Termination w.r.t. Q proof of /tmp/tmpnMIJGi/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:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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:

MARK(p(X)) → ACTIVE(p(mark(X)))
ACTIVE(p(0)) → MARK(0)
ACTIVE(diff(X, Y)) → P(X)
MARK(leq(X1, X2)) → MARK(X2)
MARK(diff(X1, X2)) → MARK(X2)
ACTIVE(p(s(X))) → MARK(X)
MARK(true) → ACTIVE(true)
MARK(if(X1, X2, X3)) → MARK(X1)
IF(X1, X2, mark(X3)) → IF(X1, X2, X3)
MARK(leq(X1, X2)) → MARK(X1)
LEQ(X1, mark(X2)) → LEQ(X1, X2)
LEQ(X1, active(X2)) → LEQ(X1, X2)
MARK(p(X)) → MARK(X)
ACTIVE(diff(X, Y)) → IF(leq(X, Y), 0, s(diff(p(X), Y)))
DIFF(X1, active(X2)) → DIFF(X1, X2)
ACTIVE(leq(s(X), s(Y))) → MARK(leq(X, Y))
S(active(X)) → S(X)
MARK(false) → ACTIVE(false)
P(active(X)) → P(X)
MARK(leq(X1, X2)) → LEQ(mark(X1), mark(X2))
ACTIVE(diff(X, Y)) → S(diff(p(X), Y))
MARK(diff(X1, X2)) → ACTIVE(diff(mark(X1), mark(X2)))
ACTIVE(leq(0, Y)) → MARK(true)
MARK(diff(X1, X2)) → MARK(X1)
DIFF(active(X1), X2) → DIFF(X1, X2)
MARK(s(X)) → MARK(X)
ACTIVE(if(false, X, Y)) → MARK(Y)
P(mark(X)) → P(X)
ACTIVE(diff(X, Y)) → LEQ(X, Y)
LEQ(mark(X1), X2) → LEQ(X1, X2)
MARK(s(X)) → ACTIVE(s(mark(X)))
DIFF(mark(X1), X2) → DIFF(X1, X2)
ACTIVE(diff(X, Y)) → MARK(if(leq(X, Y), 0, s(diff(p(X), Y))))
ACTIVE(diff(X, Y)) → DIFF(p(X), Y)
IF(X1, mark(X2), X3) → IF(X1, X2, X3)
MARK(s(X)) → S(mark(X))
S(mark(X)) → S(X)
IF(X1, X2, active(X3)) → IF(X1, X2, X3)
MARK(if(X1, X2, X3)) → IF(mark(X1), X2, X3)
ACTIVE(leq(s(X), s(Y))) → LEQ(X, Y)
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), X2, X3))
IF(mark(X1), X2, X3) → IF(X1, X2, X3)
IF(active(X1), X2, X3) → IF(X1, X2, X3)
MARK(p(X)) → P(mark(X))
ACTIVE(leq(s(X), 0)) → MARK(false)
DIFF(X1, mark(X2)) → DIFF(X1, X2)
LEQ(active(X1), X2) → LEQ(X1, X2)
IF(X1, active(X2), X3) → IF(X1, X2, X3)
MARK(0) → ACTIVE(0)
MARK(diff(X1, X2)) → DIFF(mark(X1), mark(X2))
MARK(leq(X1, X2)) → ACTIVE(leq(mark(X1), mark(X2)))
ACTIVE(if(true, X, Y)) → MARK(X)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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:

