Termination w.r.t. Q proof of /tmp/tmpREOmE0/thiemann13.trs

Termination w.r.t. Q of the following Term Rewriting System could be proven:

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
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)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

Q is empty.

↳ QTRS

  ↳ Overlay + Local Confluence

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
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)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

Q is empty.

The TRS is overlay and locally confluent. By [19] we can switch to innermost.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

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

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
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)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

Using Dependency Pairs [1,15] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

LOG2(x, y) → LE(x, 0)
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
LOG2(x, y) → INC(y)
INC(s(x)) → INC(x)
IF2(false, x, y) → QUOT(x, s(s(0)))
LE(s(x), s(y)) → LE(x, y)
QUOT(s(x), s(y)) → MINUS(x, y)
LOG2(x, y) → LE(x, s(0))
MINUS(s(x), s(y)) → MINUS(x, y)
LOG(x) → LOG2(x, 0)
IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)
IF(false, b, x, y) → IF2(b, x, y)
LOG2(x, y) → IF(le(x, 0), le(x, s(0)), x, inc(y))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
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)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

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

LOG2(x, y) → LE(x, 0)
QUOT(s(x), s(y)) → QUOT(minus(x, y), s(y))
LOG2(x, y) → INC(y)
INC(s(x)) → INC(x)
IF2(false, x, y) → QUOT(x, s(s(0)))
LE(s(x), s(y)) → LE(x, y)
QUOT(s(x), s(y)) → MINUS(x, y)
LOG2(x, y) → LE(x, s(0))
MINUS(s(x), s(y)) → MINUS(x, y)
LOG(x) → LOG2(x, 0)
IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)
IF(false, b, x, y) → IF2(b, x, y)
LOG2(x, y) → IF(le(x, 0), le(x, s(0)), x, inc(y))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
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)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 5 SCCs with 6 less nodes.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

              ↳ 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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
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)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
As all Q-normal forms are R-normal forms we are in the innermost case. Hence, by the usable rules processor [15] we can delete all non-usable rules [17] from R.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

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.
By using the subterm criterion [20] together with the size-change analysis [32] 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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ QDP

              ↳ QDP

              ↳ 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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
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)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

              ↳ QDP

              ↳ QDP

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

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: Polynomial interpretation [25]:

POL(0) = 0
POL(QUOT(x₁, x₂)) = x₁
POL(minus(x₁, x₂)) = x₁
POL(s(x₁)) = 1 + x₁

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

              ↳ QDP

              ↳ QDP

              ↳ QDP

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

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

The set Q consists of the following terms:

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

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
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)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

              ↳ QDP

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:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

              ↳ QDP

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

INC(s(x)) → INC(x)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

              ↳ 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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
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)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

              ↳ QDP

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:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ QDPSizeChangeProof

              ↳ QDP

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

LE(s(x), s(y)) → LE(x, y)

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

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

IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)
IF(false, b, x, y) → IF2(b, x, y)
LOG2(x, y) → IF(le(x, 0), le(x, s(0)), x, inc(y))

The TRS R consists of the following rules:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
inc(0) → 0
inc(s(x)) → s(inc(x))
minus(0, y) → 0
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)))
log(x) → log2(x, 0)
log2(x, y) → if(le(x, 0), le(x, s(0)), x, inc(y))
if(true, b, x, y) → log_undefined
if(false, b, x, y) → if2(b, x, y)
if2(true, x, s(y)) → y
if2(false, x, y) → log2(quot(x, s(s(0))), y)

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

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

IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)
IF(false, b, x, y) → IF2(b, x, y)
LOG2(x, y) → IF(le(x, 0), le(x, s(0)), x, inc(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

le(0, x0)
le(s(x0), 0)
le(s(x0), s(x1))
inc(0)
inc(s(x0))
minus(0, x0)
minus(x0, 0)
minus(s(x0), s(x1))
quot(0, s(x0))
quot(s(x0), s(x1))
log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

log(x0)
log2(x0, x1)
if(true, x0, x1, x2)
if(false, x0, x1, x2)
if2(true, x0, s(x1))
if2(false, x0, x1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RemovalProof

                        ↳ RemovalProof

                        ↳ Narrowing

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

IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)
IF(false, b, x, y) → IF2(b, x, y)
LOG2(x, y) → IF(le(x, 0), le(x, s(0)), x, inc(y))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
In the following pairs the term without variables s(s(0)) is replaced by the fresh variable x_removed.
Pair: IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)
Positions in right side of the pair:

[0,1]

The new variable was added to all pairs as a new argument[13].

