YES Termination w.r.t. Q proof of Secret_06_TRS_logarithm.ari

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

0 QTRS
↳1 AAECC Innermost (⇔, 0 ms)
↳2 QTRS
↳3 DependencyPairsProof (⇔, 0 ms)
↳4 QDP
↳5 DependencyGraphProof (⇔, 0 ms)
↳6 AND
    ↳7 QDP
        ↳8 UsableRulesProof (⇔, 0 ms)
        ↳9 QDP
        ↳10 QReductionProof (⇔, 0 ms)
        ↳11 QDP
        ↳12 QDPSizeChangeProof (⇔, 0 ms)
        ↳13 YES
    ↳14 QDP
        ↳15 UsableRulesProof (⇔, 0 ms)
        ↳16 QDP
        ↳17 QReductionProof (⇔, 0 ms)
        ↳18 QDP
        ↳19 QDPSizeChangeProof (⇔, 0 ms)
        ↳20 YES
    ↳21 QDP
        ↳22 UsableRulesProof (⇔, 0 ms)
        ↳23 QDP
        ↳24 QReductionProof (⇔, 0 ms)
        ↳25 QDP
        ↳26 QDPSizeChangeProof (⇔, 0 ms)
        ↳27 YES
    ↳28 QDP
        ↳29 UsableRulesProof (⇔, 0 ms)
        ↳30 QDP
        ↳31 QReductionProof (⇔, 0 ms)
        ↳32 QDP
        ↳33 TransformationProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 DependencyGraphProof (⇔, 0 ms)
        ↳36 QDP
        ↳37 TransformationProof (⇔, 0 ms)
        ↳38 QDP
        ↳39 TransformationProof (⇔, 0 ms)
        ↳40 QDP
        ↳41 TransformationProof (⇔, 0 ms)
        ↳42 QDP
        ↳43 DependencyGraphProof (⇔, 0 ms)
        ↳44 QDP
        ↳45 UsableRulesProof (⇔, 0 ms)
        ↳46 QDP
        ↳47 TransformationProof (⇔, 0 ms)
        ↳48 QDP
        ↳49 UsableRulesProof (⇔, 0 ms)
        ↳50 QDP
        ↳51 QReductionProof (⇔, 0 ms)
        ↳52 QDP
        ↳53 TransformationProof (⇔, 0 ms)
        ↳54 QDP
        ↳55 TransformationProof (⇔, 0 ms)
        ↳56 QDP
        ↳57 TransformationProof (⇔, 0 ms)
        ↳58 QDP
        ↳59 TransformationProof (⇔, 0 ms)
        ↳60 QDP
        ↳61 DependencyGraphProof (⇔, 0 ms)
        ↳62 QDP
        ↳63 MRRProof (⇔, 0 ms)
        ↳64 QDP
        ↳65 PisEmptyProof (⇔, 0 ms)
        ↳66 YES

(0) Obligation:

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

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(s(x)) → s(inc(x))
inc(0) → s(0)
logarithm(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)
f → g
f → h

Q is empty.

(1) AAECC Innermost (EQUIVALENT transformation)

We have applied [NOC,AAECCNOC] to switch to innermost. The TRS R 1 is

half(s(0)) → 0
half(s(s(x))) → s(half(x))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)
half(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
logarithm(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y) → logZeroError

The TRS R 2 is

f → g
f → h

The signature Sigma is {f, g, h}

(2) Obligation:

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

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(s(x)) → s(inc(x))
inc(0) → s(0)
logarithm(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)
f → g
f → h

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

(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:

HALF(s(s(x))) → HALF(x)
LE(s(x), s(y)) → LE(x, y)
INC(s(x)) → INC(x)
LOGARITHM(x) → LOGITER(x, 0)
LOGITER(x, y) → IF(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
LOGITER(x, y) → LE(s(0), x)
LOGITER(x, y) → LE(s(s(0)), x)
LOGITER(x, y) → HALF(x)
LOGITER(x, y) → INC(y)
IF(true, true, x, y) → LOGITER(x, y)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(s(x)) → s(inc(x))
inc(0) → s(0)
logarithm(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)
f → g
f → h

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 4 SCCs with 5 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

INC(s(x)) → INC(x)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(s(x)) → s(inc(x))
inc(0) → s(0)
logarithm(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)
f → g
f → h

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

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:

INC(s(x)) → INC(x)

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

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

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].

