Termination w.r.t. Q proof of /tmp/tmpf9MGGh/Ex1_GM03

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:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
diff(X, Y) → if(leq(X, Y), n__0, n__s(n__diff(n__p(X), Y)))
0 → n__0
s(X) → n__s(X)
diff(X1, X2) → n__diff(X1, X2)
p(X) → n__p(X)
activate(n__0) → 0
activate(n__s(X)) → s(activate(X))
activate(n__diff(X1, X2)) → diff(activate(X1), activate(X2))
activate(n__p(X)) → p(activate(X))
activate(X) → X

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
diff(X, Y) → if(leq(X, Y), n__0, n__s(n__diff(n__p(X), Y)))
0 → n__0
s(X) → n__s(X)
diff(X1, X2) → n__diff(X1, X2)
p(X) → n__p(X)
activate(n__0) → 0
activate(n__s(X)) → s(activate(X))
activate(n__diff(X1, X2)) → diff(activate(X1), activate(X2))
activate(n__p(X)) → p(activate(X))
activate(X) → 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:

ACTIVATE(n__s(X)) → S(activate(X))
IF(true, X, Y) → ACTIVATE(X)
DIFF(X, Y) → LEQ(X, Y)
ACTIVATE(n__diff(X1, X2)) → DIFF(activate(X1), activate(X2))
IF(false, X, Y) → ACTIVATE(Y)
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__diff(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__p(X)) → P(activate(X))
ACTIVATE(n__p(X)) → ACTIVATE(X)
ACTIVATE(n__0) → 0¹
LEQ(s(X), s(Y)) → LEQ(X, Y)
ACTIVATE(n__diff(X1, X2)) → ACTIVATE(X2)
DIFF(X, Y) → IF(leq(X, Y), n__0, n__s(n__diff(n__p(X), Y)))

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
diff(X, Y) → if(leq(X, Y), n__0, n__s(n__diff(n__p(X), Y)))
0 → n__0
s(X) → n__s(X)
diff(X1, X2) → n__diff(X1, X2)
p(X) → n__p(X)
activate(n__0) → 0
activate(n__s(X)) → s(activate(X))
activate(n__diff(X1, X2)) → diff(activate(X1), activate(X2))
activate(n__p(X)) → p(activate(X))
activate(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:

ACTIVATE(n__s(X)) → S(activate(X))
IF(true, X, Y) → ACTIVATE(X)
DIFF(X, Y) → LEQ(X, Y)
ACTIVATE(n__diff(X1, X2)) → DIFF(activate(X1), activate(X2))
IF(false, X, Y) → ACTIVATE(Y)
ACTIVATE(n__s(X)) → ACTIVATE(X)
ACTIVATE(n__diff(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__p(X)) → P(activate(X))
ACTIVATE(n__p(X)) → ACTIVATE(X)
ACTIVATE(n__0) → 0¹
LEQ(s(X), s(Y)) → LEQ(X, Y)
ACTIVATE(n__diff(X1, X2)) → ACTIVATE(X2)
DIFF(X, Y) → IF(leq(X, Y), n__0, n__s(n__diff(n__p(X), Y)))

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
diff(X, Y) → if(leq(X, Y), n__0, n__s(n__diff(n__p(X), Y)))
0 → n__0
s(X) → n__s(X)
diff(X1, X2) → n__diff(X1, X2)
p(X) → n__p(X)
activate(n__0) → 0
activate(n__s(X)) → s(activate(X))
activate(n__diff(X1, X2)) → diff(activate(X1), activate(X2))
activate(n__p(X)) → p(activate(X))
activate(X) → X

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] 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:

LEQ(s(X), s(Y)) → LEQ(X, Y)

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
diff(X, Y) → if(leq(X, Y), n__0, n__s(n__diff(n__p(X), Y)))
0 → n__0
s(X) → n__s(X)
diff(X1, X2) → n__diff(X1, X2)
p(X) → n__p(X)
activate(n__0) → 0
activate(n__s(X)) → s(activate(X))
activate(n__diff(X1, X2)) → diff(activate(X1), activate(X2))
activate(n__p(X)) → p(activate(X))
activate(X) → 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.

LEQ(s(X), s(Y)) → LEQ(X, Y)

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

POL(LEQ(x₁, x₂)) = (3)x_2
POL(s(x₁)) = 1 + (4)x_1

The value of delta used in the strict ordering is 3.
The following usable rules [17] 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:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
diff(X, Y) → if(leq(X, Y), n__0, n__s(n__diff(n__p(X), Y)))
0 → n__0
s(X) → n__s(X)
diff(X1, X2) → n__diff(X1, X2)
p(X) → n__p(X)
activate(n__0) → 0
activate(n__s(X)) → s(activate(X))
activate(n__diff(X1, X2)) → diff(activate(X1), activate(X2))
activate(n__p(X)) → p(activate(X))
activate(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

            ↳ QDPOrderProof

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

ACTIVATE(n__p(X)) → ACTIVATE(X)
ACTIVATE(n__diff(X1, X2)) → ACTIVATE(X2)
IF(true, X, Y) → ACTIVATE(X)
ACTIVATE(n__diff(X1, X2)) → DIFF(activate(X1), activate(X2))
DIFF(X, Y) → IF(leq(X, Y), n__0, n__s(n__diff(n__p(X), Y)))
IF(false, X, Y) → ACTIVATE(Y)
ACTIVATE(n__diff(X1, X2)) → ACTIVATE(X1)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
diff(X, Y) → if(leq(X, Y), n__0, n__s(n__diff(n__p(X), Y)))
0 → n__0
s(X) → n__s(X)
diff(X1, X2) → n__diff(X1, X2)
p(X) → n__p(X)
activate(n__0) → 0
activate(n__s(X)) → s(activate(X))
activate(n__diff(X1, X2)) → diff(activate(X1), activate(X2))
activate(n__p(X)) → p(activate(X))
activate(X) → 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.

