Term Rewriting System R:
[X, Y]
min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

MIN(s(X), s(Y)) -> MIN(X, Y)
QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))
QUOT(s(X), s(Y)) -> MIN(X, Y)
LOG(s(s(X))) -> LOG(s(quot(X, s(s(0)))))
LOG(s(s(X))) -> QUOT(X, s(s(0)))

Furthermore, R contains three SCCs.

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

Dependency Pair:

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

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

Strategy:

innermost

As we are in the innermost case, we can delete all 6 non-usable-rules.

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

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:
1. MIN(s(X), s(Y)) -> MIN(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)`
`       →DP Problem 3`
`         ↳UsableRules`

Dependency Pair:

QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

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 5`
`             ↳Negative Polynomial Order`
`       →DP Problem 3`
`         ↳UsableRules`

Dependency Pair:

QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))

Rules:

min(s(X), s(Y)) -> min(X, Y)
min(X, 0) -> X

Strategy:

innermost

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

QUOT(s(X), s(Y)) -> QUOT(min(X, Y), s(Y))

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

min(s(X), s(Y)) -> min(X, Y)
min(X, 0) -> X

Used ordering:
Polynomial Order with Interpretation:

POL( QUOT(x1, x2) ) = x1

POL( s(x1) ) = x1 + 1

POL( min(x1, x2) ) = x1

This results in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`           →DP Problem 5`
`             ↳Neg POLO`
`             ...`
`               →DP Problem 6`
`                 ↳Dependency Graph`
`       →DP Problem 3`
`         ↳UsableRules`

Dependency Pair:

Rules:

min(s(X), s(Y)) -> min(X, Y)
min(X, 0) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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

Dependency Pair:

LOG(s(s(X))) -> LOG(s(quot(X, s(s(0)))))

Rules:

min(X, 0) -> X
min(s(X), s(Y)) -> min(X, Y)
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
log(s(0)) -> 0
log(s(s(X))) -> s(log(s(quot(X, s(s(0))))))

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 3`
`         ↳UsableRules`
`           →DP Problem 7`
`             ↳Negative Polynomial Order`

Dependency Pair:

LOG(s(s(X))) -> LOG(s(quot(X, s(s(0)))))

Rules:

quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
min(s(X), s(Y)) -> min(X, Y)
min(X, 0) -> X

Strategy:

innermost

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

LOG(s(s(X))) -> LOG(s(quot(X, s(s(0)))))

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

quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
min(s(X), s(Y)) -> min(X, Y)
min(X, 0) -> X

Used ordering:
Polynomial Order with Interpretation:

POL( LOG(x1) ) = x1

POL( s(x1) ) = x1 + 1

POL( quot(x1, x2) ) = x1

POL( 0 ) = 0

POL( min(x1, x2) ) = x1

This results in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`           →DP Problem 7`
`             ↳Neg POLO`
`             ...`
`               →DP Problem 8`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(min(X, Y), s(Y)))
min(s(X), s(Y)) -> min(X, Y)
min(X, 0) -> X

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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