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

(0) Obligation:

Q restricted rewrite system:
The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

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:

TERMS(N) → SQR(N)
TERMS(N) → TERMS(s(N))
SQR(s(X)) → ADD(sqr(X), dbl(X))
SQR(s(X)) → SQR(X)
SQR(s(X)) → DBL(X)
DBL(s(X)) → DBL(X)
ADD(s(X), Y) → ADD(X, Y)
FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)
HALF(s(s(X))) → HALF(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

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 6 SCCs with 3 less nodes.

(4) Complex Obligation (AND)

(5) Obligation:

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

HALF(s(s(X))) → HALF(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

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

(6) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(7) Obligation:

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

HALF(s(s(X))) → HALF(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))
half(0)
half(s(0))
half(s(s(x0)))
half(dbl(x0))

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:

HALF(s(s(X))) → HALF(X)

R is empty.
The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))
half(0)
half(s(0))
half(s(s(x0)))
half(dbl(x0))

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].

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))
half(0)
half(s(0))
half(s(s(x0)))
half(dbl(x0))

(11) Obligation:

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

HALF(s(s(X))) → HALF(X)

R is empty.
Q is empty.
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:

• HALF(s(s(X))) → HALF(X)
  The graph contains the following edges 1 > 1

(13) TRUE

(14) Obligation:

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

FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

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

(15) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(16) Obligation:

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

FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))
half(0)
half(s(0))
half(s(s(x0)))
half(dbl(x0))

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

(17) 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.

(18) Obligation:

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

FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)

R is empty.
The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))
half(0)
half(s(0))
half(s(s(x0)))
half(dbl(x0))

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

(19) 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].

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))
half(0)
half(s(0))
half(s(s(x0)))
half(dbl(x0))

(20) Obligation:

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

FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)

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

(21) 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:

• FIRST(s(X), cons(Y, Z)) → FIRST(X, Z)
  The graph contains the following edges 1 > 1, 2 > 2

(22) TRUE

(23) Obligation:

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

ADD(s(X), Y) → ADD(X, Y)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

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

(24) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(25) Obligation:

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

ADD(s(X), Y) → ADD(X, Y)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))
half(0)
half(s(0))
half(s(s(x0)))
half(dbl(x0))

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

(26) 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.

(27) Obligation:

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

ADD(s(X), Y) → ADD(X, Y)

R is empty.
The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))
half(0)
half(s(0))
half(s(s(x0)))
half(dbl(x0))

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

(28) 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].

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))
half(0)
half(s(0))
half(s(s(x0)))
half(dbl(x0))

(29) Obligation:

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

ADD(s(X), Y) → ADD(X, Y)

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

(30) 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:

• ADD(s(X), Y) → ADD(X, Y)
  The graph contains the following edges 1 > 1, 2 >= 2

(31) TRUE

(32) Obligation:

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

DBL(s(X)) → DBL(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

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

(33) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(34) Obligation:

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

DBL(s(X)) → DBL(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))
half(0)
half(s(0))
half(s(s(x0)))
half(dbl(x0))

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

(35) 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.

(36) Obligation:

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

DBL(s(X)) → DBL(X)

R is empty.
The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))
half(0)
half(s(0))
half(s(s(x0)))
half(dbl(x0))

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

(37) 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].

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))
half(0)
half(s(0))
half(s(s(x0)))
half(dbl(x0))

(38) Obligation:

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

DBL(s(X)) → DBL(X)

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

(39) 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:

• DBL(s(X)) → DBL(X)
  The graph contains the following edges 1 > 1

(40) TRUE

(41) Obligation:

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

SQR(s(X)) → SQR(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

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

(42) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(43) Obligation:

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

SQR(s(X)) → SQR(X)

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))
half(0)
half(s(0))
half(s(s(x0)))
half(dbl(x0))

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

(44) 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.

(45) Obligation:

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

SQR(s(X)) → SQR(X)

R is empty.
The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))
half(0)
half(s(0))
half(s(s(x0)))
half(dbl(x0))

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

(46) 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].

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))
half(0)
half(s(0))
half(s(s(x0)))
half(dbl(x0))

(47) Obligation:

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

SQR(s(X)) → SQR(X)

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

(48) 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:

• SQR(s(X)) → SQR(X)
  The graph contains the following edges 1 > 1

(49) TRUE

(50) Obligation:

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

TERMS(N) → TERMS(s(N))

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

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

(51) MNOCProof (EQUIVALENT transformation)

We use the modular non-overlap check [LPAR04] to enlarge Q to all left-hand sides of R.

(52) Obligation:

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

TERMS(N) → TERMS(s(N))

The TRS R consists of the following rules:

terms(N) → cons(recip(sqr(N)), terms(s(N)))
sqr(0) → 0
sqr(s(X)) → s(add(sqr(X), dbl(X)))
dbl(0) → 0
dbl(s(X)) → s(s(dbl(X)))
add(0, X) → X
add(s(X), Y) → s(add(X, Y))
first(0, X) → nil
first(s(X), cons(Y, Z)) → cons(Y, first(X, Z))
half(0) → 0
half(s(0)) → 0
half(s(s(X))) → s(half(X))
half(dbl(X)) → X

The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))
half(0)
half(s(0))
half(s(s(x0)))
half(dbl(x0))

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

(53) 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.

(54) Obligation:

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

TERMS(N) → TERMS(s(N))

R is empty.
The set Q consists of the following terms:

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))
half(0)
half(s(0))
half(s(s(x0)))
half(dbl(x0))

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

(55) 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].

terms(x0)
sqr(0)
sqr(s(x0))
dbl(0)
dbl(s(x0))
add(0, x0)
add(s(x0), x1)
first(0, x0)
first(s(x0), cons(x1, x2))
half(0)
half(s(0))
half(s(s(x0)))
half(dbl(x0))

(56) Obligation:

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

TERMS(N) → TERMS(s(N))

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

(57) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule TERMS(N) → TERMS(s(N)) we obtained the following new rules [LPAR04]:

TERMS(s(z0)) → TERMS(s(s(z0)))

(58) Obligation:

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

TERMS(s(z0)) → TERMS(s(s(z0)))

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

(59) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule TERMS(N) → TERMS(s(N)) we obtained the following new rules [LPAR04]:

TERMS(s(z0)) → TERMS(s(s(z0)))

(60) Obligation:

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

TERMS(s(z0)) → TERMS(s(s(z0)))

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

(61) Instantiation (EQUIVALENT transformation)

By instantiating [LPAR04] the rule TERMS(s(z0)) → TERMS(s(s(z0))) we obtained the following new rules [LPAR04]:

TERMS(s(s(z0))) → TERMS(s(s(s(z0))))

(62) Obligation:

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

TERMS(s(s(z0))) → TERMS(s(s(s(z0))))

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

(63) NonTerminationProof (EQUIVALENT transformation)

We used the non-termination processor [FROCOS05] to show that the DP problem is infinite.
Found a loop by semiunifying a rule from P directly.

s = TERMS(s(s(z0))) evaluates to t =TERMS(s(s(s(z0))))

Thus s starts an infinite chain as s semiunifies with t with the following substitutions:

• Semiunifier: [ ]
• Matcher: [z0 / s(z0)]

---

Rewriting sequence

The DP semiunifies directly so there is only one rewrite step from TERMS(s(s(z0))) to TERMS(s(s(s(z0)))).

(64) FALSE