YES Termination w.r.t. Q proof of AProVE_06_logarithm.ari

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

0 QTRS
↳1 Overlay + Local Confluence (⇔, 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 QDPOrderProof (⇔, 0 ms)
        ↳34 QDP
        ↳35 PisEmptyProof (⇔, 0 ms)
        ↳36 YES
    ↳37 QDP
        ↳38 UsableRulesProof (⇔, 0 ms)
        ↳39 QDP
        ↳40 QReductionProof (⇔, 0 ms)
        ↳41 QDP
        ↳42 TransformationProof (⇔, 0 ms)
        ↳43 QDP
        ↳44 DependencyGraphProof (⇔, 0 ms)
        ↳45 QDP
        ↳46 TransformationProof (⇔, 0 ms)
        ↳47 QDP
        ↳48 TransformationProof (⇔, 0 ms)
        ↳49 QDP
        ↳50 TransformationProof (⇔, 0 ms)
        ↳51 QDP
        ↳52 TransformationProof (⇔, 0 ms)
        ↳53 QDP
        ↳54 DependencyGraphProof (⇔, 0 ms)
        ↳55 QDP
        ↳56 UsableRulesProof (⇔, 0 ms)
        ↳57 QDP
        ↳58 TransformationProof (⇔, 0 ms)
        ↳59 QDP
        ↳60 UsableRulesProof (⇔, 0 ms)
        ↳61 QDP
        ↳62 QReductionProof (⇔, 0 ms)
        ↳63 QDP
        ↳64 TransformationProof (⇔, 0 ms)
        ↳65 QDP
        ↳66 TransformationProof (⇔, 0 ms)
        ↳67 QDP
        ↳68 TransformationProof (⇔, 0 ms)
        ↳69 QDP
        ↳70 TransformationProof (⇔, 0 ms)
        ↳71 QDP
        ↳72 TransformationProof (⇔, 0 ms)
        ↳73 QDP
        ↳74 DependencyGraphProof (⇔, 0 ms)
        ↳75 QDP
        ↳76 QDPOrderProof (⇔, 0 ms)
        ↳77 QDP
        ↳78 DependencyGraphProof (⇔, 0 ms)
        ↳79 TRUE

(0) Obligation:

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

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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)
log(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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)
log(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, 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:

MINUS(s(x), s(y)) → MINUS(x, y)
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) → MINUS(x, y)
LE(s(x), s(y)) → LE(x, y)
INC(s(x)) → INC(x)
LOG(x) → LOGITER(x, 0)
LOGITER(x, y) → IF(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))
LOGITER(x, y) → LE(s(0), x)
LOGITER(x, y) → LE(s(s(0)), x)
LOGITER(x, y) → QUOT(x, s(s(0)))
LOGITER(x, y) → INC(y)
IF(true, true, x, y) → LOGITER(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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)
log(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, 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 5 SCCs with 6 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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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)
log(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, 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:

INC(s(x)) → INC(x)

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

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, 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].

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

(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:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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)
log(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, 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:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, 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].

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, 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) YES

(21) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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)
log(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, 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:

MINUS(s(x), s(y)) → MINUS(x, y)

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

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, 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].

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

(25) Obligation:

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

MINUS(s(x), s(y)) → MINUS(x, y)

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:

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

(27) YES

(28) Obligation:

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

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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)
log(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

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:

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

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].

quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

(32) Obligation:

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

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))

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

(33) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


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

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

s_1=1
dummyConstant=1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

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

(34) Obligation:

Q DP problem:
P is empty.
The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))

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

(35) PisEmptyProof (EQUIVALENT transformation)

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

(36) YES

(37) 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), quot(x, s(s(0))), inc(y))
IF(true, true, x, y) → LOGITER(x, y)

The TRS R consists of the following rules:

minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
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)
log(x) → logIter(x, 0)
logIter(x, y) → if(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))
if(false, b, x, y) → logZeroError
if(true, false, x, s(y)) → y
if(true, true, x, y) → logIter(x, y)

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

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

(38) 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.

(39) 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), quot(x, s(s(0))), 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)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(s(x0))
inc(0)
log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

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

(40) 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].

log(x0)
logIter(x0, x1)
if(false, x0, x1, x2)
if(true, false, x0, s(x1))
if(true, true, x0, x1)

(41) 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), quot(x, s(s(0))), 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)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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.

(42) TransformationProof (EQUIVALENT transformation)

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

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

(43) 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), quot(0, s(s(0))), inc(y1))
LOGITER(s(x1), y1) → IF(le(0, x1), le(s(s(0)), s(x1)), quot(s(x1), s(s(0))), inc(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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.

(44) DependencyGraphProof (EQUIVALENT transformation)

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

(45) 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)), quot(s(x1), s(s(0))), 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)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
le(0, y) → true

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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.

(46) TransformationProof (EQUIVALENT transformation)

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

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

(47) 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)), quot(s(x1), s(s(0))), inc(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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.

