Termination w.r.t. Q proof of TRS/TRCSREx1_GM03_FR.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:

p₁(0) -> 0
p₁(s₁(X)) -> X
leq₂(0, Y) -> true
leq₂(s₁(X), 0) -> false
leq₂(s₁(X), s₁(Y)) -> leq₂(X, Y)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
diff₂(X, Y) -> if₃(leq₂(X, Y), n__0, n__s₁(n__diff₂(n__p₁(X), Y)))
0 -> n__0
s₁(X) -> n__s₁(X)
diff₂(X1, X2) -> n__diff₂(X1, X2)
p₁(X) -> n__p₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__diff₂(X1, X2)) -> diff₂(activate₁(X1), activate₁(X2))
activate₁(n__p₁(X)) -> p₁(activate₁(X))
activate₁(X) -> X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

p₁(0) -> 0
p₁(s₁(X)) -> X
leq₂(0, Y) -> true
leq₂(s₁(X), 0) -> false
leq₂(s₁(X), s₁(Y)) -> leq₂(X, Y)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
diff₂(X, Y) -> if₃(leq₂(X, Y), n__0, n__s₁(n__diff₂(n__p₁(X), Y)))
0 -> n__0
s₁(X) -> n__s₁(X)
diff₂(X1, X2) -> n__diff₂(X1, X2)
p₁(X) -> n__p₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__diff₂(X1, X2)) -> diff₂(activate₁(X1), activate₁(X2))
activate₁(n__p₁(X)) -> p₁(activate₁(X))
activate₁(X) -> 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:

ACTIVATE₁(n__0) -> 0¹
DIFF₂(X, Y) -> IF₃(leq₂(X, Y), n__0, n__s₁(n__diff₂(n__p₁(X), Y)))
IF₃(true, X, Y) -> ACTIVATE₁(X)
ACTIVATE₁(n__diff₂(X1, X2)) -> DIFF₂(activate₁(X1), activate₁(X2))
DIFF₂(X, Y) -> LEQ₂(X, Y)
ACTIVATE₁(n__diff₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__s₁(X)) -> S₁(activate₁(X))
ACTIVATE₁(n__p₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__diff₂(X1, X2)) -> ACTIVATE₁(X2)
LEQ₂(s₁(X), s₁(Y)) -> LEQ₂(X, Y)
IF₃(false, X, Y) -> ACTIVATE₁(Y)
ACTIVATE₁(n__p₁(X)) -> P₁(activate₁(X))
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

p₁(0) -> 0
p₁(s₁(X)) -> X
leq₂(0, Y) -> true
leq₂(s₁(X), 0) -> false
leq₂(s₁(X), s₁(Y)) -> leq₂(X, Y)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
diff₂(X, Y) -> if₃(leq₂(X, Y), n__0, n__s₁(n__diff₂(n__p₁(X), Y)))
0 -> n__0
s₁(X) -> n__s₁(X)
diff₂(X1, X2) -> n__diff₂(X1, X2)
p₁(X) -> n__p₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__diff₂(X1, X2)) -> diff₂(activate₁(X1), activate₁(X2))
activate₁(n__p₁(X)) -> p₁(activate₁(X))
activate₁(X) -> 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:

ACTIVATE₁(n__0) -> 0¹
DIFF₂(X, Y) -> IF₃(leq₂(X, Y), n__0, n__s₁(n__diff₂(n__p₁(X), Y)))
IF₃(true, X, Y) -> ACTIVATE₁(X)
ACTIVATE₁(n__diff₂(X1, X2)) -> DIFF₂(activate₁(X1), activate₁(X2))
DIFF₂(X, Y) -> LEQ₂(X, Y)
ACTIVATE₁(n__diff₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__s₁(X)) -> S₁(activate₁(X))
ACTIVATE₁(n__p₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__diff₂(X1, X2)) -> ACTIVATE₁(X2)
LEQ₂(s₁(X), s₁(Y)) -> LEQ₂(X, Y)
IF₃(false, X, Y) -> ACTIVATE₁(Y)
ACTIVATE₁(n__p₁(X)) -> P₁(activate₁(X))
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

p₁(0) -> 0
p₁(s₁(X)) -> X
leq₂(0, Y) -> true
leq₂(s₁(X), 0) -> false
leq₂(s₁(X), s₁(Y)) -> leq₂(X, Y)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
diff₂(X, Y) -> if₃(leq₂(X, Y), n__0, n__s₁(n__diff₂(n__p₁(X), Y)))
0 -> n__0
s₁(X) -> n__s₁(X)
diff₂(X1, X2) -> n__diff₂(X1, X2)
p₁(X) -> n__p₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__diff₂(X1, X2)) -> diff₂(activate₁(X1), activate₁(X2))
activate₁(n__p₁(X)) -> p₁(activate₁(X))
activate₁(X) -> X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

