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:

function4(iszero, 0, dummy, dummy2) -> true
function4(iszero, s1(x), dummy, dummy2) -> false
function4(p, 0, dummy, dummy2) -> 0
function4(p, s1(0), dummy, dummy2) -> 0
function4(p, s1(s1(x)), dummy, dummy2) -> s1(function4(p, s1(x), x, x))
function4(plus, dummy, x, y) -> function4(if, function4(iszero, x, x, x), x, y)
function4(if, true, x, y) -> y
function4(if, false, x, y) -> function4(plus, function4(third, x, y, y), function4(p, x, x, y), s1(y))
function4(third, x, y, z) -> z

Q is empty.


↳ QTRS
  ↳ Non-Overlap Check

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

function4(iszero, 0, dummy, dummy2) -> true
function4(iszero, s1(x), dummy, dummy2) -> false
function4(p, 0, dummy, dummy2) -> 0
function4(p, s1(0), dummy, dummy2) -> 0
function4(p, s1(s1(x)), dummy, dummy2) -> s1(function4(p, s1(x), x, x))
function4(plus, dummy, x, y) -> function4(if, function4(iszero, x, x, x), x, y)
function4(if, true, x, y) -> y
function4(if, false, x, y) -> function4(plus, function4(third, x, y, y), function4(p, x, x, y), s1(y))
function4(third, x, y, z) -> z

Q is empty.

The TRS is non-overlapping. Hence, we can switch to innermost.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof

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

function4(iszero, 0, dummy, dummy2) -> true
function4(iszero, s1(x), dummy, dummy2) -> false
function4(p, 0, dummy, dummy2) -> 0
function4(p, s1(0), dummy, dummy2) -> 0
function4(p, s1(s1(x)), dummy, dummy2) -> s1(function4(p, s1(x), x, x))
function4(plus, dummy, x, y) -> function4(if, function4(iszero, x, x, x), x, y)
function4(if, true, x, y) -> y
function4(if, false, x, y) -> function4(plus, function4(third, x, y, y), function4(p, x, x, y), s1(y))
function4(third, x, y, z) -> z

The set Q consists of the following terms:

function4(iszero, 0, x0, x1)
function4(iszero, s1(x0), x1, x2)
function4(p, 0, x0, x1)
function4(p, s1(0), x0, x1)
function4(p, s1(s1(x0)), x1, x2)
function4(plus, x0, x1, x2)
function4(if, true, x0, x1)
function4(if, false, x0, x1)
function4(third, x0, x1, x2)


Using Dependency Pairs [1,13] we result in the following initial DP problem:
Q DP problem:
The TRS P consists of the following rules:

FUNCTION4(plus, dummy, x, y) -> FUNCTION4(if, function4(iszero, x, x, x), x, y)
FUNCTION4(if, false, x, y) -> FUNCTION4(plus, function4(third, x, y, y), function4(p, x, x, y), s1(y))
FUNCTION4(plus, dummy, x, y) -> FUNCTION4(iszero, x, x, x)
FUNCTION4(if, false, x, y) -> FUNCTION4(p, x, x, y)
FUNCTION4(p, s1(s1(x)), dummy, dummy2) -> FUNCTION4(p, s1(x), x, x)
FUNCTION4(if, false, x, y) -> FUNCTION4(third, x, y, y)

The TRS R consists of the following rules:

function4(iszero, 0, dummy, dummy2) -> true
function4(iszero, s1(x), dummy, dummy2) -> false
function4(p, 0, dummy, dummy2) -> 0
function4(p, s1(0), dummy, dummy2) -> 0
function4(p, s1(s1(x)), dummy, dummy2) -> s1(function4(p, s1(x), x, x))
function4(plus, dummy, x, y) -> function4(if, function4(iszero, x, x, x), x, y)
function4(if, true, x, y) -> y
function4(if, false, x, y) -> function4(plus, function4(third, x, y, y), function4(p, x, x, y), s1(y))
function4(third, x, y, z) -> z

The set Q consists of the following terms:

function4(iszero, 0, x0, x1)
function4(iszero, s1(x0), x1, x2)
function4(p, 0, x0, x1)
function4(p, s1(0), x0, x1)
function4(p, s1(s1(x0)), x1, x2)
function4(plus, x0, x1, x2)
function4(if, true, x0, x1)
function4(if, false, x0, x1)
function4(third, x0, x1, x2)

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

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof

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

FUNCTION4(plus, dummy, x, y) -> FUNCTION4(if, function4(iszero, x, x, x), x, y)
FUNCTION4(if, false, x, y) -> FUNCTION4(plus, function4(third, x, y, y), function4(p, x, x, y), s1(y))
FUNCTION4(plus, dummy, x, y) -> FUNCTION4(iszero, x, x, x)
FUNCTION4(if, false, x, y) -> FUNCTION4(p, x, x, y)
FUNCTION4(p, s1(s1(x)), dummy, dummy2) -> FUNCTION4(p, s1(x), x, x)
FUNCTION4(if, false, x, y) -> FUNCTION4(third, x, y, y)

The TRS R consists of the following rules:

function4(iszero, 0, dummy, dummy2) -> true
function4(iszero, s1(x), dummy, dummy2) -> false
function4(p, 0, dummy, dummy2) -> 0
function4(p, s1(0), dummy, dummy2) -> 0
function4(p, s1(s1(x)), dummy, dummy2) -> s1(function4(p, s1(x), x, x))
function4(plus, dummy, x, y) -> function4(if, function4(iszero, x, x, x), x, y)
function4(if, true, x, y) -> y
function4(if, false, x, y) -> function4(plus, function4(third, x, y, y), function4(p, x, x, y), s1(y))
function4(third, x, y, z) -> z