MARK(p(X)) → ACTIVE(p(mark(X)))
ACTIVE(p(0)) → MARK(0)
ACTIVE(diff(X, Y)) → P(X)
MARK(leq(X1, X2)) → MARK(X2)
MARK(diff(X1, X2)) → MARK(X2)
ACTIVE(p(s(X))) → MARK(X)
MARK(true) → ACTIVE(true)
MARK(if(X1, X2, X3)) → MARK(X1)
IF(X1, X2, mark(X3)) → IF(X1, X2, X3)
MARK(leq(X1, X2)) → MARK(X1)
LEQ(X1, mark(X2)) → LEQ(X1, X2)
LEQ(X1, active(X2)) → LEQ(X1, X2)
MARK(p(X)) → MARK(X)
ACTIVE(diff(X, Y)) → IF(leq(X, Y), 0, s(diff(p(X), Y)))
DIFF(X1, active(X2)) → DIFF(X1, X2)
ACTIVE(leq(s(X), s(Y))) → MARK(leq(X, Y))
S(active(X)) → S(X)
MARK(false) → ACTIVE(false)
P(active(X)) → P(X)
MARK(leq(X1, X2)) → LEQ(mark(X1), mark(X2))
ACTIVE(diff(X, Y)) → S(diff(p(X), Y))
MARK(diff(X1, X2)) → ACTIVE(diff(mark(X1), mark(X2)))
ACTIVE(leq(0, Y)) → MARK(true)
MARK(diff(X1, X2)) → MARK(X1)
DIFF(active(X1), X2) → DIFF(X1, X2)
MARK(s(X)) → MARK(X)
ACTIVE(if(false, X, Y)) → MARK(Y)
P(mark(X)) → P(X)
ACTIVE(diff(X, Y)) → LEQ(X, Y)
LEQ(mark(X1), X2) → LEQ(X1, X2)
MARK(s(X)) → ACTIVE(s(mark(X)))
DIFF(mark(X1), X2) → DIFF(X1, X2)
ACTIVE(diff(X, Y)) → MARK(if(leq(X, Y), 0, s(diff(p(X), Y))))
ACTIVE(diff(X, Y)) → DIFF(p(X), Y)
IF(X1, mark(X2), X3) → IF(X1, X2, X3)
MARK(s(X)) → S(mark(X))
S(mark(X)) → S(X)
IF(X1, X2, active(X3)) → IF(X1, X2, X3)
MARK(if(X1, X2, X3)) → IF(mark(X1), X2, X3)
ACTIVE(leq(s(X), s(Y))) → LEQ(X, Y)
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), X2, X3))
IF(mark(X1), X2, X3) → IF(X1, X2, X3)
IF(active(X1), X2, X3) → IF(X1, X2, X3)
MARK(p(X)) → P(mark(X))
ACTIVE(leq(s(X), 0)) → MARK(false)
DIFF(X1, mark(X2)) → DIFF(X1, X2)
LEQ(active(X1), X2) → LEQ(X1, X2)
IF(X1, active(X2), X3) → IF(X1, X2, X3)
MARK(0) → ACTIVE(0)
MARK(diff(X1, X2)) → DIFF(mark(X1), mark(X2))
MARK(leq(X1, X2)) → ACTIVE(leq(mark(X1), mark(X2)))
ACTIVE(if(true, X, Y)) → MARK(X)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 6 SCCs with 17 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

DIFF(X1, active(X2)) → DIFF(X1, X2)
DIFF(active(X1), X2) → DIFF(X1, X2)
DIFF(X1, mark(X2)) → DIFF(X1, X2)
DIFF(mark(X1), X2) → DIFF(X1, X2)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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.

DIFF(X1, active(X2)) → DIFF(X1, X2)
DIFF(active(X1), X2) → DIFF(X1, X2)
DIFF(X1, mark(X2)) → DIFF(X1, X2)
DIFF(mark(X1), X2) → DIFF(X1, X2)

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

POL(DIFF(x₁, x₂)) = (1/2)x_1 + (1/4)x_2
POL(active(x₁)) = 1/4 + (2)x_1
POL(mark(x₁)) = 13/4 + (2)x_1

The value of delta used in the strict ordering is 1/16.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

IF(X1, mark(X2), X3) → IF(X1, X2, X3)
IF(X1, X2, active(X3)) → IF(X1, X2, X3)
IF(X1, active(X2), X3) → IF(X1, X2, X3)
IF(X1, X2, mark(X3)) → IF(X1, X2, X3)
IF(active(X1), X2, X3) → IF(X1, X2, X3)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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.

IF(X1, X2, active(X3)) → IF(X1, X2, X3)
IF(X1, X2, mark(X3)) → IF(X1, X2, X3)

The remaining pairs can at least be oriented weakly.

IF(X1, mark(X2), X3) → IF(X1, X2, X3)
IF(X1, active(X2), X3) → IF(X1, X2, X3)
IF(active(X1), X2, X3) → IF(X1, X2, X3)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)

Used ordering: Polynomial interpretation [25,35]:

POL(active(x₁)) = 4 + (9/4)x_1
POL(mark(x₁)) = 9/4 + (4)x_1
POL(IF(x₁, x₂, x₃)) = (4)x_3

The value of delta used in the strict ordering is 9.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

IF(X1, mark(X2), X3) → IF(X1, X2, X3)
IF(X1, active(X2), X3) → IF(X1, X2, X3)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)
IF(active(X1), X2, X3) → IF(X1, X2, X3)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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.

IF(X1, mark(X2), X3) → IF(X1, X2, X3)
IF(mark(X1), X2, X3) → IF(X1, X2, X3)

The remaining pairs can at least be oriented weakly.

IF(X1, active(X2), X3) → IF(X1, X2, X3)
IF(active(X1), X2, X3) → IF(X1, X2, X3)

Used ordering: Polynomial interpretation [25,35]:

POL(active(x₁)) = (2)x_1
POL(mark(x₁)) = 3/2 + x_1
POL(IF(x₁, x₂, x₃)) = (4)x_1 + (3/2)x_2

The value of delta used in the strict ordering is 9/4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

IF(X1, active(X2), X3) → IF(X1, X2, X3)
IF(active(X1), X2, X3) → IF(X1, X2, X3)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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.

