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

Termination of R to be shown. 



   R
     ↳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. 


   R
     ↳DPs
       →DP Problem 1
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Inst


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





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 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. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
           →DP Problem 3
             ↳Dependency Graph
       →DP Problem 2
         ↳Inst


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





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Instantiation 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





On this DP problem, an Instantiation 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)
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Inst
           →DP Problem 4
             ↳Instantiation Transformation


Dependency Pairs:

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





On this DP problem, an Instantiation 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', s(y'''')) -> FUNCTION(plus, function(third, x', s(y''''), s(y'''')), function(p, x', x', s(y'''')), s(s(y'''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Inst
           →DP Problem 4
             ↳Inst
             ...
               →DP Problem 5
                 ↳Instantiation Transformation


Dependency Pairs:

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





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

FUNCTION(plus, dummy', x', s(y'')) -> FUNCTION(if, function(iszero, x', x', x'), x', s(y''))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Inst
           →DP Problem 4
             ↳Inst
             ...
               →DP Problem 6
                 ↳Instantiation Transformation


Dependency Pairs:

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





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

FUNCTION(if, false, x', s(y'''')) -> FUNCTION(plus, function(third, x', s(y''''), s(y'''')), function(p, x', x', s(y'''')), s(s(y'''')))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Inst
           →DP Problem 4
             ↳Inst
             ...
               →DP Problem 7
                 ↳Instantiation Transformation


Dependency Pairs:

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





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

FUNCTION(plus, dummy'', x'', s(s(y''''''))) -> FUNCTION(if, function(iszero, x'', x'', x''), x'', s(s(y'''''')))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Inst
           →DP Problem 4
             ↳Inst
             ...
               →DP Problem 8
                 ↳Instantiation Transformation


Dependency Pairs:

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





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

FUNCTION(if, false, x'', s(s(y''''''''))) -> FUNCTION(plus, function(third, x'', s(s(y'''''''')), s(s(y''''''''))), function(p, x'', x'', s(s(y''''''''))), s(s(s(y''''''''))))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Inst
           →DP Problem 4
             ↳Inst
             ...
               →DP Problem 9
                 ↳Instantiation Transformation


Dependency Pairs:

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





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

FUNCTION(plus, dummy''', x'''', s(s(s(y'''''''''')))) -> FUNCTION(if, function(iszero, x'''', x'''', x''''), x'''', s(s(s(y''''''''''))))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Inst
           →DP Problem 4
             ↳Inst
             ...
               →DP Problem 10
                 ↳Instantiation Transformation


Dependency Pairs:

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





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

FUNCTION(if, false, x''', s(s(s(y'''''''''''')))) -> FUNCTION(plus, function(third, x''', s(s(s(y''''''''''''))), s(s(s(y'''''''''''')))), function(p, x''', x''', s(s(s(y'''''''''''')))), s(s(s(s(y'''''''''''')))))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Inst
           →DP Problem 4
             ↳Inst
             ...
               →DP Problem 11
                 ↳Instantiation Transformation


Dependency Pairs:

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





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

FUNCTION(plus, dummy'''', x''''', s(s(s(s(y''''''''''''''))))) -> FUNCTION(if, function(iszero, x''''', x''''', x'''''), x''''', s(s(s(s(y'''''''''''''')))))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Inst
           →DP Problem 4
             ↳Inst
             ...
               →DP Problem 12
                 ↳Instantiation Transformation


Dependency Pairs:

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





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

FUNCTION(if, false, x'''', s(s(s(s(y''''''''''''''''))))) -> FUNCTION(plus, function(third, x'''', s(s(s(s(y'''''''''''''''')))), s(s(s(s(y''''''''''''''''))))), function(p, x'''', x'''', s(s(s(s(y''''''''''''''''))))), s(s(s(s(s(y''''''''''''''''))))))
one new Dependency Pair is created:

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

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Inst
           →DP Problem 4
             ↳Inst
             ...
               →DP Problem 13
                 ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

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




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