Termination w.r.t. Q proof of /home/nowonder/forschung/aprove/satrung/aprove/lpar16/AG01SLASHHASH3.5a.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

  ↳ Non-Overlap Check

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.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
pred₁(s₁(x0))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(x1))
if_mod₃(true, s₁(x0), s₁(x1))
if_mod₃(false, s₁(x0), s₁(x1))

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)

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
pred₁(s₁(x0))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(x1))
if_mod₃(true, s₁(x0), s₁(x1))
if_mod₃(false, s₁(x0), s₁(x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
pred₁(s₁(x0))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(x1))
if_mod₃(true, s₁(x0), s₁(x1))
if_mod₃(false, s₁(x0), s₁(x1))

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

  ↳ Non-Overlap Check

    ↳ 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)

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
pred₁(s₁(x0))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(x1))
if_mod₃(true, s₁(x0), s₁(x1))
if_mod₃(false, s₁(x0), s₁(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

MINUS₂(x, s₁(y)) -> MINUS₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MINUS₂(x1, x2) = MINUS₁(x2)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

[MINUS₁, s_1]

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
pred₁(s₁(x0))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(x1))
if_mod₃(true, s₁(x0), s₁(x1))
if_mod₃(false, s₁(x0), s₁(x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
pred₁(s₁(x0))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(x1))
if_mod₃(true, s₁(x0), s₁(x1))
if_mod₃(false, s₁(x0), s₁(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented and are deleted.

LE₂(s₁(x), s₁(y)) -> LE₂(x, y)

The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
LE₂(x1, x2) = LE₁(x1)
s₁(x1) = s₁(x1)

Lexicographic Path Order [19].
Precedence:

trivial

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
pred₁(s₁(x0))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(x1))
if_mod₃(true, s₁(x0), s₁(x1))
if_mod₃(false, s₁(x0), s₁(x1))

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
pred₁(s₁(x0))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(x1))
if_mod₃(true, s₁(x0), s₁(x1))
if_mod₃(false, s₁(x0), s₁(x1))

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].

The following pairs can be strictly oriented 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 by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
MOD₂(x1, x2) = MOD₁(x1)
s₁(x1) = s₁(x1)
IF_MOD₃(x1, x2, x3) = IF_MOD₂(x1, x2)
le₂(x1, x2) = x2
true = true
minus₂(x1, x2) = minus₁(x1)
0 = 0
false = false
pred₁(x1) = x1

Lexicographic Path Order [19].
Precedence:

s₁ > [MOD₁, IF_{MOD_2]} > true
s₁ > minus₁ > true
s₁ > [0, false] > true

The following usable rules [14] were oriented:

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

↳ QTRS

  ↳ Non-Overlap Check

    ↳ 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)
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)

The set Q consists of the following terms:

le₂(0, x0)
le₂(s₁(x0), 0)
le₂(s₁(x0), s₁(x1))
pred₁(s₁(x0))
minus₂(x0, 0)
minus₂(x0, s₁(x1))
mod₂(0, x0)
mod₂(s₁(x0), 0)
mod₂(s₁(x0), s₁(x1))
if_mod₃(true, s₁(x0), s₁(x1))
if_mod₃(false, s₁(x0), s₁(x1))

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