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)))

Innermost 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
         ↳Polynomial Ordering
       →DP Problem 2
         ↳Nar


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)))


Strategy:

innermost




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 for innermost 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. 



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


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)))


Strategy:

innermost




Using the Dependency Graph resulted in no new DP problems.


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


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)))


Strategy:

innermost




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

ACK(s(x), s(y)) -> ACK(x, plus(y, ack(s(x), y)))
five new Dependency Pairs are created:

ACK(s(x0), s(s(s(x'')))) -> ACK(x0, s(plus(x'', s(ack(s(x0), s(s(x'')))))))
ACK(s(x'), s(s(0))) -> ACK(x', s(ack(s(x'), s(0))))
ACK(s(x'), s(0)) -> ACK(x', ack(s(x'), 0))
ACK(s(x''), s(0)) -> ACK(x'', plus(0, ack(x'', s(0))))
ACK(s(x''), s(s(y''))) -> ACK(x'', plus(s(y''), ack(x'', plus(y'', ack(s(x''), y'')))))

The transformation is resulting in one new DP problem: 



   R
     ↳DPs
       →DP Problem 1
         ↳Polo
       →DP Problem 2
         ↳Nar
           →DP Problem 4
             ↳Remaining Obligation(s)




The following remains to be proven:
Dependency Pairs:

ACK(s(x''), s(s(y''))) -> ACK(x'', plus(s(y''), ack(x'', plus(y'', ack(s(x''), y'')))))
ACK(s(x''), s(0)) -> ACK(x'', plus(0, ack(x'', s(0))))
ACK(s(x'), s(s(0))) -> ACK(x', s(ack(s(x'), s(0))))
ACK(s(x0), s(s(s(x'')))) -> ACK(x0, s(plus(x'', s(ack(s(x0), s(s(x'')))))))
ACK(s(x'), s(0)) -> ACK(x', ack(s(x'), 0))
ACK(s(x), 0) -> ACK(x, s(0))
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)))


Strategy:

innermost



Innermost Termination of R could not be shown.
Duration:
0:02 minutes