Termination w.r.t. Q proof of TRS/secret05aprove3.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:

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

  ↳ 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

Q is empty.

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

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:

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

Q is empty.
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

  ↳ DependencyPairsProof

    ↳ QDP

      ↳ DependencyGraphProof

        ↳ AND

          ↳ QDP

          ↳ 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

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

↳ 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

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