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:

inc₁(s₁(x)) -> s₁(inc₁(x))
inc₁(0) -> s₁(0)
plus₂(x, y) -> ifPlus₄(eq₂(x, 0), minus₂(x, s₁(0)), x, inc₁(x))
ifPlus₄(false, x, y, z) -> plus₂(x, z)
ifPlus₄(true, x, y, z) -> y
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
minus₂(0, x) -> 0
minus₂(x, 0) -> x
minus₂(x, x) -> 0
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(0, 0) -> true
eq₂(x, x) -> true
times₂(x, y) -> timesIter₃(x, y, 0)
timesIter₃(x, y, z) -> ifTimes₅(eq₂(x, 0), minus₂(x, s₁(0)), y, z, plus₂(y, z))
ifTimes₅(true, x, y, z, u) -> z
ifTimes₅(false, x, y, z, u) -> timesIter₃(x, y, u)
f -> g
f -> h

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

inc₁(s₁(x)) -> s₁(inc₁(x))
inc₁(0) -> s₁(0)
plus₂(x, y) -> ifPlus₄(eq₂(x, 0), minus₂(x, s₁(0)), x, inc₁(x))
ifPlus₄(false, x, y, z) -> plus₂(x, z)
ifPlus₄(true, x, y, z) -> y
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
minus₂(0, x) -> 0
minus₂(x, 0) -> x
minus₂(x, x) -> 0
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(0, 0) -> true
eq₂(x, x) -> true
times₂(x, y) -> timesIter₃(x, y, 0)
timesIter₃(x, y, z) -> ifTimes₅(eq₂(x, 0), minus₂(x, s₁(0)), y, z, plus₂(y, z))
ifTimes₅(true, x, y, z, u) -> z
ifTimes₅(false, x, y, z, u) -> timesIter₃(x, y, u)
f -> g
f -> h

Q is empty.

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

PLUS₂(x, y) -> EQ₂(x, 0)
PLUS₂(x, y) -> MINUS₂(x, s₁(0))
TIMES₂(x, y) -> TIMESITER₃(x, y, 0)
TIMESITER₃(x, y, z) -> PLUS₂(y, z)
TIMESITER₃(x, y, z) -> IFTIMES₅(eq₂(x, 0), minus₂(x, s₁(0)), y, z, plus₂(y, z))
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
PLUS₂(x, y) -> IFPLUS₄(eq₂(x, 0), minus₂(x, s₁(0)), x, inc₁(x))
IFTIMES₅(false, x, y, z, u) -> TIMESITER₃(x, y, u)
TIMESITER₃(x, y, z) -> MINUS₂(x, s₁(0))
IFPLUS₄(false, x, y, z) -> PLUS₂(x, z)
EQ₂(s₁(x), s₁(y)) -> EQ₂(x, y)
TIMESITER₃(x, y, z) -> EQ₂(x, 0)
INC₁(s₁(x)) -> INC₁(x)
PLUS₂(x, y) -> INC₁(x)

The TRS R consists of the following rules:

inc₁(s₁(x)) -> s₁(inc₁(x))
inc₁(0) -> s₁(0)
plus₂(x, y) -> ifPlus₄(eq₂(x, 0), minus₂(x, s₁(0)), x, inc₁(x))
ifPlus₄(false, x, y, z) -> plus₂(x, z)
ifPlus₄(true, x, y, z) -> y
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
minus₂(0, x) -> 0
minus₂(x, 0) -> x
minus₂(x, x) -> 0
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(0, 0) -> true
eq₂(x, x) -> true
times₂(x, y) -> timesIter₃(x, y, 0)
timesIter₃(x, y, z) -> ifTimes₅(eq₂(x, 0), minus₂(x, s₁(0)), y, z, plus₂(y, z))
ifTimes₅(true, x, y, z, u) -> z
ifTimes₅(false, x, y, z, u) -> timesIter₃(x, y, u)
f -> g
f -> h

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

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

PLUS₂(x, y) -> EQ₂(x, 0)
PLUS₂(x, y) -> MINUS₂(x, s₁(0))
TIMES₂(x, y) -> TIMESITER₃(x, y, 0)
TIMESITER₃(x, y, z) -> PLUS₂(y, z)
TIMESITER₃(x, y, z) -> IFTIMES₅(eq₂(x, 0), minus₂(x, s₁(0)), y, z, plus₂(y, z))
MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)
PLUS₂(x, y) -> IFPLUS₄(eq₂(x, 0), minus₂(x, s₁(0)), x, inc₁(x))
IFTIMES₅(false, x, y, z, u) -> TIMESITER₃(x, y, u)
TIMESITER₃(x, y, z) -> MINUS₂(x, s₁(0))
IFPLUS₄(false, x, y, z) -> PLUS₂(x, z)
EQ₂(s₁(x), s₁(y)) -> EQ₂(x, y)
TIMESITER₃(x, y, z) -> EQ₂(x, 0)
INC₁(s₁(x)) -> INC₁(x)
PLUS₂(x, y) -> INC₁(x)

