Termination w.r.t. Q proof of TRS/AG01#3.5b.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)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(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)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(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(s₁(x), y) -> IF_MINUS₃(le₂(s₁(x), y), s₁(x), y)
MINUS₂(s₁(x), y) -> LE₂(s₁(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))
IF_MINUS₃(false, s₁(x), 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)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(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(s₁(x), y) -> IF_MINUS₃(le₂(s₁(x), y), s₁(x), y)
MINUS₂(s₁(x), y) -> LE₂(s₁(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))
IF_MINUS₃(false, s₁(x), 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)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(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:

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)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(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 interpretation [21]:

POL(LE₂(x₁, x₂)) = 2·x₁ + 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

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)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(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:

MINUS₂(s₁(x), y) -> IF_MINUS₃(le₂(s₁(x), y), s₁(x), y)
IF_MINUS₃(false, s₁(x), 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)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(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_MINUS₃(false, s₁(x), y) -> MINUS₂(x, y)

The remaining pairs can at least be oriented weakly.

MINUS₂(s₁(x), y) -> IF_MINUS₃(le₂(s₁(x), y), s₁(x), y)

Used ordering: Polynomial interpretation [21]:

POL(0) = 2
POL(IF_MINUS₃(x₁, x₂, x₃)) = 2 + x₂ + x₃
POL(MINUS₂(x₁, x₂)) = 1 + 2·x₁ + x₂
POL(false) = 2
POL(le₂(x₁, x₂)) = 2 + x₁
POL(s₁(x₁)) = 1 + 2·x₁
POL(true) = 2

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ DependencyGraphProof

          ↳ QDP

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

MINUS₂(s₁(x), y) -> IF_MINUS₃(le₂(s₁(x), y), s₁(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)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(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.

↳ 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)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(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.

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

POL(0) = 1
POL(IF_MOD₃(x₁, x₂, x₃)) = 1 + 2·x₂ + 2·x₃
POL(MOD₂(x₁, x₂)) = 2 + 2·x₁ + 2·x₂
POL(false) = 2
POL(if_minus₃(x₁, x₂, x₃)) = x₂
POL(le₂(x₁, x₂)) = 2 + x₁ + x₂
POL(minus₂(x₁, x₂)) = x₁
POL(s₁(x₁)) = 2 + x₁
POL(true) = 1

The following usable rules [14] were oriented:

if_minus₃(true, s₁(x), y) -> 0
minus₂(0, y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(minus₂(x, y))
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

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)
minus₂(0, y) -> 0
minus₂(s₁(x), y) -> if_minus₃(le₂(s₁(x), y), s₁(x), y)
if_minus₃(true, s₁(x), y) -> 0
if_minus₃(false, s₁(x), y) -> s₁(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.