Termination w.r.t. Q proof of /tmp/tmp4A-57y/thiemann11.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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

Q is empty.

↳ 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)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

Q is empty.

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

IF3(true, x, y) → MINUS(x, y)
MOD(x, y) → ID(y)
ID(s(x)) → ID(x)
MOD(x, y) → ZERO(x)
LE(s(x), s(y)) → LE(x, y)
MOD(x, y) → LE(y, x)
IF3(true, x, y) → MOD(minus(x, y), s(y))
MINUS(s(x), s(y)) → MINUS(x, y)
IF2(false, b2, x, y) → IF3(b2, x, y)
IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y)
MOD(x, y) → ID(x)
MOD(x, y) → IF_MOD(zero(x), zero(y), le(y, x), id(x), id(y))
MOD(x, y) → ZERO(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)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

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:

IF3(true, x, y) → MINUS(x, y)
MOD(x, y) → ID(y)
ID(s(x)) → ID(x)
MOD(x, y) → ZERO(x)
LE(s(x), s(y)) → LE(x, y)
MOD(x, y) → LE(y, x)
IF3(true, x, y) → MOD(minus(x, y), s(y))
MINUS(s(x), s(y)) → MINUS(x, y)
IF2(false, b2, x, y) → IF3(b2, x, y)
IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y)
MOD(x, y) → ID(x)
MOD(x, y) → IF_MOD(zero(x), zero(y), le(y, x), id(x), id(y))
MOD(x, y) → ZERO(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)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 4 SCCs with 6 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ 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:

le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

Q is empty.
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.

MINUS(s(x), s(y)) → MINUS(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(MINUS(x₁, x₂)) = (3)x_2
POL(s(x₁)) = 4 + (2)x_1

The value of delta used in the strict ordering is 12.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
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)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

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

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

ID(s(x)) → ID(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)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

Q is empty.
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.

ID(s(x)) → ID(x)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(ID(x₁)) = (4)x_1
POL(s(x₁)) = 1 + (4)x_1

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

Q DP problem:
P is empty.
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)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

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

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)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

Q is empty.
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.

LE(s(x), s(y)) → LE(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x₁)) = 4 + (2)x_1
POL(LE(x₁, x₂)) = (3)x_2

The value of delta used in the strict ordering is 12.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

Q DP problem:
P is empty.
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)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

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

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

IF3(true, x, y) → MOD(minus(x, y), s(y))
IF2(false, b2, x, y) → IF3(b2, x, y)
IF_MOD(false, b1, b2, x, y) → IF2(b1, b2, x, y)
MOD(x, y) → IF_MOD(zero(x), zero(y), le(y, x), id(x), id(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)
zero(0) → true
zero(s(x)) → false
id(0) → 0
id(s(x)) → s(id(x))
minus(x, 0) → x
minus(s(x), s(y)) → minus(x, y)
mod(x, y) → if_mod(zero(x), zero(y), le(y, x), id(x), id(y))
if_mod(true, b1, b2, x, y) → 0
if_mod(false, b1, b2, x, y) → if2(b1, b2, x, y)
if2(true, b2, x, y) → 0
if2(false, b2, x, y) → if3(b2, x, y)
if3(true, x, y) → mod(minus(x, y), s(y))
if3(false, x, y) → x

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