Term Rewriting System R:
[x, y]
p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

P(p(s(x))) -> P(x)
LE(p(s(x)), x) -> LE(x, x)
LE(s(x), s(y)) -> LE(x, y)
MINUS(x, y) -> IF(le(x, y), x, y)
MINUS(x, y) -> LE(x, y)
IF(false, x, y) -> MINUS(p(x), y)
IF(false, x, y) -> P(x)

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Nar


Dependency Pair:

LE(s(x), s(y)) -> LE(x, y)


Rules:


p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))


Strategy:

innermost




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

LE(s(x), s(y)) -> LE(x, y)
one new Dependency Pair is created:

LE(s(s(x'')), s(s(y''))) -> LE(s(x''), s(y''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳Forward Instantiation Transformation
       →DP Problem 2
         ↳Nar


Dependency Pair:

LE(s(s(x'')), s(s(y''))) -> LE(s(x''), s(y''))


Rules:


p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))


Strategy:

innermost




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

LE(s(s(x'')), s(s(y''))) -> LE(s(x''), s(y''))
one new Dependency Pair is created:

LE(s(s(s(x''''))), s(s(s(y'''')))) -> LE(s(s(x'''')), s(s(y'''')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 4
                 ↳Polynomial Ordering
       →DP Problem 2
         ↳Nar


Dependency Pair:

LE(s(s(s(x''''))), s(s(s(y'''')))) -> LE(s(s(x'''')), s(s(y'''')))


Rules:


p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

LE(s(s(s(x''''))), s(s(s(y'''')))) -> LE(s(s(x'''')), s(s(y'''')))


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(LE(x1, x2))=  1 + x1  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
           →DP Problem 3
             ↳FwdInst
             ...
               →DP Problem 5
                 ↳Dependency Graph
       →DP Problem 2
         ↳Nar


Dependency Pair:


Rules:


p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Narrowing Transformation


Dependency Pairs:

IF(false, x, y) -> MINUS(p(x), y)
MINUS(x, y) -> IF(le(x, y), x, y)


Rules:


p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))


Strategy:

innermost




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

MINUS(x, y) -> IF(le(x, y), x, y)
three new Dependency Pairs are created:

MINUS(0, y'') -> IF(true, 0, y'')
MINUS(s(x''), 0) -> IF(false, s(x''), 0)
MINUS(s(x''), s(y'')) -> IF(le(x'', y''), s(x''), s(y''))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Narrowing Transformation


Dependency Pairs:

MINUS(s(x''), s(y'')) -> IF(le(x'', y''), s(x''), s(y''))
MINUS(s(x''), 0) -> IF(false, s(x''), 0)
IF(false, x, y) -> MINUS(p(x), y)


Rules:


p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))


Strategy:

innermost




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

IF(false, x, y) -> MINUS(p(x), y)
two new Dependency Pairs are created:

IF(false, 0, y) -> MINUS(s(s(0)), y)
IF(false, s(x''), y) -> MINUS(x'', y)

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 7
                 ↳Narrowing Transformation


Dependency Pairs:

MINUS(s(x''), 0) -> IF(false, s(x''), 0)
IF(false, s(x''), y) -> MINUS(x'', y)
MINUS(s(x''), s(y'')) -> IF(le(x'', y''), s(x''), s(y''))


Rules:


p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))


Strategy:

innermost




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

MINUS(s(x''), s(y'')) -> IF(le(x'', y''), s(x''), s(y''))
three new Dependency Pairs are created:

MINUS(s(0), s(y''')) -> IF(true, s(0), s(y'''))
MINUS(s(s(x')), s(0)) -> IF(false, s(s(x')), s(0))
MINUS(s(s(x')), s(s(y'))) -> IF(le(x', y'), s(s(x')), s(s(y')))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 8
                 ↳Instantiation Transformation


Dependency Pairs:

MINUS(s(s(x')), s(s(y'))) -> IF(le(x', y'), s(s(x')), s(s(y')))
MINUS(s(s(x')), s(0)) -> IF(false, s(s(x')), s(0))
IF(false, s(x''), y) -> MINUS(x'', y)
MINUS(s(x''), 0) -> IF(false, s(x''), 0)


Rules:


p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))


Strategy:

innermost




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

IF(false, s(x''), y) -> MINUS(x'', y)
three new Dependency Pairs are created:

IF(false, s(x''''), 0) -> MINUS(x'''', 0)
IF(false, s(s(x'''')), s(0)) -> MINUS(s(x''''), s(0))
IF(false, s(s(x'''')), s(s(y'''))) -> MINUS(s(x''''), s(s(y''')))

The transformation is resulting in three new DP problems: 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 9
                 ↳Polynomial Ordering


Dependency Pairs:

IF(false, s(s(x'''')), s(s(y'''))) -> MINUS(s(x''''), s(s(y''')))
MINUS(s(s(x')), s(s(y'))) -> IF(le(x', y'), s(s(x')), s(s(y')))


Rules:


p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

IF(false, s(s(x'''')), s(s(y'''))) -> MINUS(s(x''''), s(s(y''')))


Additionally, the following usable rules for innermost can be oriented:

le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)


Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(false)=  0  
  POL(MINUS(x1, x2))=  x1  
  POL(true)=  0  
  POL(s(x1))=  1 + x1  
  POL(le(x1, x2))=  0  
  POL(IF(x1, x2, x3))=  x2  
  POL(p(x1))=  1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 12
                 ↳Dependency Graph


Dependency Pair:

MINUS(s(s(x')), s(s(y'))) -> IF(le(x', y'), s(s(x')), s(s(y')))


Rules:


p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 10
                 ↳Polynomial Ordering


Dependency Pairs:

IF(false, s(s(x'''')), s(0)) -> MINUS(s(x''''), s(0))
MINUS(s(s(x')), s(0)) -> IF(false, s(s(x')), s(0))


Rules:


p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

IF(false, s(s(x'''')), s(0)) -> MINUS(s(x''''), s(0))


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(false)=  0  
  POL(MINUS(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  
  POL(IF(x1, x2, x3))=  x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 13
                 ↳Dependency Graph


Dependency Pair:

MINUS(s(s(x')), s(0)) -> IF(false, s(s(x')), s(0))


Rules:


p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 11
                 ↳Polynomial Ordering


Dependency Pairs:

IF(false, s(x''''), 0) -> MINUS(x'''', 0)
MINUS(s(x''), 0) -> IF(false, s(x''), 0)


Rules:


p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))


Strategy:

innermost




The following dependency pair can be strictly oriented:

IF(false, s(x''''), 0) -> MINUS(x'''', 0)


There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(0)=  0  
  POL(false)=  0  
  POL(MINUS(x1, x2))=  x1  
  POL(s(x1))=  1 + x1  
  POL(IF(x1, x2, x3))=  x2  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳FwdInst
       →DP Problem 2
         ↳Nar
           →DP Problem 6
             ↳Nar
             ...
               →DP Problem 14
                 ↳Dependency Graph


Dependency Pair:

MINUS(s(x''), 0) -> IF(false, s(x''), 0)


Rules:


p(0) -> s(s(0))
p(s(x)) -> x
p(p(s(x))) -> p(x)
le(p(s(x)), x) -> le(x, x)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
minus(x, y) -> if(le(x, y), x, y)
if(true, x, y) -> 0
if(false, x, y) -> s(minus(p(x), y))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:00 minutes