Termination Proof using AProVE

Term Rewriting System R:
[dummy, dummy2, x, y, z]
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

Innermost Termination of R to be shown.

     ↳Dependency Pair Analysis

R contains the following Dependency Pairs:

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)

Furthermore, R contains two SCCs.

     ↳DPs

       →DP Problem 1

         ↳Polynomial Ordering

       →DP Problem 2

         ↳Rw

Dependency Pair:

FUNCTION(p, s(s(x)), dummy, dummy2) -> FUNCTION(p, s(x), x, x)

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

FUNCTION(p, s(s(x)), dummy, dummy2) -> FUNCTION(p, s(x), x, x)

There are no usable rules for innermost w.r.t. to the implicit AFS that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:

POL(FUNCTION(x₁, x₂, x₃, x₄)) = 1 + x₂

POL(s(x₁)) = 1 + x₁

POL(p) = 0

resulting in one new DP problem.

     ↳DPs

       →DP Problem 1

         ↳Polo

           →DP Problem 3

             ↳Dependency Graph

       →DP Problem 2

         ↳Rw

Dependency Pair:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Rewriting Transformation

Dependency Pairs:

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)

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

Strategy:

innermost

On this DP problem, a Rewriting SCC transformation can be performed.
As a result of transforming the rule

FUNCTION(if, false, x, y) -> FUNCTION(plus, function(third, x, y, y), function(p, x, x, y), s(y))

one new Dependency Pair is created:

FUNCTION(if, false, x, y) -> FUNCTION(plus, y, function(p, x, x, y), s(y))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Rw

           →DP Problem 4

             ↳Narrowing Transformation

Dependency Pairs:

FUNCTION(if, false, x, y) -> FUNCTION(plus, y, function(p, x, x, y), s(y))
FUNCTION(plus, dummy, x, y) -> FUNCTION(if, function(iszero, x, x, x), x, y)

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

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

FUNCTION(plus, dummy, x, y) -> FUNCTION(if, function(iszero, x, x, x), x, y)

two new Dependency Pairs are created:

FUNCTION(plus, dummy, 0, y) -> FUNCTION(if, true, 0, y)
FUNCTION(plus, dummy, s(x''), y) -> FUNCTION(if, false, s(x''), y)

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Rw

           →DP Problem 4

             ↳Nar

...

               →DP Problem 5

                 ↳Narrowing Transformation

Dependency Pairs:

FUNCTION(plus, dummy, s(x''), y) -> FUNCTION(if, false, s(x''), y)
FUNCTION(if, false, x, y) -> FUNCTION(plus, y, function(p, x, x, y), s(y))

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

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

FUNCTION(if, false, x, y) -> FUNCTION(plus, y, function(p, x, x, y), s(y))

three new Dependency Pairs are created:

FUNCTION(if, false, 0, y') -> FUNCTION(plus, y', 0, s(y'))
FUNCTION(if, false, s(0), y') -> FUNCTION(plus, y', 0, s(y'))
FUNCTION(if, false, s(s(x'')), y') -> FUNCTION(plus, y', s(function(p, s(x''), x'', x'')), s(y'))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Rw

           →DP Problem 4

             ↳Nar

...

               →DP Problem 6

                 ↳Instantiation Transformation

Dependency Pairs:

FUNCTION(if, false, s(s(x'')), y') -> FUNCTION(plus, y', s(function(p, s(x''), x'', x'')), s(y'))
FUNCTION(plus, dummy, s(x''), y) -> FUNCTION(if, false, s(x''), y)

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

Strategy:

innermost

On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

FUNCTION(plus, dummy, s(x''), y) -> FUNCTION(if, false, s(x''), y)

one new Dependency Pair is created:

FUNCTION(plus, dummy', s(x'''), s(dummy')) -> FUNCTION(if, false, s(x'''), s(dummy'))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Rw

           →DP Problem 4

             ↳Nar

...

               →DP Problem 7

                 ↳Instantiation Transformation

Dependency Pairs:

FUNCTION(plus, dummy', s(x'''), s(dummy')) -> FUNCTION(if, false, s(x'''), s(dummy'))
FUNCTION(if, false, s(s(x'')), y') -> FUNCTION(plus, y', s(function(p, s(x''), x'', x'')), s(y'))

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

Strategy:

innermost

On this DP problem, an Instantiation SCC transformation can be performed.
As a result of transforming the rule

FUNCTION(if, false, s(s(x'')), y') -> FUNCTION(plus, y', s(function(p, s(x''), x'', x'')), s(y'))

one new Dependency Pair is created:

FUNCTION(if, false, s(s(x''')), s(dummy''')) -> FUNCTION(plus, s(dummy'''), s(function(p, s(x'''), x''', x''')), s(s(dummy''')))

The transformation is resulting in one new DP problem:

     ↳DPs

       →DP Problem 1

         ↳Polo

       →DP Problem 2

         ↳Rw

           →DP Problem 4

             ↳Nar

...

               →DP Problem 8

                 ↳Remaining Obligation(s)

The following remains to be proven:
Dependency Pairs:

FUNCTION(if, false, s(s(x''')), s(dummy''')) -> FUNCTION(plus, s(dummy'''), s(function(p, s(x'''), x''', x''')), s(s(dummy''')))
FUNCTION(plus, dummy', s(x'''), s(dummy')) -> FUNCTION(if, false, s(x'''), s(dummy'))

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

Strategy:

innermost

Innermost Termination of R could not be shown.
Duration:
0:00 minutes

POL(FUNCTION(x₁, x₂, x₃, x₄))	= 1 + x₂
POL(s(x₁))	= 1 + x₁
POL(p)	= 0