(48) TransformationProof (EQUIVALENT transformation)

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

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

(49) 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), quot(s(x1), s(s(0))), inc(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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.

(50) TransformationProof (EQUIVALENT transformation)

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

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

(51) 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), s(quot(minus(x1, s(0)), s(s(0)))), inc(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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.

(52) TransformationProof (EQUIVALENT transformation)

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

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

(53) 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, s(quot(minus(0, s(0)), s(s(0)))), inc(y1))
LOGITER(s(s(x1)), y1) → IF(true, le(0, x1), s(quot(minus(s(x1), s(0)), s(s(0)))), inc(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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.

(54) DependencyGraphProof (EQUIVALENT transformation)

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

(55) Obligation:

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

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

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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.

(56) 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.

(57) Obligation:

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

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

The TRS R consists of the following rules:

le(0, y) → true
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
minus(x, 0) → x

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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.

(58) TransformationProof (EQUIVALENT transformation)

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

LOGITER(s(s(x1)), y1) → IF(true, true, s(quot(minus(s(x1), s(0)), s(s(0)))), inc(y1)) → LOGITER(s(s(x1)), y1) → IF(true, true, s(quot(minus(s(x1), s(0)), s(s(0)))), inc(y1))

(59) 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(quot(minus(s(x1), s(0)), s(s(0)))), inc(y1))

The TRS R consists of the following rules:

le(0, y) → true
minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
minus(x, 0) → x

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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.

(60) 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.

(61) 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(quot(minus(s(x1), s(0)), s(s(0)))), inc(y1))

The TRS R consists of the following rules:

minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
minus(x, 0) → x

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
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.

(62) 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))

(63) 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(quot(minus(s(x1), s(0)), s(s(0)))), inc(y1))

The TRS R consists of the following rules:

minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
minus(x, 0) → x

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)

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

(64) TransformationProof (EQUIVALENT transformation)

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

LOGITER(s(s(x1)), y1) → IF(true, true, s(quot(minus(x1, 0), s(s(0)))), inc(y1)) → LOGITER(s(s(x1)), y1) → IF(true, true, s(quot(minus(x1, 0), s(s(0)))), inc(y1))

(65) 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(quot(minus(x1, 0), s(s(0)))), inc(y1))

The TRS R consists of the following rules:

minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
minus(x, 0) → x

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)

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

(66) TransformationProof (EQUIVALENT transformation)

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

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

(67) 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(quot(x1, s(s(0)))), inc(y1))

The TRS R consists of the following rules:

minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
minus(x, 0) → x

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)

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

(68) 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)

(69) Obligation:

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

LOGITER(s(s(x1)), y1) → IF(true, true, s(quot(x1, s(s(0)))), inc(y1))
IF(true, true, s(y_0), y_1) → LOGITER(s(y_0), y_1)

The TRS R consists of the following rules:

minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
minus(x, 0) → x

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)

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

(70) 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)

(71) Obligation:

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

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

The TRS R consists of the following rules:

minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
minus(x, 0) → x

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)

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

(72) TransformationProof (EQUIVALENT transformation)

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

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

(73) 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(0)), y1) → IF(true, true, s(0), inc(y1))
LOGITER(s(s(s(x0))), y1) → IF(true, true, s(s(quot(minus(x0, s(0)), s(s(0))))), inc(y1))

The TRS R consists of the following rules:

minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
minus(x, 0) → x

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)

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

(74) DependencyGraphProof (EQUIVALENT transformation)

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

(75) Obligation:

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

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

The TRS R consists of the following rules:

minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
minus(x, 0) → x

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)

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

(76) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04,JAR06].


The following pairs can be oriented strictly and are deleted.


LOGITER(s(s(s(x0))), y1) → IF(true, true, s(s(quot(minus(x0, s(0)), s(s(0))))), inc(y1))
The remaining pairs can at least be oriented weakly.
Used ordering: Combined order from the following AFS and order.
LOGITER(x1, x2)  =  x1
s(x1)  =  s(x1)
IF(x1, x2, x3, x4)  =  x3
quot(x1, x2)  =  x1
minus(x1, x2)  =  x1
0  =  0

Knuth-Bendix order [KBO] with precedence:
trivial

and weight map:

s_1=1
0=1

The following usable rules [FROCOS05] with respect to the argument filtering of the ordering [JAR06] were oriented:

minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
minus(x, 0) → x

(77) 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)

The TRS R consists of the following rules:

minus(s(x), s(y)) → minus(x, y)
quot(0, s(y)) → 0
quot(s(x), s(y)) → s(quot(minus(x, y), s(y)))
inc(s(x)) → s(inc(x))
inc(0) → s(0)
minus(x, 0) → x

The set Q consists of the following terms:

minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
inc(s(x0))
inc(0)

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

(78) DependencyGraphProof (EQUIVALENT transformation)

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

(79) TRUE