IF(X1, active(X2), X3) → IF(X1, X2, X3)

The remaining pairs can at least be oriented weakly.

IF(active(X1), X2, X3) → IF(X1, X2, X3)

Used ordering: Polynomial interpretation [25,35]:

POL(active(x₁)) = 4 + (5/4)x_1
POL(IF(x₁, x₂, x₃)) = x_2

The value of delta used in the strict ordering is 4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

IF(active(X1), X2, X3) → IF(X1, X2, X3)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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.

IF(active(X1), X2, X3) → IF(X1, X2, X3)

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

POL(active(x₁)) = 1/4 + (7/2)x_1
POL(IF(x₁, x₂, x₃)) = (2)x_1

The value of delta used in the strict ordering is 1/2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ QDPOrderProof

                      ↳ QDP

                        ↳ QDPOrderProof

                          ↳ QDP

                            ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

LEQ(active(X1), X2) → LEQ(X1, X2)
LEQ(X1, mark(X2)) → LEQ(X1, X2)
LEQ(X1, active(X2)) → LEQ(X1, X2)
LEQ(mark(X1), X2) → LEQ(X1, X2)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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(active(X1), X2) → LEQ(X1, X2)
LEQ(X1, active(X2)) → LEQ(X1, X2)

The remaining pairs can at least be oriented weakly.

LEQ(X1, mark(X2)) → LEQ(X1, X2)
LEQ(mark(X1), X2) → LEQ(X1, X2)

Used ordering: Polynomial interpretation [25,35]:

POL(active(x₁)) = 3/2 + (11/4)x_1
POL(LEQ(x₁, x₂)) = (3/2)x_1 + (4)x_2
POL(mark(x₁)) = (3/2)x_1

The value of delta used in the strict ordering is 9/4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

LEQ(X1, mark(X2)) → LEQ(X1, X2)
LEQ(mark(X1), X2) → LEQ(X1, X2)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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(X1, mark(X2)) → LEQ(X1, X2)
LEQ(mark(X1), X2) → LEQ(X1, X2)

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

POL(LEQ(x₁, x₂)) = (3)x_1 + (1/2)x_2
POL(mark(x₁)) = 4 + (9/4)x_1

The value of delta used in the strict ordering is 2.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ QDPOrderProof

                  ↳ QDP

                    ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

          ↳ QDP

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

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

          ↳ QDP

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

S(mark(X)) → S(X)
S(active(X)) → S(X)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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.

S(mark(X)) → S(X)
S(active(X)) → S(X)

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

POL(active(x₁)) = 1/2 + (3/2)x_1
POL(mark(x₁)) = 9/4 + x_1
POL(S(x₁)) = (1/2)x_1

The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

          ↳ QDP

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

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

          ↳ QDP

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

P(active(X)) → P(X)
P(mark(X)) → P(X)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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.

P(active(X)) → P(X)
P(mark(X)) → P(X)

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

POL(active(x₁)) = 9/4 + x_1
POL(P(x₁)) = (1/2)x_1
POL(mark(x₁)) = 1/2 + (3/2)x_1

The value of delta used in the strict ordering is 1/4.
The following usable rules [17] were oriented: none

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

                ↳ PisEmptyProof

          ↳ QDP

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

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

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

MARK(diff(X1, X2)) → MARK(X1)
ACTIVE(leq(s(X), s(Y))) → MARK(leq(X, Y))
MARK(s(X)) → MARK(X)
ACTIVE(if(false, X, Y)) → MARK(Y)
MARK(p(X)) → ACTIVE(p(mark(X)))
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), X2, X3))
MARK(diff(X1, X2)) → MARK(X2)
MARK(leq(X1, X2)) → MARK(X2)
ACTIVE(p(s(X))) → MARK(X)
MARK(diff(X1, X2)) → ACTIVE(diff(mark(X1), mark(X2)))
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(s(X)) → ACTIVE(s(mark(X)))
MARK(leq(X1, X2)) → MARK(X1)
ACTIVE(diff(X, Y)) → MARK(if(leq(X, Y), 0, s(diff(p(X), Y))))
MARK(leq(X1, X2)) → ACTIVE(leq(mark(X1), mark(X2)))
MARK(p(X)) → MARK(X)
ACTIVE(if(true, X, Y)) → MARK(X)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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.

MARK(s(X)) → ACTIVE(s(mark(X)))

The remaining pairs can at least be oriented weakly.

