Termination w.r.t. Q proof of /tmp/tmpgreZth/thiemann40.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:

nonZero(0) → false
nonZero(s(x)) → true
p(0) → 0
p(s(x)) → x
id_inc(x) → x
id_inc(x) → s(x)
random(x) → rand(x, 0)
rand(x, y) → if(nonZero(x), x, y)
if(false, x, y) → y
if(true, x, y) → rand(p(x), id_inc(y))

Q is empty.

↳ QTRS

  ↳ DependencyPairsProof

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

nonZero(0) → false
nonZero(s(x)) → true
p(0) → 0
p(s(x)) → x
id_inc(x) → x
id_inc(x) → s(x)
random(x) → rand(x, 0)
rand(x, y) → if(nonZero(x), x, y)
if(false, x, y) → y
if(true, x, y) → rand(p(x), id_inc(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:

RAND(x, y) → IF(nonZero(x), x, y)
IF(true, x, y) → ID_INC(y)
IF(true, x, y) → P(x)
IF(true, x, y) → RAND(p(x), id_inc(y))
RAND(x, y) → NONZERO(x)
RANDOM(x) → RAND(x, 0)

The TRS R consists of the following rules:

nonZero(0) → false
nonZero(s(x)) → true
p(0) → 0
p(s(x)) → x
id_inc(x) → x
id_inc(x) → s(x)
random(x) → rand(x, 0)
rand(x, y) → if(nonZero(x), x, y)
if(false, x, y) → y
if(true, x, y) → rand(p(x), id_inc(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:

RAND(x, y) → IF(nonZero(x), x, y)
IF(true, x, y) → ID_INC(y)
IF(true, x, y) → P(x)
IF(true, x, y) → RAND(p(x), id_inc(y))
RAND(x, y) → NONZERO(x)
RANDOM(x) → RAND(x, 0)

The TRS R consists of the following rules:

nonZero(0) → false
nonZero(s(x)) → true
p(0) → 0
p(s(x)) → x
id_inc(x) → x
id_inc(x) → s(x)
random(x) → rand(x, 0)
rand(x, y) → if(nonZero(x), x, y)
if(false, x, y) → y
if(true, x, y) → rand(p(x), id_inc(y))

Q is empty.
We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [15,17,22] contains 1 SCC with 4 less nodes.

↳ QTRS

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ QDP

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

RAND(x, y) → IF(nonZero(x), x, y)
IF(true, x, y) → RAND(p(x), id_inc(y))

The TRS R consists of the following rules:

nonZero(0) → false
nonZero(s(x)) → true
p(0) → 0
p(s(x)) → x
id_inc(x) → x
id_inc(x) → s(x)
random(x) → rand(x, 0)
rand(x, y) → if(nonZero(x), x, y)
if(false, x, y) → y
if(true, x, y) → rand(p(x), id_inc(y))

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