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RemovalProof

                          ↳ QDP

                        ↳ RemovalProof

                        ↳ Narrowing

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

IF2(false, x, y, x_removed) → LOG2(quot(x, x_removed), y, x_removed)
LOG2(x, y, x_removed) → IF(le(x, 0), le(x, s(0)), x, inc(y), x_removed)
IF(false, b, x, y, x_removed) → IF2(b, x, y, x_removed)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

[0,1]

The new variable was added to all pairs as a new argument[13].

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RemovalProof

                        ↳ RemovalProof

                          ↳ QDP

                        ↳ Narrowing

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

IF2(false, x, y, x_removed) → LOG2(quot(x, x_removed), y, x_removed)
LOG2(x, y, x_removed) → IF(le(x, 0), le(x, s(0)), x, inc(y), x_removed)
IF(false, b, x, y, x_removed) → IF2(b, x, y, x_removed)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule LOG2(x, y) → IF(le(x, 0), le(x, s(0)), x, inc(y)) at position [0] we obtained the following new rules:

LOG2(s(x0), y1) → IF(false, le(s(x0), s(0)), s(x0), inc(y1))
LOG2(0, y1) → IF(true, le(0, s(0)), 0, inc(y1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RemovalProof

                        ↳ RemovalProof

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

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

LOG2(s(x0), y1) → IF(false, le(s(x0), s(0)), s(x0), inc(y1))
LOG2(0, y1) → IF(true, le(0, s(0)), 0, inc(y1))
IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)
IF(false, b, x, y) → IF2(b, x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RemovalProof

                        ↳ RemovalProof

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

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

LOG2(s(x0), y1) → IF(false, le(s(x0), s(0)), s(x0), inc(y1))
IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)
IF(false, b, x, y) → IF2(b, x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule LOG2(s(x0), y1) → IF(false, le(s(x0), s(0)), s(x0), inc(y1)) at position [1] we obtained the following new rules:

LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RemovalProof

                        ↳ RemovalProof

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ UsableRulesProof

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

LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1))
IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)
IF(false, b, x, y) → IF2(b, x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RemovalProof

                        ↳ RemovalProof

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ Narrowing

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

LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1))
IF2(false, x, y) → LOG2(quot(x, s(s(0))), y)
IF(false, b, x, y) → IF2(b, x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule IF2(false, x, y) → LOG2(quot(x, s(s(0))), y) at position [0] we obtained the following new rules:

IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)
IF2(false, 0, y1) → LOG2(0, y1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RemovalProof

                        ↳ RemovalProof

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

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

IF2(false, 0, y1) → LOG2(0, y1)
IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)
LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1))
IF(false, b, x, y) → IF2(b, x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RemovalProof

                        ↳ RemovalProof

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

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

IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)
LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1))
IF(false, b, x, y) → IF2(b, x, y)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF(false, b, x, y) → IF2(b, x, y) we obtained the following new rules:

IF(false, y_0, s(z0), y_1) → IF2(y_0, s(z0), y_1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RemovalProof

                        ↳ RemovalProof

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ ForwardInstantiation

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

IF(false, y_0, s(z0), y_1) → IF2(y_0, s(z0), y_1)
IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)
LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By forward instantiating [14] the rule IF(false, y_0, s(z0), y_1) → IF2(y_0, s(z0), y_1) we obtained the following new rules:

IF(false, false, s(x1), x2) → IF2(false, s(x1), x2)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RemovalProof

