Term Rewriting System R:
[x, y, z]
times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

TIMES(x, plus(y, s(z))) -> PLUS(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
TIMES(x, plus(y, s(z))) -> TIMES(x, plus(y, times(s(z), 0)))
TIMES(x, plus(y, s(z))) -> PLUS(y, times(s(z), 0))
TIMES(x, plus(y, s(z))) -> TIMES(s(z), 0)
TIMES(x, plus(y, s(z))) -> TIMES(x, s(z))
TIMES(x, s(y)) -> PLUS(times(x, y), x)
TIMES(x, s(y)) -> TIMES(x, y)
PLUS(x, s(y)) -> PLUS(x, y)

Furthermore, R contains two SCCs. 


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


Dependency Pair:

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


Rules:


times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))





The following dependency pair can be strictly oriented:

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


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(PLUS(x1, x2))=  x2  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



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


Dependency Pair:


Rules:


times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))





Using the Dependency Graph resulted in no new DP problems.


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


Dependency Pairs:

TIMES(x, s(y)) -> TIMES(x, y)
TIMES(x, plus(y, s(z))) -> TIMES(x, s(z))
TIMES(x, plus(y, s(z))) -> TIMES(x, plus(y, times(s(z), 0)))


Rules:


times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))





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

TIMES(x, plus(y, s(z))) -> TIMES(x, plus(y, times(s(z), 0)))
one new Dependency Pair is created:

TIMES(x, plus(y, s(z'))) -> TIMES(x, plus(y, 0))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Narrowing Transformation


Dependency Pairs:

TIMES(x, plus(y, s(z'))) -> TIMES(x, plus(y, 0))
TIMES(x, plus(y, s(z))) -> TIMES(x, s(z))
TIMES(x, s(y)) -> TIMES(x, y)


Rules:


times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))





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

TIMES(x, plus(y, s(z'))) -> TIMES(x, plus(y, 0))
one new Dependency Pair is created:

TIMES(x, plus(y', s(z'))) -> TIMES(x, y')

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Nar
             ...
               →DP Problem 5
                 ↳Polynomial Ordering


Dependency Pairs:

TIMES(x, plus(y', s(z'))) -> TIMES(x, y')
TIMES(x, s(y)) -> TIMES(x, y)
TIMES(x, plus(y, s(z))) -> TIMES(x, s(z))


Rules:


times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))





The following dependency pairs can be strictly oriented:

TIMES(x, plus(y', s(z'))) -> TIMES(x, y')
TIMES(x, plus(y, s(z))) -> TIMES(x, s(z))


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(TIMES(x1, x2))=  1 + x2  
  POL(plus(x1, x2))=  1 + x1 + x2  
  POL(s(x1))=  x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Nar
             ...
               →DP Problem 6
                 ↳Polynomial Ordering


Dependency Pair:

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


Rules:


times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))





The following dependency pair can be strictly oriented:

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


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(TIMES(x1, x2))=  x2  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem. 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Nar
             ...
               →DP Problem 7
                 ↳Dependency Graph


Dependency Pair:


Rules:


times(x, plus(y, s(z))) -> plus(times(x, plus(y, times(s(z), 0))), times(x, s(z)))
times(x, 0) -> 0
times(x, s(y)) -> plus(times(x, y), x)
plus(x, 0) -> x
plus(x, s(y)) -> s(plus(x, y))





Using the Dependency Graph resulted in no new DP problems.

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