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) → X
if(false, X, Y) → Y
diff(X, Y) → if(leq(X, Y), 0, s(diff(p(X), Y)))

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) → X
if(false, X, Y) → Y
diff(X, Y) → if(leq(X, Y), 0, s(diff(p(X), Y)))

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:

DIFF(X, Y) → DIFF(p(X), Y)
LEQ(s(X), s(Y)) → LEQ(X, Y)
DIFF(X, Y) → P(X)
DIFF(X, Y) → LEQ(X, Y)
DIFF(X, Y) → IF(leq(X, Y), 0, s(diff(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) → X
if(false, X, Y) → Y
diff(X, Y) → if(leq(X, Y), 0, s(diff(p(X), Y)))

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:

DIFF(X, Y) → DIFF(p(X), Y)
LEQ(s(X), s(Y)) → LEQ(X, Y)
DIFF(X, Y) → P(X)
DIFF(X, Y) → LEQ(X, Y)
DIFF(X, Y) → IF(leq(X, Y), 0, s(diff(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) → X
if(false, X, Y) → Y
diff(X, Y) → if(leq(X, Y), 0, s(diff(p(X), Y)))

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 3 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) → X
if(false, X, Y) → Y
diff(X, Y) → if(leq(X, Y), 0, s(diff(p(X), Y)))

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)) = x_1 + (13/4)x_2   
POL(s(x1)) = 5/4 + (15/4)x_1   
The value of delta used in the strict ordering is 85/16.
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) → X
if(false, X, Y) → Y
diff(X, Y) → if(leq(X, Y), 0, s(diff(p(X), Y)))

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:

DIFF(X, Y) → DIFF(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) → X
if(false, X, Y) → Y
diff(X, Y) → if(leq(X, Y), 0, s(diff(p(X), Y)))

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