Termination w.r.t. Q of the given QTRS could not be shown:

(0) Obligation:

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.

(1) Overlay + Local Confluence (EQUIVALENT transformation)

The TRS is overlay and locally confluent. By [NOC] we can switch to innermost.

(2) Obligation:

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

The set Q consists of the following terms:

function(iszero, 0, x0, x1)
function(iszero, s(x0), x1, x2)
function(p, 0, x0, x1)
function(p, s(0), x0, x1)
function(p, s(s(x0)), x1, x2)
function(plus, x0, x1, x2)
function(if, true, x0, x1)
function(if, false, x0, x1)
function(third, x0, x1, x2)

(3) DependencyPairsProof (EQUIVALENT transformation)

Using Dependency Pairs [AG00,LPAR04] we result in the following initial DP problem.

(4) Obligation:

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

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

The set Q consists of the following terms:

function(iszero, 0, x0, x1)
function(iszero, s(x0), x1, x2)
function(p, 0, x0, x1)
function(p, s(0), x0, x1)
function(p, s(s(x0)), x1, x2)
function(plus, x0, x1, x2)
function(if, true, x0, x1)
function(if, false, x0, x1)
function(third, x0, x1, x2)

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

(5) DependencyGraphProof (EQUIVALENT transformation)

The approximation of the Dependency Graph [LPAR04,FROCOS05,EDGSTAR] contains 2 SCCs with 3 less nodes.

(6) Complex Obligation (AND)

(7) Obligation:

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

The set Q consists of the following terms:

function(iszero, 0, x0, x1)
function(iszero, s(x0), x1, x2)
function(p, 0, x0, x1)
function(p, s(0), x0, x1)
function(p, s(s(x0)), x1, x2)
function(plus, x0, x1, x2)
function(if, true, x0, x1)
function(if, false, x0, x1)
function(third, x0, x1, x2)

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

(8) QDPOrderProof (EQUIVALENT transformation)

We use the reduction pair processor [LPAR04].

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.
Used ordering: Combined order from the following AFS and order.
FUNCTION(x1, x2, x3, x4)  =  FUNCTION(x2)
p  =  p
s(x1)  =  s(x1)

Lexicographic path order with status [LPO].
Quasi-Precedence:

p > [FUNCTION₁, s₁]

Status:

p: []
FUNCTION₁: [1]
s₁: [1]

The following usable rules [FROCOS05] were oriented: none

(9) Obligation:

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

The set Q consists of the following terms:

function(iszero, 0, x0, x1)
function(iszero, s(x0), x1, x2)
function(p, 0, x0, x1)
function(p, s(0), x0, x1)
function(p, s(s(x0)), x1, x2)
function(plus, x0, x1, x2)
function(if, true, x0, x1)
function(if, false, x0, x1)
function(third, x0, x1, x2)

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

(10) PisEmptyProof (EQUIVALENT transformation)

The TRS P is empty. Hence, there is no (P,Q,R) chain.

(11) TRUE

(12) Obligation:

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

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

The set Q consists of the following terms:

function(iszero, 0, x0, x1)
function(iszero, s(x0), x1, x2)
function(p, 0, x0, x1)
function(p, s(0), x0, x1)
function(p, s(s(x0)), x1, x2)
function(plus, x0, x1, x2)
function(if, true, x0, x1)
function(if, false, x0, x1)
function(third, x0, x1, x2)

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