LEQ₂(s₁(X), s₁(Y)) -> LEQ₂(X, Y)

The TRS R consists of the following rules:

p₁(0) -> 0
p₁(s₁(X)) -> X
leq₂(0, Y) -> true
leq₂(s₁(X), 0) -> false
leq₂(s₁(X), s₁(Y)) -> leq₂(X, Y)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
diff₂(X, Y) -> if₃(leq₂(X, Y), n__0, n__s₁(n__diff₂(n__p₁(X), Y)))
0 -> n__0
s₁(X) -> n__s₁(X)
diff₂(X1, X2) -> n__diff₂(X1, X2)
p₁(X) -> n__p₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__diff₂(X1, X2)) -> diff₂(activate₁(X1), activate₁(X2))
activate₁(n__p₁(X)) -> p₁(activate₁(X))
activate₁(X) -> 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.

LEQ₂(s₁(X), s₁(Y)) -> LEQ₂(X, Y)

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

POL(LEQ₂(x₁, x₂)) = x₁·x₂
POL(s₁(x₁)) = 1 + x₁

The following usable rules [14] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

p₁(0) -> 0
p₁(s₁(X)) -> X
leq₂(0, Y) -> true
leq₂(s₁(X), 0) -> false
leq₂(s₁(X), s₁(Y)) -> leq₂(X, Y)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
diff₂(X, Y) -> if₃(leq₂(X, Y), n__0, n__s₁(n__diff₂(n__p₁(X), Y)))
0 -> n__0
s₁(X) -> n__s₁(X)
diff₂(X1, X2) -> n__diff₂(X1, X2)
p₁(X) -> n__p₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__diff₂(X1, X2)) -> diff₂(activate₁(X1), activate₁(X2))
activate₁(n__p₁(X)) -> p₁(activate₁(X))
activate₁(X) -> 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

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

ACTIVATE₁(n__diff₂(X1, X2)) -> ACTIVATE₁(X1)
ACTIVATE₁(n__p₁(X)) -> ACTIVATE₁(X)
ACTIVATE₁(n__diff₂(X1, X2)) -> ACTIVATE₁(X2)
DIFF₂(X, Y) -> IF₃(leq₂(X, Y), n__0, n__s₁(n__diff₂(n__p₁(X), Y)))
IF₃(false, X, Y) -> ACTIVATE₁(Y)
IF₃(true, X, Y) -> ACTIVATE₁(X)
ACTIVATE₁(n__diff₂(X1, X2)) -> DIFF₂(activate₁(X1), activate₁(X2))
ACTIVATE₁(n__s₁(X)) -> ACTIVATE₁(X)

The TRS R consists of the following rules:

p₁(0) -> 0
p₁(s₁(X)) -> X
leq₂(0, Y) -> true
leq₂(s₁(X), 0) -> false
leq₂(s₁(X), s₁(Y)) -> leq₂(X, Y)
if₃(true, X, Y) -> activate₁(X)
if₃(false, X, Y) -> activate₁(Y)
diff₂(X, Y) -> if₃(leq₂(X, Y), n__0, n__s₁(n__diff₂(n__p₁(X), Y)))
0 -> n__0
s₁(X) -> n__s₁(X)
diff₂(X1, X2) -> n__diff₂(X1, X2)
p₁(X) -> n__p₁(X)
activate₁(n__0) -> 0
activate₁(n__s₁(X)) -> s₁(activate₁(X))
activate₁(n__diff₂(X1, X2)) -> diff₂(activate₁(X1), activate₁(X2))
activate₁(n__p₁(X)) -> p₁(activate₁(X))
activate₁(X) -> X

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