Termination w.r.t. Q of the following Term Rewriting System could be proven:

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
          ↳ QDPOrderProof

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.
We use the reduction pair processor [15].


The following pairs can be oriented strictly and are deleted.


IF(true, x, y) → RAND(p(x), id_inc(y))
The remaining pairs can at least be oriented weakly.

RAND(x, y) → IF(nonZero(x), x, y)
Used ordering: Polynomial interpretation [25,35]:

POL(RAND(x1, x2)) = x_1   
POL(true) = 1/2   
POL(id_inc(x1)) = 3   
POL(false) = 0   
POL(p(x1)) = (1/2)x_1   
POL(s(x1)) = 3/4 + (4)x_1   
POL(IF(x1, x2, x3)) = (1/4)x_1 + (1/2)x_2   
POL(0) = 0   
POL(nonZero(x1)) = (3/4)x_1   
The value of delta used in the strict ordering is 1/8.
The following usable rules [17] were oriented:

nonZero(0) → false
nonZero(s(x)) → true
p(0) → 0
p(s(x)) → x



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

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

RAND(x, y) → IF(nonZero(x), x, 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.
The approximation of the Dependency Graph [15,17,22] contains 0 SCCs with 1 less node.