MARK(diff(X1, X2)) → MARK(X1)
ACTIVE(leq(s(X), s(Y))) → MARK(leq(X, Y))
MARK(s(X)) → MARK(X)
ACTIVE(if(false, X, Y)) → MARK(Y)
MARK(p(X)) → ACTIVE(p(mark(X)))
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), X2, X3))
MARK(diff(X1, X2)) → MARK(X2)
MARK(leq(X1, X2)) → MARK(X2)
ACTIVE(p(s(X))) → MARK(X)
MARK(diff(X1, X2)) → ACTIVE(diff(mark(X1), mark(X2)))
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(leq(X1, X2)) → MARK(X1)
ACTIVE(diff(X, Y)) → MARK(if(leq(X, Y), 0, s(diff(p(X), Y))))
MARK(leq(X1, X2)) → ACTIVE(leq(mark(X1), mark(X2)))
MARK(p(X)) → MARK(X)
ACTIVE(if(true, X, Y)) → MARK(X)

Used ordering: Polynomial interpretation [25,35]:

POL(active(x₁)) = 1
POL(MARK(x₁)) = 2
POL(if(x₁, x₂, x₃)) = 1/2
POL(diff(x₁, x₂)) = 1/2
POL(true) = 4
POL(false) = 2
POL(mark(x₁)) = 4
POL(s(x₁)) = 0
POL(p(x₁)) = 1/2
POL(0) = 3/2
POL(ACTIVE(x₁)) = 1 + (2)x_1
POL(leq(x₁, x₂)) = 1/2

The value of delta used in the strict ordering is 1.
The following usable rules [17] were oriented:

p(active(X)) → p(X)
p(mark(X)) → p(X)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
s(active(X)) → s(X)
s(mark(X)) → s(X)
leq(X1, active(X2)) → leq(X1, X2)
leq(mark(X1), X2) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(mark(X1), X2) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

          ↳ QDP

            ↳ QDPOrderProof

              ↳ QDP

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

MARK(diff(X1, X2)) → MARK(X1)
ACTIVE(leq(s(X), s(Y))) → MARK(leq(X, Y))
MARK(s(X)) → MARK(X)
ACTIVE(if(false, X, Y)) → MARK(Y)
MARK(p(X)) → ACTIVE(p(mark(X)))
MARK(if(X1, X2, X3)) → ACTIVE(if(mark(X1), X2, X3))
MARK(diff(X1, X2)) → MARK(X2)
MARK(leq(X1, X2)) → MARK(X2)
ACTIVE(p(s(X))) → MARK(X)
MARK(diff(X1, X2)) → ACTIVE(diff(mark(X1), mark(X2)))
MARK(if(X1, X2, X3)) → MARK(X1)
MARK(leq(X1, X2)) → MARK(X1)
ACTIVE(diff(X, Y)) → MARK(if(leq(X, Y), 0, s(diff(p(X), Y))))
MARK(leq(X1, X2)) → ACTIVE(leq(mark(X1), mark(X2)))
MARK(p(X)) → MARK(X)
ACTIVE(if(true, X, Y)) → MARK(X)

The TRS R consists of the following rules:

active(p(0)) → mark(0)
active(p(s(X))) → mark(X)
active(leq(0, Y)) → mark(true)
active(leq(s(X), 0)) → mark(false)
active(leq(s(X), s(Y))) → mark(leq(X, Y))
active(if(true, X, Y)) → mark(X)
active(if(false, X, Y)) → mark(Y)
active(diff(X, Y)) → mark(if(leq(X, Y), 0, s(diff(p(X), Y))))
mark(p(X)) → active(p(mark(X)))
mark(0) → active(0)
mark(s(X)) → active(s(mark(X)))
mark(leq(X1, X2)) → active(leq(mark(X1), mark(X2)))
mark(true) → active(true)
mark(false) → active(false)
mark(if(X1, X2, X3)) → active(if(mark(X1), X2, X3))
mark(diff(X1, X2)) → active(diff(mark(X1), mark(X2)))
p(mark(X)) → p(X)
p(active(X)) → p(X)
s(mark(X)) → s(X)
s(active(X)) → s(X)
leq(mark(X1), X2) → leq(X1, X2)
leq(X1, mark(X2)) → leq(X1, X2)
leq(active(X1), X2) → leq(X1, X2)
leq(X1, active(X2)) → leq(X1, X2)
if(mark(X1), X2, X3) → if(X1, X2, X3)
if(X1, mark(X2), X3) → if(X1, X2, X3)
if(X1, X2, mark(X3)) → if(X1, X2, X3)
if(active(X1), X2, X3) → if(X1, X2, X3)
if(X1, active(X2), X3) → if(X1, X2, X3)
if(X1, X2, active(X3)) → if(X1, X2, X3)
diff(mark(X1), X2) → diff(X1, X2)
diff(X1, mark(X2)) → diff(X1, X2)
diff(active(X1), X2) → diff(X1, X2)
diff(X1, active(X2)) → diff(X1, X2)

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