ACTIVATE(n__diff(X1, X2)) → ACTIVATE(X2)
ACTIVATE(n__diff(X1, X2)) → ACTIVATE(X1)

The remaining pairs can at least be oriented weakly.

ACTIVATE(n__p(X)) → ACTIVATE(X)
IF(true, X, Y) → ACTIVATE(X)
ACTIVATE(n__diff(X1, X2)) → DIFF(activate(X1), activate(X2))
DIFF(X, Y) → IF(leq(X, Y), n__0, n__s(n__diff(n__p(X), Y)))
IF(false, X, Y) → ACTIVATE(Y)
ACTIVATE(n__s(X)) → ACTIVATE(X)

Used ordering: Polynomial interpretation [25,35]:

POL(DIFF(x₁, x₂)) = 4 + (4)x_1 + (4)x_2
POL(n__p(x₁)) = x_1
POL(true) = 0
POL(activate(x₁)) = x_1
POL(p(x₁)) = x_1
POL(n__s(x₁)) = x_1
POL(IF(x₁, x₂, x₃)) = (4)x_2 + (4)x_3
POL(0) = 0
POL(if(x₁, x₂, x₃)) = x_2 + x_3
POL(diff(x₁, x₂)) = 1 + x_1 + x_2
POL(n__0) = 0
POL(false) = 0
POL(s(x₁)) = x_1
POL(ACTIVATE(x₁)) = (4)x_1
POL(leq(x₁, x₂)) = x_2
POL(n__diff(x₁, x₂)) = 1 + x_1 + x_2

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented:

p(0) → 0
p(s(X)) → X
s(X) → n__s(X)
0 → n__0
p(X) → n__p(X)
diff(X1, X2) → n__diff(X1, X2)
activate(n__s(X)) → s(activate(X))
activate(n__0) → 0
activate(n__p(X)) → p(activate(X))
if(false, X, Y) → activate(Y)
activate(n__diff(X1, X2)) → diff(activate(X1), activate(X2))
diff(X, Y) → if(leq(X, Y), n__0, n__s(n__diff(n__p(X), Y)))
if(true, X, Y) → activate(X)
activate(X) → X

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

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

ACTIVATE(n__p(X)) → ACTIVATE(X)
IF(true, X, Y) → ACTIVATE(X)
ACTIVATE(n__diff(X1, X2)) → DIFF(activate(X1), activate(X2))
DIFF(X, Y) → IF(leq(X, Y), n__0, n__s(n__diff(n__p(X), Y)))
IF(false, X, Y) → ACTIVATE(Y)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
diff(X, Y) → if(leq(X, Y), n__0, n__s(n__diff(n__p(X), Y)))
0 → n__0
s(X) → n__s(X)
diff(X1, X2) → n__diff(X1, X2)
p(X) → n__p(X)
activate(n__0) → 0
activate(n__s(X)) → s(activate(X))
activate(n__diff(X1, X2)) → diff(activate(X1), activate(X2))
activate(n__p(X)) → p(activate(X))
activate(X) → 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.

ACTIVATE(n__p(X)) → ACTIVATE(X)

The remaining pairs can at least be oriented weakly.

IF(true, X, Y) → ACTIVATE(X)
ACTIVATE(n__diff(X1, X2)) → DIFF(activate(X1), activate(X2))
DIFF(X, Y) → IF(leq(X, Y), n__0, n__s(n__diff(n__p(X), Y)))
IF(false, X, Y) → ACTIVATE(Y)
ACTIVATE(n__s(X)) → ACTIVATE(X)

Used ordering: Polynomial interpretation [25,35]:

POL(DIFF(x₁, x₂)) = 0
POL(n__p(x₁)) = 1 + (4)x_1
POL(true) = 0
POL(activate(x₁)) = 3
POL(p(x₁)) = 3 + (3)x_1
POL(n__s(x₁)) = (2)x_1
POL(IF(x₁, x₂, x₃)) = x_2 + x_3
POL(0) = 4
POL(if(x₁, x₂, x₃)) = 1 + (4)x_1 + (2)x_2 + (4)x_3
POL(diff(x₁, x₂)) = 1 + (3)x_1 + (4)x_2
POL(n__0) = 0
POL(false) = 0
POL(s(x₁)) = 3
POL(ACTIVATE(x₁)) = x_1
POL(leq(x₁, x₂)) = (2)x_2
POL(n__diff(x₁, x₂)) = 0

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

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

IF(true, X, Y) → ACTIVATE(X)
ACTIVATE(n__diff(X1, X2)) → DIFF(activate(X1), activate(X2))
DIFF(X, Y) → IF(leq(X, Y), n__0, n__s(n__diff(n__p(X), Y)))
IF(false, X, Y) → ACTIVATE(Y)
ACTIVATE(n__s(X)) → ACTIVATE(X)

The TRS R consists of the following rules:

p(0) → 0
p(s(X)) → X
leq(0, Y) → true
leq(s(X), 0) → false
leq(s(X), s(Y)) → leq(X, Y)
if(true, X, Y) → activate(X)
if(false, X, Y) → activate(Y)
diff(X, Y) → if(leq(X, Y), n__0, n__s(n__diff(n__p(X), Y)))
0 → n__0
s(X) → n__s(X)
diff(X1, X2) → n__diff(X1, X2)
p(X) → n__p(X)
activate(n__0) → 0
activate(n__s(X)) → s(activate(X))
activate(n__diff(X1, X2)) → diff(activate(X1), activate(X2))
activate(n__p(X)) → p(activate(X))
activate(X) → X

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