                        ↳ RemovalProof

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ ForwardInstantiation

                                                      ↳ QDP

                                                        ↳ Narrowing

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

IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)
LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1))
IF(false, false, s(x1), x2) → IF2(false, s(x1), x2)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By narrowing [15] the rule LOG2(s(x0), y1) → IF(false, le(x0, 0), s(x0), inc(y1)) at position [1] we obtained the following new rules:

LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))
LOG2(s(0), y1) → IF(false, true, s(0), inc(y1))

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RemovalProof

                        ↳ RemovalProof

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ ForwardInstantiation

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

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

IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)
LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))
IF(false, false, s(x1), x2) → IF2(false, s(x1), x2)
LOG2(s(0), y1) → IF(false, true, s(0), inc(y1))

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 1 less node.

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RemovalProof

                        ↳ RemovalProof

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ ForwardInstantiation

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ UsableRulesProof

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

IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)
LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))
IF(false, false, s(x1), x2) → IF2(false, s(x1), x2)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RemovalProof

                        ↳ RemovalProof

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ ForwardInstantiation

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ UsableRulesProof

                                                                  ↳ QDP

                                                                    ↳ QReductionProof

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

IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)
LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))
IF(false, false, s(x1), x2) → IF2(false, s(x1), x2)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
We deleted the following terms from Q as each root-symbol of these terms does neither occur in P nor in R.

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RemovalProof

                        ↳ RemovalProof

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ ForwardInstantiation

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ UsableRulesProof

                                                                  ↳ QDP

                                                                    ↳ QReductionProof

                                                                      ↳ QDP

                                                                        ↳ Instantiation

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

IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)
LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))
IF(false, false, s(x1), x2) → IF2(false, s(x1), x2)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF(false, false, s(x1), x2) → IF2(false, s(x1), x2) we obtained the following new rules:

IF(false, false, s(s(z0)), y_0) → IF2(false, s(s(z0)), y_0)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RemovalProof

                        ↳ RemovalProof

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ ForwardInstantiation

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ UsableRulesProof

                                                                  ↳ QDP

                                                                    ↳ QReductionProof

                                                                      ↳ QDP

                                                                        ↳ Instantiation

                                                                          ↳ QDP

                                                                            ↳ Instantiation

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

IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1)
LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))
IF(false, false, s(s(z0)), y_0) → IF2(false, s(s(z0)), y_0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By instantiating [15] the rule IF2(false, s(x0), y1) → LOG2(s(quot(minus(x0, s(0)), s(s(0)))), y1) we obtained the following new rules:

IF2(false, s(s(z0)), z1) → LOG2(s(quot(minus(s(z0), s(0)), s(s(0)))), z1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RemovalProof

                        ↳ RemovalProof

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ ForwardInstantiation

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ UsableRulesProof

                                                                  ↳ QDP

                                                                    ↳ QReductionProof

                                                                      ↳ QDP

                                                                        ↳ Instantiation

                                                                          ↳ QDP

                                                                            ↳ Instantiation

                                                                              ↳ QDP

                                                                                ↳ Rewriting

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

LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))
IF2(false, s(s(z0)), z1) → LOG2(s(quot(minus(s(z0), s(0)), s(s(0)))), z1)
IF(false, false, s(s(z0)), y_0) → IF2(false, s(s(z0)), y_0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule IF2(false, s(s(z0)), z1) → LOG2(s(quot(minus(s(z0), s(0)), s(s(0)))), z1) at position [0,0,0] we obtained the following new rules:

IF2(false, s(s(z0)), z1) → LOG2(s(quot(minus(z0, 0), s(s(0)))), z1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RemovalProof

                        ↳ RemovalProof

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ ForwardInstantiation

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ UsableRulesProof

                                                                  ↳ QDP

                                                                    ↳ QReductionProof

                                                                      ↳ QDP

                                                                        ↳ Instantiation

                                                                          ↳ QDP

                                                                            ↳ Instantiation

                                                                              ↳ QDP

                                                                                ↳ Rewriting

                                                                                  ↳ QDP

                                                                                    ↳ Rewriting

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

LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))
IF(false, false, s(s(z0)), y_0) → IF2(false, s(s(z0)), y_0)
IF2(false, s(s(z0)), z1) → LOG2(s(quot(minus(z0, 0), s(s(0)))), z1)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
By rewriting [15] the rule IF2(false, s(s(z0)), z1) → LOG2(s(quot(minus(z0, 0), s(s(0)))), z1) at position [0,0,0] we obtained the following new rules:

