Termination w.r.t. Q proof of TRS/secret06times.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:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
f → g
f → h

Q is empty.

↳ QTRS

  ↳ AAECC Innermost

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

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
f → g
f → h

Q is empty.

We have applied [15,7] to switch to innermost. The TRS R 1 is

inc(0) → s(0)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
inc(s(x)) → s(inc(x))
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))

The TRS R 2 is

f → g
f → h

The signature Sigma is {f, g, h}

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

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

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
f → g
f → h

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f

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

PLUS(x, y) → EQ(x, 0)
PLUS(x, y) → MINUS(x, s(0))
TIMES(x, y) → TIMESITER(x, y, 0)
TIMESITER(x, y, z) → PLUS(y, z)
TIMESITER(x, y, z) → IFTIMES(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
MINUS(s(x), s(y)) → MINUS(x, y)
PLUS(x, y) → IFPLUS(eq(x, 0), minus(x, s(0)), x, inc(x))
IFTIMES(false, x, y, z, u) → TIMESITER(x, y, u)
TIMESITER(x, y, z) → MINUS(x, s(0))
IFPLUS(false, x, y, z) → PLUS(x, z)
EQ(s(x), s(y)) → EQ(x, y)
TIMESITER(x, y, z) → EQ(x, 0)
INC(s(x)) → INC(x)
PLUS(x, y) → INC(x)

The TRS R consists of the following rules:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
f → g
f → h

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

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

PLUS(x, y) → EQ(x, 0)
PLUS(x, y) → MINUS(x, s(0))
TIMES(x, y) → TIMESITER(x, y, 0)
TIMESITER(x, y, z) → PLUS(y, z)
TIMESITER(x, y, z) → IFTIMES(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
MINUS(s(x), s(y)) → MINUS(x, y)
PLUS(x, y) → IFPLUS(eq(x, 0), minus(x, s(0)), x, inc(x))
IFTIMES(false, x, y, z, u) → TIMESITER(x, y, u)
TIMESITER(x, y, z) → MINUS(x, s(0))
IFPLUS(false, x, y, z) → PLUS(x, z)
EQ(s(x), s(y)) → EQ(x, y)
TIMESITER(x, y, z) → EQ(x, 0)
INC(s(x)) → INC(x)
PLUS(x, y) → INC(x)

The TRS R consists of the following rules:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
f → g
f → h

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f

We have to consider all minimal (P,Q,R)-chains.
We deleted some edges using various graph approximations

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

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

PLUS(x, y) → MINUS(x, s(0))
PLUS(x, y) → EQ(x, 0)
TIMES(x, y) → TIMESITER(x, y, 0)
TIMESITER(x, y, z) → PLUS(y, z)
TIMESITER(x, y, z) → IFTIMES(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
MINUS(s(x), s(y)) → MINUS(x, y)
PLUS(x, y) → IFPLUS(eq(x, 0), minus(x, s(0)), x, inc(x))
IFTIMES(false, x, y, z, u) → TIMESITER(x, y, u)
TIMESITER(x, y, z) → MINUS(x, s(0))
IFPLUS(false, x, y, z) → PLUS(x, z)
EQ(s(x), s(y)) → EQ(x, y)
TIMESITER(x, y, z) → EQ(x, 0)
INC(s(x)) → INC(x)
PLUS(x, y) → INC(x)

The TRS R consists of the following rules:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
f → g
f → h

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 5 SCCs with 7 less nodes.

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

EQ(s(x), s(y)) → EQ(x, y)

The TRS R consists of the following rules:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
f → g
f → h

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f

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.

EQ(s(x), s(y)) → EQ(x, y)

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

Recursive path order with status [2].
Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ AAECC Innermost

    ↳ 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:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
f → g
f → h

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ 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:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
f → g
f → h

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f

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) = x2
s(x1) = s(x1)

Recursive path order with status [2].
Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ AAECC Innermost

    ↳ 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:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
f → g
f → h

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                    ↳ QDPOrderProof

                  ↳ 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:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
f → g
f → h

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f

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) = x1
s(x1) = s(x1)

Recursive path order with status [2].
Precedence:

trivial

Status:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ AAECC Innermost

    ↳ 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:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
f → g
f → h

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

IFPLUS(false, x, y, z) → PLUS(x, z)
PLUS(x, y) → IFPLUS(eq(x, 0), minus(x, s(0)), x, inc(x))

The TRS R consists of the following rules:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
f → g
f → h

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f

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

↳ QTRS

  ↳ AAECC Innermost

    ↳ QTRS

      ↳ DependencyPairsProof

        ↳ QDP

          ↳ EdgeDeletionProof

            ↳ QDP

              ↳ DependencyGraphProof

                ↳ AND

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

                  ↳ QDP

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

TIMESITER(x, y, z) → IFTIMES(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
IFTIMES(false, x, y, z, u) → TIMESITER(x, y, u)

The TRS R consists of the following rules:

inc(s(x)) → s(inc(x))
inc(0) → s(0)
plus(x, y) → ifPlus(eq(x, 0), minus(x, s(0)), x, inc(x))
ifPlus(false, x, y, z) → plus(x, z)
ifPlus(true, x, y, z) → y
minus(s(x), s(y)) → minus(x, y)
minus(0, x) → 0
minus(x, 0) → x
minus(x, x) → 0
eq(s(x), s(y)) → eq(x, y)
eq(0, s(x)) → false
eq(s(x), 0) → false
eq(0, 0) → true
eq(x, x) → true
times(x, y) → timesIter(x, y, 0)
timesIter(x, y, z) → ifTimes(eq(x, 0), minus(x, s(0)), y, z, plus(y, z))
ifTimes(true, x, y, z, u) → z
ifTimes(false, x, y, z, u) → timesIter(x, y, u)
f → g
f → h

The set Q consists of the following terms:

inc(s(x0))
inc(0)
plus(x0, x1)
ifPlus(false, x0, x1, x2)
ifPlus(true, x0, x1, x2)
minus(s(x0), s(x1))
minus(0, x0)
minus(x0, 0)
minus(x0, x0)
eq(s(x0), s(x1))
eq(0, s(x0))
eq(s(x0), 0)
eq(x0, x0)
times(x0, x1)
timesIter(x0, x1, x2)
ifTimes(true, x0, x1, x2, x3)
ifTimes(false, x0, x1, x2, x3)
f

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