Term Rewriting System R:
[X, Y]
minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

MINUS(s(X), s(Y)) -> P(minus(X, Y))
MINUS(s(X), s(Y)) -> MINUS(X, Y)
DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))
DIV(s(X), s(Y)) -> MINUS(X, Y)

Furthermore, R contains two SCCs.

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

Dependency Pair:

MINUS(s(X), s(Y)) -> MINUS(X, Y)

Rules:

minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))

Strategy:

innermost

As we are in the innermost case, we can delete all 5 non-usable-rules.

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

Dependency Pair:

MINUS(s(X), s(Y)) -> MINUS(X, Y)

Rule:

none

Strategy:

innermost

We number the DPs as follows:
1. MINUS(s(X), s(Y)) -> MINUS(X, Y)
and get the following Size-Change Graph(s):
{1} , {1}
1>1
2>2

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

Dependency Pair:

DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))

Rules:

minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))
p(s(X)) -> X
div(0, s(Y)) -> 0
div(s(X), s(Y)) -> s(div(minus(X, Y), s(Y)))

Strategy:

innermost

As we are in the innermost case, we can delete all 2 non-usable-rules.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`           →DP Problem 4`
`             ↳Negative Polynomial Order`

Dependency Pair:

DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))

Rules:

p(s(X)) -> X
minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))

Strategy:

innermost

The following Dependency Pair can be strictly oriented using the given order.

DIV(s(X), s(Y)) -> DIV(minus(X, Y), s(Y))

Moreover, the following usable rules (regarding the implicit AFS) are oriented.

p(s(X)) -> X
minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))

Used ordering:
Polynomial Order with Interpretation:

POL( DIV(x1, x2) ) = x1

POL( s(x1) ) = x1 + 1

POL( minus(x1, x2) ) = x1

POL( p(x1) ) = x1

This results in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`           →DP Problem 4`
`             ↳Neg POLO`
`             ...`
`               →DP Problem 5`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

p(s(X)) -> X
minus(X, 0) -> X
minus(s(X), s(Y)) -> p(minus(X, Y))

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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