The set Q consists of the following terms:

function4(iszero, 0, x0, x1)
function4(iszero, s1(x0), x1, x2)
function4(p, 0, x0, x1)
function4(p, s1(0), x0, x1)
function4(p, s1(s1(x0)), x1, x2)
function4(plus, x0, x1, x2)
function4(if, true, x0, x1)
function4(if, false, x0, x1)
function4(third, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
The approximation of the Dependency Graph [13,14,18] contains 2 SCCs with 3 less nodes.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
              ↳ QDP

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

FUNCTION4(p, s1(s1(x)), dummy, dummy2) -> FUNCTION4(p, s1(x), x, x)

The TRS R consists of the following rules:

function4(iszero, 0, dummy, dummy2) -> true
function4(iszero, s1(x), dummy, dummy2) -> false
function4(p, 0, dummy, dummy2) -> 0
function4(p, s1(0), dummy, dummy2) -> 0
function4(p, s1(s1(x)), dummy, dummy2) -> s1(function4(p, s1(x), x, x))
function4(plus, dummy, x, y) -> function4(if, function4(iszero, x, x, x), x, y)
function4(if, true, x, y) -> y
function4(if, false, x, y) -> function4(plus, function4(third, x, y, y), function4(p, x, x, y), s1(y))
function4(third, x, y, z) -> z

The set Q consists of the following terms:

function4(iszero, 0, x0, x1)
function4(iszero, s1(x0), x1, x2)
function4(p, 0, x0, x1)
function4(p, s1(0), x0, x1)
function4(p, s1(s1(x0)), x1, x2)
function4(plus, x0, x1, x2)
function4(if, true, x0, x1)
function4(if, false, x0, x1)
function4(third, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
We use the reduction pair processor [13].


The following pairs can be strictly oriented and are deleted.


FUNCTION4(p, s1(s1(x)), dummy, dummy2) -> FUNCTION4(p, s1(x), x, x)
The remaining pairs can at least by weakly be oriented.
none
Used ordering: Combined order from the following AFS and order.
FUNCTION4(x1, x2, x3, x4)  =  FUNCTION4(x1, x2, x3, x4)
p  =  p
s1(x1)  =  s1(x1)

Lexicographic Path Order [19].
Precedence:
s1 > FUNCTION4
s1 > p

The following usable rules [14] were oriented: none



↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
                ↳ QDPOrderProof
                  ↳ QDP
                    ↳ PisEmptyProof
              ↳ QDP

Q DP problem:
P is empty.
The TRS R consists of the following rules:

function4(iszero, 0, dummy, dummy2) -> true
function4(iszero, s1(x), dummy, dummy2) -> false
function4(p, 0, dummy, dummy2) -> 0
function4(p, s1(0), dummy, dummy2) -> 0
function4(p, s1(s1(x)), dummy, dummy2) -> s1(function4(p, s1(x), x, x))
function4(plus, dummy, x, y) -> function4(if, function4(iszero, x, x, x), x, y)
function4(if, true, x, y) -> y
function4(if, false, x, y) -> function4(plus, function4(third, x, y, y), function4(p, x, x, y), s1(y))
function4(third, x, y, z) -> z

The set Q consists of the following terms:

function4(iszero, 0, x0, x1)
function4(iszero, s1(x0), x1, x2)
function4(p, 0, x0, x1)
function4(p, s1(0), x0, x1)
function4(p, s1(s1(x0)), x1, x2)
function4(plus, x0, x1, x2)
function4(if, true, x0, x1)
function4(if, false, x0, x1)
function4(third, x0, x1, x2)

We have to consider all minimal (P,Q,R)-chains.
The TRS P is empty. Hence, there is no (P,Q,R) chain.

↳ QTRS
  ↳ Non-Overlap Check
    ↳ QTRS
      ↳ DependencyPairsProof
        ↳ QDP
          ↳ DependencyGraphProof
            ↳ AND
              ↳ QDP
              ↳ QDP

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

FUNCTION4(plus, dummy, x, y) -> FUNCTION4(if, function4(iszero, x, x, x), x, y)
FUNCTION4(if, false, x, y) -> FUNCTION4(plus, function4(third, x, y, y), function4(p, x, x, y), s1(y))

The TRS R consists of the following rules:

function4(iszero, 0, dummy, dummy2) -> true
function4(iszero, s1(x), dummy, dummy2) -> false
function4(p, 0, dummy, dummy2) -> 0
function4(p, s1(0), dummy, dummy2) -> 0
function4(p, s1(s1(x)), dummy, dummy2) -> s1(function4(p, s1(x), x, x))
function4(plus, dummy, x, y) -> function4(if, function4(iszero, x, x, x), x, y)
function4(if, true, x, y) -> y
function4(if, false, x, y) -> function4(plus, function4(third, x, y, y), function4(p, x, x, y), s1(y))
function4(third, x, y, z) -> z

The set Q consists of the following terms:

function4(iszero, 0, x0, x1)
function4(iszero, s1(x0), x1, x2)
function4(p, 0, x0, x1)
function4(p, s1(0), x0, x1)
function4(p, s1(s1(x0)), x1, x2)
function4(plus, x0, x1, x2)
function4(if, true, x0, x1)
function4(if, false, x0, x1)
function4(third, x0, x1, x2)

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