The TRS R consists of the following rules:

inc₁(s₁(x)) -> s₁(inc₁(x))
inc₁(0) -> s₁(0)
plus₂(x, y) -> ifPlus₄(eq₂(x, 0), minus₂(x, s₁(0)), x, inc₁(x))
ifPlus₄(false, x, y, z) -> plus₂(x, z)
ifPlus₄(true, x, y, z) -> y
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
minus₂(0, x) -> 0
minus₂(x, 0) -> x
minus₂(x, x) -> 0
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(0, 0) -> true
eq₂(x, x) -> true
times₂(x, y) -> timesIter₃(x, y, 0)
timesIter₃(x, y, z) -> ifTimes₅(eq₂(x, 0), minus₂(x, s₁(0)), y, z, plus₂(y, z))
ifTimes₅(true, x, y, z, u) -> z
ifTimes₅(false, x, y, z, u) -> timesIter₃(x, y, u)
f -> g
f -> h

Q is empty.
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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

EQ₂(s₁(x), s₁(y)) -> EQ₂(x, y)

The TRS R consists of the following rules:

inc₁(s₁(x)) -> s₁(inc₁(x))
inc₁(0) -> s₁(0)
plus₂(x, y) -> ifPlus₄(eq₂(x, 0), minus₂(x, s₁(0)), x, inc₁(x))
ifPlus₄(false, x, y, z) -> plus₂(x, z)
ifPlus₄(true, x, y, z) -> y
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
minus₂(0, x) -> 0
minus₂(x, 0) -> x
minus₂(x, x) -> 0
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(0, 0) -> true
eq₂(x, x) -> true
times₂(x, y) -> timesIter₃(x, y, 0)
timesIter₃(x, y, z) -> ifTimes₅(eq₂(x, 0), minus₂(x, s₁(0)), y, z, plus₂(y, z))
ifTimes₅(true, x, y, z, u) -> z
ifTimes₅(false, x, y, z, u) -> timesIter₃(x, y, u)
f -> g
f -> h

Q is empty.
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.

EQ₂(s₁(x), s₁(y)) -> EQ₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(EQ₂(x₁, x₂)) = 2·x₂
POL(s₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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:

inc₁(s₁(x)) -> s₁(inc₁(x))
inc₁(0) -> s₁(0)
plus₂(x, y) -> ifPlus₄(eq₂(x, 0), minus₂(x, s₁(0)), x, inc₁(x))
ifPlus₄(false, x, y, z) -> plus₂(x, z)
ifPlus₄(true, x, y, z) -> y
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
minus₂(0, x) -> 0
minus₂(x, 0) -> x
minus₂(x, x) -> 0
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(0, 0) -> true
eq₂(x, x) -> true
times₂(x, y) -> timesIter₃(x, y, 0)
timesIter₃(x, y, z) -> ifTimes₅(eq₂(x, 0), minus₂(x, s₁(0)), y, z, plus₂(y, z))
ifTimes₅(true, x, y, z, u) -> z
ifTimes₅(false, x, y, z, u) -> timesIter₃(x, y, u)
f -> g
f -> h

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

The TRS R consists of the following rules:

inc₁(s₁(x)) -> s₁(inc₁(x))
inc₁(0) -> s₁(0)
plus₂(x, y) -> ifPlus₄(eq₂(x, 0), minus₂(x, s₁(0)), x, inc₁(x))
ifPlus₄(false, x, y, z) -> plus₂(x, z)
ifPlus₄(true, x, y, z) -> y
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
minus₂(0, x) -> 0
minus₂(x, 0) -> x
minus₂(x, x) -> 0
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(0, 0) -> true
eq₂(x, x) -> true
times₂(x, y) -> timesIter₃(x, y, 0)
timesIter₃(x, y, z) -> ifTimes₅(eq₂(x, 0), minus₂(x, s₁(0)), y, z, plus₂(y, z))
ifTimes₅(true, x, y, z, u) -> z
ifTimes₅(false, x, y, z, u) -> timesIter₃(x, y, u)
f -> g
f -> h

Q is empty.
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.

MINUS₂(s₁(x), s₁(y)) -> MINUS₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(MINUS₂(x₁, x₂)) = 2·x₂
POL(s₁(x₁)) = 2 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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:

inc₁(s₁(x)) -> s₁(inc₁(x))
inc₁(0) -> s₁(0)
plus₂(x, y) -> ifPlus₄(eq₂(x, 0), minus₂(x, s₁(0)), x, inc₁(x))
ifPlus₄(false, x, y, z) -> plus₂(x, z)
ifPlus₄(true, x, y, z) -> y
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
minus₂(0, x) -> 0
minus₂(x, 0) -> x
minus₂(x, x) -> 0
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(0, 0) -> true
eq₂(x, x) -> true
times₂(x, y) -> timesIter₃(x, y, 0)
timesIter₃(x, y, z) -> ifTimes₅(eq₂(x, 0), minus₂(x, s₁(0)), y, z, plus₂(y, z))
ifTimes₅(true, x, y, z, u) -> z
ifTimes₅(false, x, y, z, u) -> timesIter₃(x, y, u)
f -> g
f -> h

