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

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

+'(x, s(y)) -> +'(x, y)
+'(s(x), y) -> +'(x, y)
+'(x, +(y, z)) -> +'(+(x, y), z)
+'(x, +(y, z)) -> +'(x, y)
F(g(f(x))) -> F(h(s(0), x))
F(g(h(x, y))) -> F(h(s(x), y))
F(h(x, h(y, z))) -> F(h(+(x, y), z))
F(h(x, h(y, z))) -> +'(x, y)

Furthermore, R contains two SCCs.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳Usable Rules (Innermost)`
`       →DP Problem 2`
`         ↳UsableRules`

Dependency Pairs:

+'(x, +(y, z)) -> +'(x, y)
+'(x, +(y, z)) -> +'(+(x, y), z)
+'(s(x), y) -> +'(x, y)
+'(x, s(y)) -> +'(x, y)

Rules:

+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(x, +(y, z)) -> +(+(x, y), z)
f(g(f(x))) -> f(h(s(0), x))
f(g(h(x, y))) -> f(h(s(x), y))
f(h(x, h(y, z))) -> f(h(+(x, y), z))

Strategy:

innermost

As we are in the innermost case, we can delete all 3 non-usable-rules.

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

Dependency Pairs:

+'(x, +(y, z)) -> +'(x, y)
+'(x, +(y, z)) -> +'(+(x, y), z)
+'(s(x), y) -> +'(x, y)
+'(x, s(y)) -> +'(x, y)

Rules:

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

Strategy:

innermost

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

which lead(s) to this/these maximal multigraph(s):
{1} , {1}
1=1
2>2
{3} , {3}
1>1
2=2
{4} , {4}
1=1
2>2
{3} , {2}
2>2
{3} , {4}
1>1
2>2
{4} , {3}
1>1
2>2
{2} , {3}
2>2
{1} , {3}
1>1
2>2
{3} , {1}
1>1
2>2
{2} , {1}
2>2
{1} , {1}
1>1
2>2
{3} , {3}
2>2
{4} , {4}
1>1
2>2
{1} , {4}
1>1
2>2
{4} , {3}
2>2
{3} , {3}
1>1
2>2
{1} , {3}
2>2
{2} , {4}
2>2
{4} , {4}
2>2
{1} , {4}
2>2
{1} , {1}
2>2
{3} , {1}
2>2
{3} , {4}
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`
`         ↳Usable Rules (Innermost)`

Dependency Pair:

F(h(x, h(y, z))) -> F(h(+(x, y), z))

Rules:

+(x, 0) -> x
+(x, s(y)) -> s(+(x, y))
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(x, +(y, z)) -> +(+(x, y), z)
f(g(f(x))) -> f(h(s(0), x))
f(g(h(x, y))) -> f(h(s(x), y))
f(h(x, h(y, z))) -> f(h(+(x, y), z))

Strategy:

innermost

As we are in the innermost case, we can delete all 3 non-usable-rules.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`           →DP Problem 4`
`             ↳Modular Removal of Rules`

Dependency Pair:

F(h(x, h(y, z))) -> F(h(+(x, y), z))

Rules:

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

Strategy:

innermost

We have the following set of usable rules:

+(x, 0) -> x
+(0, y) -> y
+(s(x), y) -> s(+(x, y))
+(x, s(y)) -> s(+(x, y))
+(x, +(y, z)) -> +(+(x, y), z)
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
 POL(0) =  1 POL(h(x1, x2)) =  x1 + x2 POL(s(x1)) =  x1 POL(F(x1)) =  1 + x1 POL(+(x1, x2)) =  x1 + x2

We have the following set D of usable symbols: {h, s, F, +}
No Dependency Pairs can be deleted.
The following rules can be deleted as they contain symbols in their lhs which do not occur in D:

+(x, 0) -> x
+(0, y) -> y

The result of this processor delivers one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`           →DP Problem 4`
`             ↳MRR`
`             ...`
`               →DP Problem 5`
`                 ↳Modular Removal of Rules`

Dependency Pair:

F(h(x, h(y, z))) -> F(h(+(x, y), z))

Rules:

+(s(x), y) -> s(+(x, y))
+(x, s(y)) -> s(+(x, y))
+(x, +(y, z)) -> +(+(x, y), z)

Strategy:

innermost

We have the following set of usable rules:

+(s(x), y) -> s(+(x, y))
+(x, s(y)) -> s(+(x, y))
+(x, +(y, z)) -> +(+(x, y), z)
To remove rules and DPs from this DP problem we used the following monotonic and CE-compatible order: Polynomial ordering.
Polynomial interpretation:
 POL(h(x1, x2)) =  1 + x1 + x2 POL(s(x1)) =  x1 POL(F(x1)) =  1 + x1 POL(+(x1, x2)) =  x1 + x2

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

F(h(x, h(y, z))) -> F(h(+(x, y), z))

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.

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