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:

digits → d(0)
d(x) → if(le(x, s(s(s(s(s(s(s(s(s(0)))))))))), x)
if(true, x) → cons(x, d(s(x)))
if(false, x) → nil
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

digits → d(0)
d(x) → if(le(x, s(s(s(s(s(s(s(s(s(0)))))))))), x)
if(true, x) → cons(x, d(s(x)))
if(false, x) → nil
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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:

D(x) → LE(x, s(s(s(s(s(s(s(s(s(0))))))))))
D(x) → IF(le(x, s(s(s(s(s(s(s(s(s(0)))))))))), x)
DIGITS → D(0)
IF(true, x) → D(s(x))
LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

digits → d(0)
d(x) → if(le(x, s(s(s(s(s(s(s(s(s(0)))))))))), x)
if(true, x) → cons(x, d(s(x)))
if(false, x) → nil
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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:

D(x) → LE(x, s(s(s(s(s(s(s(s(s(0))))))))))
D(x) → IF(le(x, s(s(s(s(s(s(s(s(s(0)))))))))), x)
DIGITS → D(0)
IF(true, x) → D(s(x))
LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

digits → d(0)
d(x) → if(le(x, s(s(s(s(s(s(s(s(s(0)))))))))), x)
if(true, x) → cons(x, d(s(x)))
if(false, x) → nil
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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 2 less nodes.

↳ QTRS
  ↳ DependencyPairsProof
    ↳ QDP
      ↳ DependencyGraphProof
        ↳ AND
          ↳ QDP
            ↳ QDPOrderProof
          ↳ QDP

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

LE(s(x), s(y)) → LE(x, y)

The TRS R consists of the following rules:

digits → d(0)
d(x) → if(le(x, s(s(s(s(s(s(s(s(s(0)))))))))), x)
if(true, x) → cons(x, d(s(x)))
if(false, x) → nil
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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.


LE(s(x), s(y)) → LE(x, y)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(s(x1)) = 5/4 + (15/4)x_1   
POL(LE(x1, x2)) = x_1 + (13/4)x_2   
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:

digits → d(0)
d(x) → if(le(x, s(s(s(s(s(s(s(s(s(0)))))))))), x)
if(true, x) → cons(x, d(s(x)))
if(false, x) → nil
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(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:

D(x) → IF(le(x, s(s(s(s(s(s(s(s(s(0)))))))))), x)
IF(true, x) → D(s(x))

The TRS R consists of the following rules:

digits → d(0)
d(x) → if(le(x, s(s(s(s(s(s(s(s(s(0)))))))))), x)
if(true, x) → cons(x, d(s(x)))
if(false, x) → nil
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)

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