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

(11) Obligation:

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

INC(s(x)) → INC(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:

  • INC(s(x)) → INC(x)
    The graph contains the following edges 1 > 1

(13) YES

(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:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(s(x)) → s(inc(x))
inc(0) → s(0)
logarithm(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)
f → g
f → h

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

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:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

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].

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

(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) YES

(21) 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:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(s(x)) → s(inc(x))
inc(0) → s(0)
logarithm(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)
f → g
f → h

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

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:

HALF(s(s(x))) → HALF(x)

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

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

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].

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

(25) 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.

(26) 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

(27) YES

(28) Obligation:

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

LOGITER(x, y) → IF(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
IF(true, true, x, y) → LOGITER(x, y)

The TRS R consists of the following rules:

half(0) → 0
half(s(0)) → 0
half(s(s(x))) → s(half(x))
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(s(x)) → s(inc(x))
inc(0) → s(0)
logarithm(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)
f → g
f → h

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

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

(29) 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.

(30) Obligation:

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

LOGITER(x, y) → IF(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
IF(true, true, x, y) → LOGITER(x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

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

(31) 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].

logarithm(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)
f

(32) Obligation:

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

LOGITER(x, y) → IF(le(s(0), x), le(s(s(0)), x), half(x), inc(y))
IF(true, true, x, y) → LOGITER(x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)

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

(33) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule LOGITER(x, y) → IF(le(s(0), x), le(s(s(0)), x), half(x), inc(y)) at position [0] we obtained the following new rules [LPAR04]:

LOGITER(0, y1) → IF(false, le(s(s(0)), 0), half(0), inc(y1)) → LOGITER(0, y1) → IF(false, le(s(s(0)), 0), half(0), inc(y1))
LOGITER(s(x1), y1) → IF(le(0, x1), le(s(s(0)), s(x1)), half(s(x1)), inc(y1)) → LOGITER(s(x1), y1) → IF(le(0, x1), le(s(s(0)), s(x1)), half(s(x1)), inc(y1))

(34) Obligation:

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

IF(true, true, x, y) → LOGITER(x, y)
LOGITER(0, y1) → IF(false, le(s(s(0)), 0), half(0), inc(y1))
LOGITER(s(x1), y1) → IF(le(0, x1), le(s(s(0)), s(x1)), half(s(x1)), inc(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)

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

(35) DependencyGraphProof (EQUIVALENT transformation)

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

(36) Obligation:

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

LOGITER(s(x1), y1) → IF(le(0, x1), le(s(s(0)), s(x1)), half(s(x1)), inc(y1))
IF(true, true, x, y) → LOGITER(x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)

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

(37) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LOGITER(s(x1), y1) → IF(le(0, x1), le(s(s(0)), s(x1)), half(s(x1)), inc(y1)) at position [0] we obtained the following new rules [LPAR04]:

LOGITER(s(x1), y1) → IF(true, le(s(s(0)), s(x1)), half(s(x1)), inc(y1)) → LOGITER(s(x1), y1) → IF(true, le(s(s(0)), s(x1)), half(s(x1)), inc(y1))

(38) Obligation:

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

IF(true, true, x, y) → LOGITER(x, y)
LOGITER(s(x1), y1) → IF(true, le(s(s(0)), s(x1)), half(s(x1)), inc(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)

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

(39) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LOGITER(s(x1), y1) → IF(true, le(s(s(0)), s(x1)), half(s(x1)), inc(y1)) at position [1] we obtained the following new rules [LPAR04]:

LOGITER(s(x1), y1) → IF(true, le(s(0), x1), half(s(x1)), inc(y1)) → LOGITER(s(x1), y1) → IF(true, le(s(0), x1), half(s(x1)), inc(y1))

(40) Obligation:

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

IF(true, true, x, y) → LOGITER(x, y)
LOGITER(s(x1), y1) → IF(true, le(s(0), x1), half(s(x1)), inc(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)

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

(41) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule LOGITER(s(x1), y1) → IF(true, le(s(0), x1), half(s(x1)), inc(y1)) at position [1] we obtained the following new rules [LPAR04]:

LOGITER(s(0), y1) → IF(true, false, half(s(0)), inc(y1)) → LOGITER(s(0), y1) → IF(true, false, half(s(0)), inc(y1))
LOGITER(s(s(x1)), y1) → IF(true, le(0, x1), half(s(s(x1))), inc(y1)) → LOGITER(s(s(x1)), y1) → IF(true, le(0, x1), half(s(s(x1))), inc(y1))

(42) Obligation:

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

IF(true, true, x, y) → LOGITER(x, y)
LOGITER(s(0), y1) → IF(true, false, half(s(0)), inc(y1))
LOGITER(s(s(x1)), y1) → IF(true, le(0, x1), half(s(s(x1))), inc(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)

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

(43) DependencyGraphProof (EQUIVALENT transformation)

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

(44) Obligation:

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

LOGITER(s(s(x1)), y1) → IF(true, le(0, x1), half(s(s(x1))), inc(y1))
IF(true, true, x, y) → LOGITER(x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)

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

(45) 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.

(46) Obligation:

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

LOGITER(s(s(x1)), y1) → IF(true, le(0, x1), half(s(s(x1))), inc(y1))
IF(true, true, x, y) → LOGITER(x, y)

The TRS R consists of the following rules:

le(0, y) → true
half(s(s(x))) → s(half(x))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
half(0) → 0
half(s(0)) → 0

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)

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

(47) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LOGITER(s(s(x1)), y1) → IF(true, le(0, x1), half(s(s(x1))), inc(y1)) at position [1] we obtained the following new rules [LPAR04]:

LOGITER(s(s(x1)), y1) → IF(true, true, half(s(s(x1))), inc(y1)) → LOGITER(s(s(x1)), y1) → IF(true, true, half(s(s(x1))), inc(y1))

(48) Obligation:

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

IF(true, true, x, y) → LOGITER(x, y)
LOGITER(s(s(x1)), y1) → IF(true, true, half(s(s(x1))), inc(y1))

The TRS R consists of the following rules:

le(0, y) → true
half(s(s(x))) → s(half(x))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
half(0) → 0
half(s(0)) → 0

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)

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

(49) 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.

(50) Obligation:

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

IF(true, true, x, y) → LOGITER(x, y)
LOGITER(s(s(x1)), y1) → IF(true, true, half(s(s(x1))), inc(y1))

The TRS R consists of the following rules:

half(s(s(x))) → s(half(x))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
half(0) → 0
half(s(0)) → 0

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)

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

(51) 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].

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

(52) Obligation:

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

IF(true, true, x, y) → LOGITER(x, y)
LOGITER(s(s(x1)), y1) → IF(true, true, half(s(s(x1))), inc(y1))

The TRS R consists of the following rules:

half(s(s(x))) → s(half(x))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
half(0) → 0
half(s(0)) → 0

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(s(x0))
inc(0)

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

(53) TransformationProof (EQUIVALENT transformation)

By rewriting [LPAR04] the rule LOGITER(s(s(x1)), y1) → IF(true, true, half(s(s(x1))), inc(y1)) at position [2] we obtained the following new rules [LPAR04]:

LOGITER(s(s(x1)), y1) → IF(true, true, s(half(x1)), inc(y1)) → LOGITER(s(s(x1)), y1) → IF(true, true, s(half(x1)), inc(y1))

(54) Obligation:

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

IF(true, true, x, y) → LOGITER(x, y)
LOGITER(s(s(x1)), y1) → IF(true, true, s(half(x1)), inc(y1))

The TRS R consists of the following rules:

half(s(s(x))) → s(half(x))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
half(0) → 0
half(s(0)) → 0

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(s(x0))
inc(0)

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

(55) TransformationProof (EQUIVALENT transformation)

By instantiating [LPAR04] the rule IF(true, true, x, y) → LOGITER(x, y) we obtained the following new rules [LPAR04]:

IF(true, true, s(y_0), y_1) → LOGITER(s(y_0), y_1) → IF(true, true, s(y_0), y_1) → LOGITER(s(y_0), y_1)

(56) Obligation:

