Termination w.r.t. Q proof of /tmp/tmpW26_De/thiemann06.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:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
average(x, y) → if(le(x, 0), le(y, 0), le(y, s(0)), le(y, s(s(0))), x, y)
if(true, b1, b2, b3, x, y) → if2(b1, b2, b3, x, y)
if(false, b1, b2, b3, x, y) → average(p(x), s(y))
if2(true, b2, b3, x, y) → 0
if2(false, b2, b3, x, y) → if3(b2, b3, x, y)
if3(true, b3, x, y) → 0
if3(false, b3, x, y) → if4(b3, x, y)
if4(true, x, y) → s(0)
if4(false, x, y) → average(s(x), p(p(y)))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
average(x, y) → if(le(x, 0), le(y, 0), le(y, s(0)), le(y, s(s(0))), x, y)
if(true, b1, b2, b3, x, y) → if2(b1, b2, b3, x, y)
if(false, b1, b2, b3, x, y) → average(p(x), s(y))
if2(true, b2, b3, x, y) → 0
if2(false, b2, b3, x, y) → if3(b2, b3, x, y)
if3(true, b3, x, y) → 0
if3(false, b3, x, y) → if4(b3, x, y)
if4(true, x, y) → s(0)
if4(false, x, y) → average(s(x), p(p(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:

AVERAGE(x, y) → LE(y, s(0))
IF(false, b1, b2, b3, x, y) → P(x)
IF(true, b1, b2, b3, x, y) → IF2(b1, b2, b3, x, y)
IF2(false, b2, b3, x, y) → IF3(b2, b3, x, y)
AVERAGE(x, y) → IF(le(x, 0), le(y, 0), le(y, s(0)), le(y, s(s(0))), x, y)
IF(false, b1, b2, b3, x, y) → AVERAGE(p(x), s(y))
LE(s(x), s(y)) → LE(x, y)
IF4(false, x, y) → P(y)
IF4(false, x, y) → P(p(y))
IF4(false, x, y) → AVERAGE(s(x), p(p(y)))
IF3(false, b3, x, y) → IF4(b3, x, y)
AVERAGE(x, y) → LE(x, 0)
AVERAGE(x, y) → LE(y, 0)
AVERAGE(x, y) → LE(y, s(s(0)))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
average(x, y) → if(le(x, 0), le(y, 0), le(y, s(0)), le(y, s(s(0))), x, y)
if(true, b1, b2, b3, x, y) → if2(b1, b2, b3, x, y)
if(false, b1, b2, b3, x, y) → average(p(x), s(y))
if2(true, b2, b3, x, y) → 0
if2(false, b2, b3, x, y) → if3(b2, b3, x, y)
if3(true, b3, x, y) → 0
if3(false, b3, x, y) → if4(b3, x, y)
if4(true, x, y) → s(0)
if4(false, x, y) → average(s(x), p(p(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:

AVERAGE(x, y) → LE(y, s(0))
IF(false, b1, b2, b3, x, y) → P(x)
IF(true, b1, b2, b3, x, y) → IF2(b1, b2, b3, x, y)
IF2(false, b2, b3, x, y) → IF3(b2, b3, x, y)
AVERAGE(x, y) → IF(le(x, 0), le(y, 0), le(y, s(0)), le(y, s(s(0))), x, y)
IF(false, b1, b2, b3, x, y) → AVERAGE(p(x), s(y))
LE(s(x), s(y)) → LE(x, y)
IF4(false, x, y) → P(y)
IF4(false, x, y) → P(p(y))
IF4(false, x, y) → AVERAGE(s(x), p(p(y)))
IF3(false, b3, x, y) → IF4(b3, x, y)
AVERAGE(x, y) → LE(x, 0)
AVERAGE(x, y) → LE(y, 0)
AVERAGE(x, y) → LE(y, s(s(0)))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
average(x, y) → if(le(x, 0), le(y, 0), le(y, s(0)), le(y, s(s(0))), x, y)
if(true, b1, b2, b3, x, y) → if2(b1, b2, b3, x, y)
if(false, b1, b2, b3, x, y) → average(p(x), s(y))
if2(true, b2, b3, x, y) → 0
if2(false, b2, b3, x, y) → if3(b2, b3, x, y)
if3(true, b3, x, y) → 0
if3(false, b3, x, y) → if4(b3, x, y)
if4(true, x, y) → s(0)
if4(false, x, y) → average(s(x), p(p(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 7 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:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
average(x, y) → if(le(x, 0), le(y, 0), le(y, s(0)), le(y, s(s(0))), x, y)
if(true, b1, b2, b3, x, y) → if2(b1, b2, b3, x, y)
if(false, b1, b2, b3, x, y) → average(p(x), s(y))
if2(true, b2, b3, x, y) → 0
if2(false, b2, b3, x, y) → if3(b2, b3, x, y)
if3(true, b3, x, y) → 0
if3(false, b3, x, y) → if4(b3, x, y)
if4(true, x, y) → s(0)
if4(false, x, y) → average(s(x), p(p(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)) = 1 + (4)x_1
POL(LE(x₁, x₂)) = (3)x_2

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(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
average(x, y) → if(le(x, 0), le(y, 0), le(y, s(0)), le(y, s(s(0))), x, y)
if(true, b1, b2, b3, x, y) → if2(b1, b2, b3, x, y)
if(false, b1, b2, b3, x, y) → average(p(x), s(y))
if2(true, b2, b3, x, y) → 0
if2(false, b2, b3, x, y) → if3(b2, b3, x, y)
if3(true, b3, x, y) → 0
if3(false, b3, x, y) → if4(b3, x, y)
if4(true, x, y) → s(0)
if4(false, x, y) → average(s(x), p(p(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:

IF(true, b1, b2, b3, x, y) → IF2(b1, b2, b3, x, y)
IF4(false, x, y) → AVERAGE(s(x), p(p(y)))
IF2(false, b2, b3, x, y) → IF3(b2, b3, x, y)
IF3(false, b3, x, y) → IF4(b3, x, y)
AVERAGE(x, y) → IF(le(x, 0), le(y, 0), le(y, s(0)), le(y, s(s(0))), x, y)
IF(false, b1, b2, b3, x, y) → AVERAGE(p(x), s(y))

The TRS R consists of the following rules:

p(s(x)) → x
p(0) → 0
le(0, y) → true
le(s(x), 0) → false
le(s(x), s(y)) → le(x, y)
average(x, y) → if(le(x, 0), le(y, 0), le(y, s(0)), le(y, s(s(0))), x, y)
if(true, b1, b2, b3, x, y) → if2(b1, b2, b3, x, y)
if(false, b1, b2, b3, x, y) → average(p(x), s(y))
if2(true, b2, b3, x, y) → 0
if2(false, b2, b3, x, y) → if3(b2, b3, x, y)
if3(true, b3, x, y) → 0
if3(false, b3, x, y) → if4(b3, x, y)
if4(true, x, y) → s(0)
if4(false, x, y) → average(s(x), p(p(y)))

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