s(0), 0))))))), x) at position [0,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(0, ack(s(0), ack(s(0), 0))))))), x)

(57) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(0, ack(s(0), ack(s(0), 0))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(58) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), ack(0, ack(s(0), ack(s(0), 0))))))), x) at position [0,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), s(ack(s(0), ack(s(0), 0))))))), x)

(59) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), s(ack(s(0), ack(s(0), 0))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(60) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(s(0)), s(ack(s(0), ack(s(0), 0))))))), x) at position [0,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), ack(s(0), 0))))))), x)

(61) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), ack(s(0), 0))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(62) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), ack(s(0), 0))))))), x) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), ack(0, s(0)))))))), x)

(63) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), ack(0, s(0)))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(64) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), ack(0, s(0)))))))), x) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), s(s(0)))))))), x)

(65) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), s(s(0)))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(66) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(s(0), s(s(0)))))))), x) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(0, ack(s(0), s(0)))))))), x)

(67) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(0, ack(s(0), s(0)))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(68) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), ack(0, ack(s(0), s(0)))))))), x) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), s(ack(s(0), s(0)))))))), x)

(69) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), s(ack(s(0), s(0)))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(70) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(s(0)), s(ack(s(0), s(0)))))))), x) at position [0,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(s(0)), ack(s(0), s(0)))))))), x)

(71) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(s(0)), ack(s(0), s(0)))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(72) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(s(0)), ack(s(0), s(0)))))))), x) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(s(0)), ack(0, ack(s(0), 0)))))))), x)

(73) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(s(0)), ack(0, ack(s(0), 0)))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(74) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(s(0)), ack(0, ack(s(0), 0)))))))), x) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(s(0)), s(ack(s(0), 0)))))))), x)

(75) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(s(0)), s(ack(s(0), 0)))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(76) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(s(0)), s(ack(s(0), 0)))))))), x) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), ack(s(0), 0)))))))), x)

(77) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), ack(s(0), 0)))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(78) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), ack(s(0), 0)))))))), x) at position [0,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), ack(0, s(0))))))))), x)

(79) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), ack(0, s(0))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(80) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), ack(0, s(0))))))))), x) at position [0,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), s(s(0))))))))), x)

(81) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), s(s(0))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(82) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), s(s(0))))))))), x) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), s(0))))))))), x)

(83) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), s(0))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(84) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), s(0))))))))), x) at position [0,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), 0))))))))), x)

(85) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), 0))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(86) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(s(0)), 0))))))))), x) at position [0,1,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(0)))))))))), x)

(87) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(0)))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(88) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(0)))))))))), x) at position [0,1,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), 0)))))))))), x)

(89) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), 0)))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(90) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), 0)))))))))), x) at position [0,1,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), 0)))))))))), x)

(91) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), 0)))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(92) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), 0)))))))))), x) at position [0,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), 0)))))))))), x)

(93) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), 0)))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(94) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), 0)))))))))), x) at position [0,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), 0)))))))))), x)

(95) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), 0)))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(96) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), 0)))))))))), x) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x)

(97) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(98) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x)

(99) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(100) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x)

(101) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(102) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x)

(103) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(104) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x) at position [0,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x)

(105) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(106) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x) at position [0,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x)

(107) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(108) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x) at position [0,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x)

(109) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(110) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x) at position [0,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), s(ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x)

(111) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), s(ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(112) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(s(0))), s(ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x) at position [0,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x)

(113) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(114) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), 0)))))))))), x) at position [0,1,1,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, s(0))))))))))), x)

(115) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, s(0))))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(116) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, s(0))))))))))), x) at position [0,1,1,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(s(0))))))))))), x)

(117) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(s(0))))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(118) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(s(0))))))))))), x) at position [0,1,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), s(0))))))))))), x)

(119) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), s(0))))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(120) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), s(0))))))))))), x) at position [0,1,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), s(0))))))))))), x)

(121) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), s(0))))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(122) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), s(0))))))))))), x) at position [0,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), s(0))))))))))), x)

(123) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), s(0))))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(124) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), s(0))))))))))), x) at position [0,1,1,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), s(0))))))))))), x)

(125) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), s(0))))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(126) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), s(0))))))))))), x) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), s(0))))))))))), x)

(127) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), s(0))))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(128) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), s(0))))))))))), x) at position [0,1,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), s(0))))))))))), x)

(129) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), s(0))))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(130) Rewriting (EQUIVALENT transformation)

By rewriting [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(s(0), s(ack(s(0), ack(s(0), ack(s(0), s(0))))))))))), x) at position [0,1,1,1,1,1] we obtained the following new rules [LPAR04]:

D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(0))))))))))), x)

(131) Obligation:

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

IF(true, x) → D(s(x))
D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(0))))))))))), x)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(132) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(0))))))))))), x) we obtained the following new rules [LPAR04]:

D(s(z0)) → IF(le(s(z0), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(0))))))))))), s(z0))

(133) Obligation:

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

IF(true, x) → D(s(x))
D(s(z0)) → IF(le(s(z0), ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(0))))))))))), s(z0))

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(134) RemovalProof (SOUND transformation)

In the following pairs the term without variables ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(0)))))))))) is replaced by the fresh variable x_removed.
Pair: D(x) → IF(le(x, ack(s(s(s(s(0)))), ack(s(s(0)), ack(s(s(s(0))), ack(s(0), ack(0, ack(s(0), ack(s(0), ack(s(0), ack(s(0), s(0))))))))))), x)
Positions in right side of the pair:

• [0,1]

The following rules were checked for non-overlappingness:

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

The new variable was added to all pairs as a new argument[CONREM].

