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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

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

Furthermore, R contains one SCC. 


   R
     ↳DPs
       →DP Problem 1
         ↳Narrowing Transformation


Dependency Pairs:

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


Rules:


ack(0, y) -> s(y)
ack(s(x), 0) -> ack(x, s(0))
ack(s(x), s(y)) -> ack(x, 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, ack(s(x), y))
two new Dependency Pairs are created:

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

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




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

ACK(s(x), 0) -> ACK(x, s(0))
two new Dependency Pairs are created:

ACK(s(s(x'')), 0) -> ACK(s(x''), s(0))
ACK(s(s(x'''')), 0) -> ACK(s(x''''), s(0))

The transformation is resulting in one new DP problem: 



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


Dependency Pairs:

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


Rules:


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


Strategy:

innermost




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

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

ACK(s(x''), s(s(y''))) -> ACK(s(x''), s(y''))
ACK(s(x'), s(s(0))) -> ACK(s(x'), s(0))
ACK(s(x'), s(s(s(y'''')))) -> ACK(s(x'), s(s(y'''')))
ACK(s(s(x'''')), s(0)) -> ACK(s(s(x'''')), 0)
ACK(s(s(x'''''')), s(0)) -> ACK(s(s(x'''''')), 0)

The transformation is resulting in one new DP problem: 



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




The following remains to be proven:
Dependency Pairs:

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


Rules:


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


Strategy:

innermost



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