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

LOGITER(s(s(x1)), y1) → IF(true, true, s(half(x1)), inc(y1))
IF(true, true, s(y_0), y_1) → LOGITER(s(y_0), y_1)

The TRS R consists of the following rules:

half(s(s(x))) → s(half(x))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
half(0) → 0
half(s(0)) → 0

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(s(x0))
inc(0)

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

(57) TransformationProof (EQUIVALENT transformation)

By forward instantiating [JAR06] the rule IF(true, true, s(y_0), y_1) → LOGITER(s(y_0), y_1) we obtained the following new rules [LPAR04]:

IF(true, true, s(s(y_0)), x1) → LOGITER(s(s(y_0)), x1) → IF(true, true, s(s(y_0)), x1) → LOGITER(s(s(y_0)), x1)

(58) Obligation:

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

LOGITER(s(s(x1)), y1) → IF(true, true, s(half(x1)), inc(y1))
IF(true, true, s(s(y_0)), x1) → LOGITER(s(s(y_0)), x1)

The TRS R consists of the following rules:

half(s(s(x))) → s(half(x))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
half(0) → 0
half(s(0)) → 0

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(s(x0))
inc(0)

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

(59) TransformationProof (EQUIVALENT transformation)

By narrowing [LPAR04] the rule LOGITER(s(s(x1)), y1) → IF(true, true, s(half(x1)), inc(y1)) at position [2,0] we obtained the following new rules [LPAR04]:

LOGITER(s(s(s(s(x0)))), y1) → IF(true, true, s(s(half(x0))), inc(y1)) → LOGITER(s(s(s(s(x0)))), y1) → IF(true, true, s(s(half(x0))), inc(y1))
LOGITER(s(s(0)), y1) → IF(true, true, s(0), inc(y1)) → LOGITER(s(s(0)), y1) → IF(true, true, s(0), inc(y1))
LOGITER(s(s(s(0))), y1) → IF(true, true, s(0), inc(y1)) → LOGITER(s(s(s(0))), y1) → IF(true, true, s(0), inc(y1))

(60) Obligation:

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

IF(true, true, s(s(y_0)), x1) → LOGITER(s(s(y_0)), x1)
LOGITER(s(s(s(s(x0)))), y1) → IF(true, true, s(s(half(x0))), inc(y1))
LOGITER(s(s(0)), y1) → IF(true, true, s(0), inc(y1))
LOGITER(s(s(s(0))), y1) → IF(true, true, s(0), inc(y1))

The TRS R consists of the following rules:

half(s(s(x))) → s(half(x))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
half(0) → 0
half(s(0)) → 0

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(s(x0))
inc(0)

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

(61) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 1 SCC with 2 less nodes.

(62) Obligation:

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

LOGITER(s(s(s(s(x0)))), y1) → IF(true, true, s(s(half(x0))), inc(y1))
IF(true, true, s(s(y_0)), x1) → LOGITER(s(s(y_0)), x1)

The TRS R consists of the following rules:

half(s(s(x))) → s(half(x))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
half(0) → 0
half(s(0)) → 0

The set Q consists of the following terms:

half(0)
half(s(0))
half(s(s(x0)))
inc(s(x0))
inc(0)

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

(63) MRRProof (EQUIVALENT transformation)

By using the rule removal processor [LPAR04] with the following ordering, at least one Dependency Pair or term rewrite system rule of this QDP problem can be strictly oriented.
Strictly oriented dependency pairs:

LOGITER(s(s(s(s(x0)))), y1) → IF(true, true, s(s(half(x0))), inc(y1))
IF(true, true, s(s(y_0)), x1) → LOGITER(s(s(y_0)), x1)

Strictly oriented rules of the TRS R:

half(s(s(x))) → s(half(x))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
half(0) → 0
half(s(0)) → 0

Used ordering: Knuth-Bendix order [KBO] with precedence:
half1 > true > IF4 > LOGITER2 > 0 > inc1 > s1

and weight map:

0=1
true=10
half_1=0
s_1=2
inc_1=3
LOGITER_2=20
IF_4=0

The variable weight is 1

(64) Obligation:

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

half(0)
half(s(0))
half(s(s(x0)))
inc(s(x0))
inc(0)

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

(65) PisEmptyProof (EQUIVALENT transformation)

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

(66) YES