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

INC₁(s₁(x)) -> INC₁(x)

The TRS R consists of the following rules:

inc₁(s₁(x)) -> s₁(inc₁(x))
inc₁(0) -> s₁(0)
plus₂(x, y) -> ifPlus₄(eq₂(x, 0), minus₂(x, s₁(0)), x, inc₁(x))
ifPlus₄(false, x, y, z) -> plus₂(x, z)
ifPlus₄(true, x, y, z) -> y
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
minus₂(0, x) -> 0
minus₂(x, 0) -> x
minus₂(x, x) -> 0
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(0, 0) -> true
eq₂(x, x) -> true
times₂(x, y) -> timesIter₃(x, y, 0)
timesIter₃(x, y, z) -> ifTimes₅(eq₂(x, 0), minus₂(x, s₁(0)), y, z, plus₂(y, z))
ifTimes₅(true, x, y, z, u) -> z
ifTimes₅(false, x, y, z, u) -> timesIter₃(x, y, u)
f -> g
f -> h

Q is empty.
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.

INC₁(s₁(x)) -> INC₁(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [21]:

POL(INC₁(x₁)) = 2·x₁
POL(s₁(x₁)) = 1 + 2·x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ 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:

inc₁(s₁(x)) -> s₁(inc₁(x))
inc₁(0) -> s₁(0)
plus₂(x, y) -> ifPlus₄(eq₂(x, 0), minus₂(x, s₁(0)), x, inc₁(x))
ifPlus₄(false, x, y, z) -> plus₂(x, z)
ifPlus₄(true, x, y, z) -> y
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
minus₂(0, x) -> 0
minus₂(x, 0) -> x
minus₂(x, x) -> 0
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(0, 0) -> true
eq₂(x, x) -> true
times₂(x, y) -> timesIter₃(x, y, 0)
timesIter₃(x, y, z) -> ifTimes₅(eq₂(x, 0), minus₂(x, s₁(0)), y, z, plus₂(y, z))
ifTimes₅(true, x, y, z, u) -> z
ifTimes₅(false, x, y, z, u) -> timesIter₃(x, y, u)
f -> g
f -> h

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

IFPLUS₄(false, x, y, z) -> PLUS₂(x, z)
PLUS₂(x, y) -> IFPLUS₄(eq₂(x, 0), minus₂(x, s₁(0)), x, inc₁(x))

The TRS R consists of the following rules:

inc₁(s₁(x)) -> s₁(inc₁(x))
inc₁(0) -> s₁(0)
plus₂(x, y) -> ifPlus₄(eq₂(x, 0), minus₂(x, s₁(0)), x, inc₁(x))
ifPlus₄(false, x, y, z) -> plus₂(x, z)
ifPlus₄(true, x, y, z) -> y
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
minus₂(0, x) -> 0
minus₂(x, 0) -> x
minus₂(x, x) -> 0
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(0, 0) -> true
eq₂(x, x) -> true
times₂(x, y) -> timesIter₃(x, y, 0)
timesIter₃(x, y, z) -> ifTimes₅(eq₂(x, 0), minus₂(x, s₁(0)), y, z, plus₂(y, z))
ifTimes₅(true, x, y, z, u) -> z
ifTimes₅(false, x, y, z, u) -> timesIter₃(x, y, u)
f -> g
f -> h

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

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

TIMESITER₃(x, y, z) -> IFTIMES₅(eq₂(x, 0), minus₂(x, s₁(0)), y, z, plus₂(y, z))
IFTIMES₅(false, x, y, z, u) -> TIMESITER₃(x, y, u)

The TRS R consists of the following rules:

inc₁(s₁(x)) -> s₁(inc₁(x))
inc₁(0) -> s₁(0)
plus₂(x, y) -> ifPlus₄(eq₂(x, 0), minus₂(x, s₁(0)), x, inc₁(x))
ifPlus₄(false, x, y, z) -> plus₂(x, z)
ifPlus₄(true, x, y, z) -> y
minus₂(s₁(x), s₁(y)) -> minus₂(x, y)
minus₂(0, x) -> 0
minus₂(x, 0) -> x
minus₂(x, x) -> 0
eq₂(s₁(x), s₁(y)) -> eq₂(x, y)
eq₂(0, s₁(x)) -> false
eq₂(s₁(x), 0) -> false
eq₂(0, 0) -> true
eq₂(x, x) -> true
times₂(x, y) -> timesIter₃(x, y, 0)
timesIter₃(x, y, z) -> ifTimes₅(eq₂(x, 0), minus₂(x, s₁(0)), y, z, plus₂(y, z))
ifTimes₅(true, x, y, z, u) -> z
ifTimes₅(false, x, y, z, u) -> timesIter₃(x, y, u)
f -> g
f -> h

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