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
         ↳Modular Removal of Rules
       →DP Problem 2
         ↳SCP


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)))





We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(PLUS(x1, x2))=  x1 + x2  
  POL(s(x1))=  x1  

We have the following set D of usable symbols: {PLUS, s}
No Dependency Pairs can be deleted.
7 non usable rules have been deleted.

The result of this processor delivers one new DP problem.



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
           →DP Problem 3
             ↳Modular Removal of Rules
       →DP Problem 2
         ↳SCP


Dependency Pairs:

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


Rule:

none





We have the following set of usable rules: none
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
  POL(PLUS(x1, x2))=  1 + x1 + x2  
  POL(s(x1))=  1 + x1  

We have the following set D of usable symbols: {PLUS, s}
The following Dependency Pairs can be deleted as the lhs is strictly greater than the corresponding rhs:

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

No Rules can be deleted.

After the removal, there are no SCCs in the dependency graph which results in no DP problems which have to be solved. 



   R
     ↳DPs
       →DP Problem 1
         ↳MRR
       →DP Problem 2
         ↳Size-Change Principle


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)))





We number the DPs as follows:
  1. ACK(s(x), s(y)) -> ACK(s(x), y)
  2. ACK(s(x), s(y)) -> ACK(x, plus(y, ack(s(x), y)))
  3. ACK(s(x), 0) -> ACK(x, s(0))
and get the following Size-Change Graph(s):
{2, 1} , {2, 1}
1=1
2>2
{2, 1} , {2, 1}
1>1
{3} , {3}
1>1

which lead(s) to this/these maximal multigraph(s):
{2, 1} , {2, 1}
1>1
{2, 1} , {2, 1}
1=1
2>2
{3} , {2, 1}
1>1
{2, 1} , {3}
1>1

DP: empty set
Oriented Rules: none

We used the order Homeomorphic Embedding Order with Non-Strict Precedence.
trivial

with Argument Filtering System:
s(x1) -> s(x1)

We obtain no new DP problems.

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