(135) Obligation:

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

IF(true, x, x_removed) → D(s(x), x_removed)
D(x, x_removed) → IF(le(x, x_removed), x, x_removed)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(136) NonInfProof (EQUIVALENT transformation)

The DP Problem is simplified using the Induction Calculus [NONINF] with the following steps:
Note that final constraints are written in bold face.


For Pair IF(true, x, x_removed) → D(s(x), x_removed) the following chains were created:

• We consider the chain D(x, x_removed) → IF(le(x, x_removed), x, x_removed), IF(true, x, x_removed) → D(s(x), x_removed) which results in the following constraint:
  

  (1)    (IF(le(x2, x3), x2, x3)=IF(true, x4, x5) ⇒ IF(true, x4, x5)≥D(s(x4), x5))

  
  
  We simplified constraint (1) using rules (I), (II), (III) which results in the following new constraint:
  

  (2)    (le(x2, x3)=true ⇒ IF(true, x2, x3)≥D(s(x2), x3))

  
  
  We simplified constraint (2) using rule (V) (with possible (I) afterwards) using induction on le(x2, x3)=true which results in the following new constraints:
  

  (3)    (true=true ⇒ IF(true, 0, x12)≥D(s(0), x12))

  
  

  (4)    (le(x15, x14)=true∧(le(x15, x14)=true ⇒ IF(true, x15, x14)≥D(s(x15), x14)) ⇒ IF(true, s(x15), s(x14))≥D(s(s(x15)), s(x14)))

  
  
  We simplified constraint (3) using rules (I), (II) which results in the following new constraint:
  

  (5)    (IF(true, 0, x12)≥D(s(0), x12))

  
  
  We simplified constraint (4) using rule (VI) where we applied the induction hypothesis (le(x15, x14)=true ⇒ IF(true, x15, x14)≥D(s(x15), x14)) with σ = [ ] which results in the following new constraint:
  

  (6)    (IF(true, x15, x14)≥D(s(x15), x14) ⇒ IF(true, s(x15), s(x14))≥D(s(s(x15)), s(x14)))

  
  
  

For Pair D(x, x_removed) → IF(le(x, x_removed), x, x_removed) the following chains were created:

• We consider the chain IF(true, x, x_removed) → D(s(x), x_removed), D(x, x_removed) → IF(le(x, x_removed), x, x_removed) which results in the following constraint:
  

  (7)    (D(s(x6), x7)=D(x8, x9) ⇒ D(x8, x9)≥IF(le(x8, x9), x8, x9))

  
  
  We simplified constraint (7) using rules (I), (II), (III) which results in the following new constraint:
  

  (8)    (D(s(x6), x7)≥IF(le(s(x6), x7), s(x6), x7))

  
  
  

To summarize, we get the following constraints P_≥ for the following pairs.

• IF(true, x, x_removed) → D(s(x), x_removed)
  
  • (IF(true, 0, x12)≥D(s(0), x12))
    
  • (IF(true, x15, x14)≥D(s(x15), x14) ⇒ IF(true, s(x15), s(x14))≥D(s(s(x15)), s(x14)))
    
  
  
• D(x, x_removed) → IF(le(x, x_removed), x, x_removed)
  
  • (D(s(x6), x7)≥IF(le(s(x6), x7), s(x6), x7))
    
  
  

The constraints for P_> respective P_bound are constructed from P_≥ where we just replace every occurence of "t ≥ s" in P_≥ by "t > s" respective "t ≥ c". Here c stands for the fresh constant used for P_bound.
Using the following integer polynomial ordering the resulting constraints can be solved
Polynomial interpretation [NONINF]:

POL(0) = 0   
POL(D(x₁, x₂)) = -1 - x₁ + x₂   
POL(IF(x₁, x₂, x₃)) = -1 - x₁ - x₂ + x₃   
POL(c) = -2   
POL(false) = 0   
POL(le(x₁, x₂)) = 0   
POL(s(x₁)) = 1 + x₁   
POL(true) = 0   

The following pairs are in P_>:

IF(true, x, x_removed) → D(s(x), x_removed)

The following pairs are in P_bound:

IF(true, x, x_removed) → D(s(x), x_removed)

The following rules are usable:

true → le(0, y)
false → le(s(x), 0)
le(x, y) → le(s(x), s(y))

(137) Obligation:

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

D(x, x_removed) → IF(le(x, x_removed), x, x_removed)

The TRS R consists of the following rules:

ack(s(x), s(y)) → ack(x, ack(s(x), y))
ack(s(x), 0) → ack(x, s(0))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
ack(0, x) → s(x)

The set Q consists of the following terms:

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

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

(138) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 0 SCCs with 1 less node.

(139) TRUE