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

Innermost Termination of R to be shown. 



   R
     ↳Dependency Pair Analysis



R contains the following Dependency Pairs:

F(s(0), y, z) -> F(0, s(y), s(z))
F(s(x), s(y), 0) -> F(x, y, s(0))
F(s(x), 0, s(z)) -> F(x, s(0), z)
F(s(x), s(y), s(z)) -> F(x, y, f(s(x), s(y), z))
F(s(x), s(y), s(z)) -> F(s(x), s(y), z)
F(0, s(s(y)), s(0)) -> F(0, y, s(0))
F(0, s(0), s(s(z))) -> F(0, s(0), z)
F(0, s(s(y)), s(s(z))) -> F(0, y, f(0, s(s(y)), s(z)))
F(0, s(s(y)), s(s(z))) -> F(0, s(s(y)), s(z))

Furthermore, R contains four SCCs. 


   R
     ↳DPs
       →DP Problem 1
         ↳Usable Rules (Innermost)
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳SCP


Dependency Pair:

F(0, s(s(y)), s(0)) -> F(0, y, s(0))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




As we are in the innermost case, we can delete all 11 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
           →DP Problem 5
             ↳Size-Change Principle
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳SCP


Dependency Pair:

F(0, s(s(y)), s(0)) -> F(0, y, s(0))


Rule:

none


Strategy:

innermost




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

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

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. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳Usable Rules (Innermost)
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳SCP


Dependency Pair:

F(0, s(0), s(s(z))) -> F(0, s(0), z)


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




As we are in the innermost case, we can delete all 11 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
           →DP Problem 6
             ↳Size-Change Principle
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳SCP


Dependency Pair:

F(0, s(0), s(s(z))) -> F(0, s(0), z)


Rule:

none


Strategy:

innermost




We number the DPs as follows:
  1. F(0, s(0), s(s(z))) -> F(0, s(0), z)
and get the following Size-Change Graph(s):
{1} , {1}
1=1
2>1
2=2
3>3

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

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. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳Usable Rules (Innermost)
       →DP Problem 4
         ↳SCP


Dependency Pairs:

F(0, s(s(y)), s(s(z))) -> F(0, s(s(y)), s(z))
F(0, s(s(y)), s(s(z))) -> F(0, y, f(0, s(s(y)), s(z)))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




As we are in the innermost case, we can delete all 4 non-usable-rules.


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
           →DP Problem 7
             ↳Size-Change Principle
       →DP Problem 4
         ↳SCP


Dependency Pairs:

F(0, s(s(y)), s(s(z))) -> F(0, s(s(y)), s(z))
F(0, s(s(y)), s(s(z))) -> F(0, y, f(0, s(s(y)), s(z)))


Rules:


f(0, s(0), s(0)) -> s(s(0))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))
f(0, 0, z) -> s(z)
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, y, 0) -> s(y)
f(x, 0, 0) -> s(x)


Strategy:

innermost




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

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

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. 


   R
     ↳DPs
       →DP Problem 1
         ↳UsableRules
       →DP Problem 2
         ↳UsableRules
       →DP Problem 3
         ↳UsableRules
       →DP Problem 4
         ↳Size-Change Principle


Dependency Pairs:

F(s(x), s(y), s(z)) -> F(s(x), s(y), z)
F(s(x), s(y), s(z)) -> F(x, y, f(s(x), s(y), z))
F(s(x), 0, s(z)) -> F(x, s(0), z)
F(s(x), s(y), 0) -> F(x, y, s(0))


Rules:


f(x, 0, 0) -> s(x)
f(0, y, 0) -> s(y)
f(0, 0, z) -> s(z)
f(s(0), y, z) -> f(0, s(y), s(z))
f(s(x), s(y), 0) -> f(x, y, s(0))
f(s(x), 0, s(z)) -> f(x, s(0), z)
f(0, s(0), s(0)) -> s(s(0))
f(s(x), s(y), s(z)) -> f(x, y, f(s(x), s(y), z))
f(0, s(s(y)), s(0)) -> f(0, y, s(0))
f(0, s(0), s(s(z))) -> f(0, s(0), z)
f(0, s(s(y)), s(s(z))) -> f(0, y, f(0, s(s(y)), s(z)))


Strategy:

innermost




We number the DPs as follows:
  1. F(s(x), s(y), s(z)) -> F(s(x), s(y), z)
  2. F(s(x), s(y), s(z)) -> F(x, y, f(s(x), s(y), z))
  3. F(s(x), 0, s(z)) -> F(x, s(0), z)
  4. F(s(x), s(y), 0) -> F(x, y, s(0))
and get the following Size-Change Graph(s):
{1} , {1}
1=1
2=2
3>3
{2} , {2}
1>1
2>2
{3} , {3}
1>1
3>3
{4} , {4}
1>1
2>2

which lead(s) to this/these maximal multigraph(s):
{2} , {2}
1>1
2>2
{1} , {1}
1=1
2=2
3>3
{2} , {3}
1>1
{4} , {1}
1>1
2>2
{1} , {4}
1>1
2>2
{2} , {1}
1>1
2>2
{4} , {3}
1>1
{1} , {2}
1>1
2>2
{3} , {2}
1>1
{3} , {4}
1>1
{2} , {4}
1>1
{2} , {1}
1>1
{4} , {1}
1>1
{1} , {1}
1>1
2>2
{1} , {3}
1>1
{2} , {2}
1>1
{2} , {4}
1>1
2>2
{1} , {2}
1>1
{1} , {4}
1>1
{1} , {1}
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.

Innermost Termination of R successfully shown.
Duration:
0:01 minutes