IF2(false, s(s(z0)), z1) → LOG2(s(quot(z0, s(s(0)))), z1)

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RemovalProof

                        ↳ RemovalProof

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ ForwardInstantiation

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ UsableRulesProof

                                                                  ↳ QDP

                                                                    ↳ QReductionProof

                                                                      ↳ QDP

                                                                        ↳ Instantiation

                                                                          ↳ QDP

                                                                            ↳ Instantiation

                                                                              ↳ QDP

                                                                                ↳ Rewriting

                                                                                  ↳ QDP

                                                                                    ↳ Rewriting

                                                                                      ↳ QDP

                                                                                        ↳ QDPOrderProof

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

IF2(false, s(s(z0)), z1) → LOG2(s(quot(z0, s(s(0)))), z1)
LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))
IF(false, false, s(s(z0)), y_0) → IF2(false, s(s(z0)), y_0)

The TRS R consists of the following rules:

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [15].

The following pairs can be oriented strictly and are deleted.

IF2(false, s(s(z0)), z1) → LOG2(s(quot(z0, s(s(0)))), z1)
LOG2(s(s(x0)), y1) → IF(false, false, s(s(x0)), inc(y1))
IF(false, false, s(s(z0)), y_0) → IF2(false, s(s(z0)), y_0)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Combined order from the following AFS and order.
IF2(x1, x2, x3) = IF2(x1, x2)
false = false
s(x1) = s(x1)
LOG2(x1, x2) = LOG2(x1)
quot(x1, x2) = x1
0 = 0
IF(x1, x2, x3, x4) = IF(x2, x3)
inc(x1) = inc
minus(x1, x2) = x1

Recursive path order with status [2].
Quasi-Precedence:

[s₁, LOG2₁, 0, inc] > false > [IF2₂, IF₂]

Status:

IF2₂: [1,2]
false: multiset
inc: multiset
s₁: [1]
IF₂: [2,1]
0: multiset
LOG2₁: [1]

The following usable rules [17] were oriented:

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

↳ QTRS

  ↳ Overlay + Local Confluence

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ DependencyGraphProof

            ↳ AND

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

              ↳ QDP

                ↳ UsableRulesProof

                  ↳ QDP

                    ↳ QReductionProof

                      ↳ QDP

                        ↳ RemovalProof

                        ↳ RemovalProof

                        ↳ Narrowing

                          ↳ QDP

                            ↳ DependencyGraphProof

                              ↳ QDP

                                ↳ Rewriting

                                  ↳ QDP

                                    ↳ UsableRulesProof

                                      ↳ QDP

                                        ↳ Narrowing

                                          ↳ QDP

                                            ↳ DependencyGraphProof

                                              ↳ QDP

                                                ↳ Instantiation

                                                  ↳ QDP

                                                    ↳ ForwardInstantiation

                                                      ↳ QDP

                                                        ↳ Narrowing

                                                          ↳ QDP

                                                            ↳ DependencyGraphProof

                                                              ↳ QDP

                                                                ↳ UsableRulesProof

                                                                  ↳ QDP

                                                                    ↳ QReductionProof

                                                                      ↳ QDP

                                                                        ↳ Instantiation

                                                                          ↳ QDP

                                                                            ↳ Instantiation

                                                                              ↳ QDP

                                                                                ↳ Rewriting

                                                                                  ↳ QDP

                                                                                    ↳ Rewriting

                                                                                      ↳ QDP

                                                                                        ↳ QDPOrderProof

                                                                                          ↳ QDP

                                                                                            ↳ PisEmptyProof

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

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

The set Q consists of the following terms:

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

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.