Termination w.r.t. Q proof of TRS/Thiemannlogarithm.trs

Termination w.r.t. Q of the following Term Rewriting System could not be shown:

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) → 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.

↳ QTRS

  ↳ Overlay + Local Confluence

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) → 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.

The TRS is overlay and locally confluent. By [15] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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) → 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)

Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

IF(true, true, x, y) → LOGITER(x, y)
LOGITER(x, y) → LE(s(0), x)
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
LOG(x) → LOGITER(x, 0)
LOGITER(x, y) → LE(s(s(0)), x)
LOGITER(x, y) → IF(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))
LE(s(x), s(y)) → LE(x, y)
LOGITER(x, y) → QUOT(x, s(s(0)))
MINUS(s(x), s(y)) → MINUS(x, y)
LOGITER(x, y) → INC(y)
INC(s(x)) → INC(x)
QUOT(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) → 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.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
IF(true, true, x, y) → LOGITER(x, y)
MINUS(s(x), s(y)) → MINUS(x, y)
QUOT(s(x), s(y)) → MINUS(x, y)
LOGITER(x, y) → LE(s(0), x)
LOG(x) → LOGITER(x, 0)
LOGITER(x, y) → LE(s(s(0)), x)
LOGITER(x, y) → IF(le(s(0), x), le(s(s(0)), x), quot(x, s(s(0))), inc(y))
LE(s(x), s(y)) → LE(x, y)
LOGITER(x, y) → QUOT(x, s(s(0)))
INC(s(x)) → INC(x)
LOGITER(x, y) → INC(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) → 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.
The approximation of the Dependency Graph [13,14,18] contains 5 SCCs with 6 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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) → 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

INC(s(x)) → INC(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
INC(x1) = INC(x1)
s(x1) = s(x1)

Lexicographic path order with status [19].
Quasi-Precedence:

[INC₁, s_1]

Status:

INC₁: [1]
s₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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) → 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

LE(s(x), s(y)) → LE(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
LE(x1, x2) = LE(x2)
s(x1) = s(x1)

Lexicographic path order with status [19].
Quasi-Precedence:

s₁ > LE₁

Status:

LE₁: [1]
s₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

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) → 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.
We use the reduction pair processor [13].

The following pairs can be oriented strictly and are deleted.

MINUS(s(x), s(y)) → MINUS(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
MINUS(x1, x2) = MINUS(x2)
s(x1) = s(x1)

Lexicographic path order with status [19].
Quasi-Precedence:

s₁ > MINUS₁

Status:

MINUS₁: [1]
s₁: [1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

                  ↳ QDP

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

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) → 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.
We use the reduction pair processor [13].

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.
none
Used ordering: Combined order from the following AFS and order.
QUOT(x1, x2) = QUOT(x1, x2)
s(x1) = s(x1)
minus(x1, x2) = minus(x1)
0 = 0

Lexicographic path order with status [19].
Quasi-Precedence:

[QUOT₂, s_1] > minus₁
0 > minus₁

Status:

minus₁: [1]
QUOT₂: [2,1]
0: multiset
s₁: [1]

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ PisEmptyProof

                  ↳ QDP

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

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