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:

function(iszero, 0, dummy, dummy2) → true
function(iszero, s(x), dummy, dummy2) → false
function(p, 0, dummy, dummy2) → 0
function(p, s(0), dummy, dummy2) → 0
function(p, s(s(x)), dummy, dummy2) → s(function(p, s(x), x, x))
function(plus, dummy, x, y) → function(if, function(iszero, x, x, x), x, y)
function(if, true, x, y) → y
function(if, false, x, y) → function(plus, function(third, x, y, y), function(p, x, x, y), s(y))
function(third, x, y, z) → z

Q is empty.


↳ QTRS
  ↳ DependencyPairsProof

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

function(iszero, 0, dummy, dummy2) → true
function(iszero, s(x), dummy, dummy2) → false
function(p, 0, dummy, dummy2) → 0
function(p, s(0), dummy, dummy2) → 0
function(p, s(s(x)), dummy, dummy2) → s(function(p, s(x), x, x))
function(plus, dummy, x, y) → function(if, function(iszero, x, x, x), x, y)
function(if, true, x, y) → y
function(if, false, x, y) → function(plus, function(third, x, y, y), function(p, x, x, y), s(y))
function(third, x, y, z) → z

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:

FUNCTION(plus, dummy, x, y) → FUNCTION(if, function(iszero, x, x, x), x, y)
FUNCTION(p, s(s(x)), dummy, dummy2) → FUNCTION(p, s(x), x, x)
FUNCTION(plus, dummy, x, y) → FUNCTION(iszero, x, x, x)
FUNCTION(if, false, x, y) → FUNCTION(p, x, x, y)
FUNCTION(if, false, x, y) → FUNCTION(plus, function(third, x, y, y), function(p, x, x, y), s(y))
FUNCTION(if, false, x, y) → FUNCTION(third, x, y, y)

The TRS R consists of the following rules:

function(iszero, 0, dummy, dummy2) → true
function(iszero, s(x), dummy, dummy2) → false
function(p, 0, dummy, dummy2) → 0
function(p, s(0), dummy, dummy2) → 0
function(p, s(s(x)), dummy, dummy2) → s(function(p, s(x), x, x))
function(plus, dummy, x, y) → function(if, function(iszero, x, x, x), x, y)
function(if, true, x, y) → y
function(if, false, x, y) → function(plus, function(third, x, y, y), function(p, x, x, y), s(y))
function(third, x, y, z) → z

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:

FUNCTION(plus, dummy, x, y) → FUNCTION(if, function(iszero, x, x, x), x, y)
FUNCTION(p, s(s(x)), dummy, dummy2) → FUNCTION(p, s(x), x, x)
FUNCTION(plus, dummy, x, y) → FUNCTION(iszero, x, x, x)
FUNCTION(if, false, x, y) → FUNCTION(p, x, x, y)
FUNCTION(if, false, x, y) → FUNCTION(plus, function(third, x, y, y), function(p, x, x, y), s(y))
FUNCTION(if, false, x, y) → FUNCTION(third, x, y, y)

The TRS R consists of the following rules:

function(iszero, 0, dummy, dummy2) → true
function(iszero, s(x), dummy, dummy2) → false
function(p, 0, dummy, dummy2) → 0
function(p, s(0), dummy, dummy2) → 0
function(p, s(s(x)), dummy, dummy2) → s(function(p, s(x), x, x))
function(plus, dummy, x, y) → function(if, function(iszero, x, x, x), x, y)
function(if, true, x, y) → y
function(if, false, x, y) → function(plus, function(third, x, y, y), function(p, x, x, y), s(y))
function(third, x, y, z) → z

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:

FUNCTION(p, s(s(x)), dummy, dummy2) → FUNCTION(p, s(x), x, x)

The TRS R consists of the following rules:

function(iszero, 0, dummy, dummy2) → true
function(iszero, s(x), dummy, dummy2) → false
function(p, 0, dummy, dummy2) → 0
function(p, s(0), dummy, dummy2) → 0
function(p, s(s(x)), dummy, dummy2) → s(function(p, s(x), x, x))
function(plus, dummy, x, y) → function(if, function(iszero, x, x, x), x, y)
function(if, true, x, y) → y
function(if, false, x, y) → function(plus, function(third, x, y, y), function(p, x, x, y), s(y))
function(third, x, y, z) → z

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.


FUNCTION(p, s(s(x)), dummy, dummy2) → FUNCTION(p, s(x), x, x)
The remaining pairs can at least be oriented weakly.
none
Used ordering: Polynomial interpretation [25,35]:

POL(p) = 0   
POL(s(x1)) = 4 + (2)x_1   
POL(FUNCTION(x1, x2, x3, x4)) = x_2 + (1/4)x_3 + (1/4)x_4   
The value of delta used in the strict ordering is 8.
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:

function(iszero, 0, dummy, dummy2) → true
function(iszero, s(x), dummy, dummy2) → false
function(p, 0, dummy, dummy2) → 0
function(p, s(0), dummy, dummy2) → 0
function(p, s(s(x)), dummy, dummy2) → s(function(p, s(x), x, x))
function(plus, dummy, x, y) → function(if, function(iszero, x, x, x), x, y)
function(if, true, x, y) → y
function(if, false, x, y) → function(plus, function(third, x, y, y), function(p, x, x, y), s(y))
function(third, x, y, z) → z

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:

FUNCTION(plus, dummy, x, y) → FUNCTION(if, function(iszero, x, x, x), x, y)
FUNCTION(if, false, x, y) → FUNCTION(plus, function(third, x, y, y), function(p, x, x, y), s(y))

The TRS R consists of the following rules:

function(iszero, 0, dummy, dummy2) → true
function(iszero, s(x), dummy, dummy2) → false
function(p, 0, dummy, dummy2) → 0
function(p, s(0), dummy, dummy2) → 0
function(p, s(s(x)), dummy, dummy2) → s(function(p, s(x), x, x))
function(plus, dummy, x, y) → function(if, function(iszero, x, x, x), x, y)
function(if, true, x, y) → y
function(if, false, x, y) → function(plus, function(third, x, y, y), function(p, x, x, y), s(y))
function(third, x, y, z) → z

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