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

Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

PLUS(s(s(x)), y) -> PLUS(x, s(y))
PLUS(x, s(s(y))) -> PLUS(s(x), y)
ACK(s(x), 0) -> ACK(x, s(0))
ACK(s(x), s(y)) -> ACK(x, plus(y, ack(s(x), y)))
ACK(s(x), s(y)) -> PLUS(y, ack(s(x), y))
ACK(s(x), s(y)) -> ACK(s(x), y)

Furthermore, R contains two SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Argument Filtering and Ordering
       →DP Problem 2
         ↳AFS


Dependency Pairs:

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


Rules:


plus(s(s(x)), y) -> s(plus(x, s(y)))
plus(x, s(s(y))) -> s(plus(s(x), y))
plus(s(0), y) -> s(y)
plus(0, y) -> y
ack(0, y) -> s(y)
ack(s(x), 0) -> ack(x, s(0))
ack(s(x), s(y)) -> ack(x, plus(y, ack(s(x), y)))





The following dependency pairs can be strictly oriented:

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


There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(PLUS(x1, x2))=  1 + x1 + x2  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.
Used Argument Filtering System:
PLUS(x1, x2) -> PLUS(x1, x2)
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
           →DP Problem 3
             ↳Dependency Graph
       →DP Problem 2
         ↳AFS


Dependency Pair:


Rules:


plus(s(s(x)), y) -> s(plus(x, s(y)))
plus(x, s(s(y))) -> s(plus(s(x), y))
plus(s(0), y) -> s(y)
plus(0, y) -> y
ack(0, y) -> s(y)
ack(s(x), 0) -> ack(x, s(0))
ack(s(x), s(y)) -> ack(x, plus(y, ack(s(x), y)))





Using the Dependency Graph resulted in no new DP problems.


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳Argument Filtering and Ordering


Dependency Pairs:

ACK(s(x), s(y)) -> ACK(s(x), y)
ACK(s(x), s(y)) -> ACK(x, plus(y, ack(s(x), y)))
ACK(s(x), 0) -> ACK(x, s(0))


Rules:


plus(s(s(x)), y) -> s(plus(x, s(y)))
plus(x, s(s(y))) -> s(plus(s(x), y))
plus(s(0), y) -> s(y)
plus(0, y) -> y
ack(0, y) -> s(y)
ack(s(x), 0) -> ack(x, s(0))
ack(s(x), s(y)) -> ack(x, plus(y, ack(s(x), y)))





The following dependency pairs can be strictly oriented:

ACK(s(x), s(y)) -> ACK(x, plus(y, ack(s(x), y)))
ACK(s(x), 0) -> ACK(x, s(0))


There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.
Used Argument Filtering System:
ACK(x1, x2) -> x1
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
           →DP Problem 4
             ↳Argument Filtering and Ordering


Dependency Pair:

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


Rules:


plus(s(s(x)), y) -> s(plus(x, s(y)))
plus(x, s(s(y))) -> s(plus(s(x), y))
plus(s(0), y) -> s(y)
plus(0, y) -> y
ack(0, y) -> s(y)
ack(s(x), 0) -> ack(x, s(0))
ack(s(x), s(y)) -> ack(x, plus(y, ack(s(x), y)))





The following dependency pair can be strictly oriented:

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


There are no usable rules w.r.t. to the AFS that need to be oriented.
Used ordering: Polynomial ordering with Polynomial interpretation:
  POL(ACK(x1, x2))=  x1 + x2  
  POL(s(x1))=  1 + x1  

resulting in one new DP problem.
Used Argument Filtering System:
ACK(x1, x2) -> ACK(x1, x2)
s(x1) -> s(x1)


   R
     ↳DPs
       →DP Problem 1
         ↳AFS
       →DP Problem 2
         ↳AFS
           →DP Problem 4
             ↳AFS
             ...
               →DP Problem 5
                 ↳Dependency Graph


Dependency Pair:


Rules:


plus(s(s(x)), y) -> s(plus(x, s(y)))
plus(x, s(s(y))) -> s(plus(s(x), y))
plus(s(0), y) -> s(y)
plus(0, y) -> y
ack(0, y) -> s(y)
ack(s(x), 0) -> ack(x, s(0))
ack(s(x), s(y)) -> ack(x, plus(y, ack(s(x), y)))





Using the Dependency Graph resulted in no new DP problems.

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