Termination w.r.t. Q proof of TRS/AG01#3.5a.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:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if_mod₃(le₂(y, x), s₁(x), s₁(y))
if_mod₃(true, s₁(x), s₁(y)) -> mod₂(minus₂(x, y), s₁(y))
if_mod₃(false, s₁(x), s₁(y)) -> s₁(x)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if_mod₃(le₂(y, x), s₁(x), s₁(y))
if_mod₃(true, s₁(x), s₁(y)) -> mod₂(minus₂(x, y), s₁(y))
if_mod₃(false, s₁(x), s₁(y)) -> s₁(x)

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:

MINUS₂(x, s₁(y)) -> PRED₁(minus₂(x, y))
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
IF_MOD₃(true, s₁(x), s₁(y)) -> MINUS₂(x, y)
MOD₂(s₁(x), s₁(y)) -> IF_MOD₃(le₂(y, x), s₁(x), s₁(y))
MINUS₂(x, s₁(y)) -> MINUS₂(x, y)
IF_MOD₃(true, s₁(x), s₁(y)) -> MOD₂(minus₂(x, y), s₁(y))
MOD₂(s₁(x), s₁(y)) -> LE₂(y, x)

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if_mod₃(le₂(y, x), s₁(x), s₁(y))
if_mod₃(true, s₁(x), s₁(y)) -> mod₂(minus₂(x, y), s₁(y))
if_mod₃(false, s₁(x), s₁(y)) -> s₁(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:

MINUS₂(x, s₁(y)) -> PRED₁(minus₂(x, y))
LE₂(s₁(x), s₁(y)) -> LE₂(x, y)
IF_MOD₃(true, s₁(x), s₁(y)) -> MINUS₂(x, y)
MOD₂(s₁(x), s₁(y)) -> IF_MOD₃(le₂(y, x), s₁(x), s₁(y))
MINUS₂(x, s₁(y)) -> MINUS₂(x, y)
IF_MOD₃(true, s₁(x), s₁(y)) -> MOD₂(minus₂(x, y), s₁(y))
MOD₂(s₁(x), s₁(y)) -> LE₂(y, x)

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if_mod₃(le₂(y, x), s₁(x), s₁(y))
if_mod₃(true, s₁(x), s₁(y)) -> mod₂(minus₂(x, y), s₁(y))
if_mod₃(false, s₁(x), s₁(y)) -> s₁(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 3 SCCs with 3 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

MINUS₂(x, s₁(y)) -> MINUS₂(x, y)

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if_mod₃(le₂(y, x), s₁(x), s₁(y))
if_mod₃(true, s₁(x), s₁(y)) -> mod₂(minus₂(x, y), s₁(y))
if_mod₃(false, s₁(x), s₁(y)) -> s₁(x)

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(x, s₁(y)) -> MINUS₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( MINUS₂(x₁, x₂) ) = 2x₁ + 2x₂ + 2

POL( s₁(x₁) ) = 3x₁ + 1

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if_mod₃(le₂(y, x), s₁(x), s₁(y))
if_mod₃(true, s₁(x), s₁(y)) -> mod₂(minus₂(x, y), s₁(y))
if_mod₃(false, s₁(x), s₁(y)) -> s₁(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

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

LE₂(s₁(x), s₁(y)) -> LE₂(x, y)

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if_mod₃(le₂(y, x), s₁(x), s₁(y))
if_mod₃(true, s₁(x), s₁(y)) -> mod₂(minus₂(x, y), s₁(y))
if_mod₃(false, s₁(x), s₁(y)) -> s₁(x)

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.

LE₂(s₁(x), s₁(y)) -> LE₂(x, y)

The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial Order [17,21] with Interpretation:

POL( LE₂(x₁, x₂) ) = 3x₂ + 3

POL( s₁(x₁) ) = x₁ + 3

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if_mod₃(le₂(y, x), s₁(x), s₁(y))
if_mod₃(true, s₁(x), s₁(y)) -> mod₂(minus₂(x, y), s₁(y))
if_mod₃(false, s₁(x), s₁(y)) -> s₁(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

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

MOD₂(s₁(x), s₁(y)) -> IF_MOD₃(le₂(y, x), s₁(x), s₁(y))
IF_MOD₃(true, s₁(x), s₁(y)) -> MOD₂(minus₂(x, y), s₁(y))

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if_mod₃(le₂(y, x), s₁(x), s₁(y))
if_mod₃(true, s₁(x), s₁(y)) -> mod₂(minus₂(x, y), s₁(y))
if_mod₃(false, s₁(x), s₁(y)) -> s₁(x)

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.

IF_MOD₃(true, s₁(x), s₁(y)) -> MOD₂(minus₂(x, y), s₁(y))

The remaining pairs can at least be oriented weakly.

MOD₂(s₁(x), s₁(y)) -> IF_MOD₃(le₂(y, x), s₁(x), s₁(y))

Used ordering: Polynomial Order [17,21] with Interpretation:

POL( minus₂(x₁, x₂) ) = x₁

POL( IF_MOD₃(x₁, ..., x₃) ) = max{0, 2x₁ + x₂ - 1}

POL( true ) = 1

POL( false ) = max{0, -1}

POL( 0 ) = 1

POL( s₁(x₁) ) = 2x₁ + 1

POL( le₂(x₁, x₂) ) = x₂ + 1

POL( pred₁(x₁) ) = x₁

POL( MOD₂(x₁, x₂) ) = 2x₁

The following usable rules [14] were oriented:

minus₂(x, 0) -> x
le₂(s₁(x), 0) -> false
pred₁(s₁(x)) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
le₂(0, y) -> true
le₂(s₁(x), s₁(y)) -> le₂(x, y)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

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

MOD₂(s₁(x), s₁(y)) -> IF_MOD₃(le₂(y, x), s₁(x), s₁(y))

The TRS R consists of the following rules:

le₂(0, y) -> true
le₂(s₁(x), 0) -> false
le₂(s₁(x), s₁(y)) -> le₂(x, y)
pred₁(s₁(x)) -> x
minus₂(x, 0) -> x
minus₂(x, s₁(y)) -> pred₁(minus₂(x, y))
mod₂(0, y) -> 0
mod₂(s₁(x), 0) -> 0
mod₂(s₁(x), s₁(y)) -> if_mod₃(le₂(y, x), s₁(x), s₁(y))
if_mod₃(true, s₁(x), s₁(y)) -> mod₂(minus₂(x, y), s₁(y))
if_mod₃(false, s₁(x), s₁(y)) -> s₁(x)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 0 SCCs with 1 less node.