Term Rewriting System R:
[x, y]
pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
log(s(0)) -> 0
log(s(s(x))) -> s(log(s(quot(x, s(s(0))))))

Termination of R to be shown.

`   R`
`     ↳Overlay and local confluence Check`

The TRS is overlay and locally confluent (all critical pairs are trivially joinable).Hence, we can switch to innermost.

`   R`
`     ↳OC`
`       →TRS2`
`         ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

MINUS(x, s(y)) -> PRED(minus(x, y))
MINUS(x, s(y)) -> MINUS(x, y)
QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))
QUOT(s(x), s(y)) -> MINUS(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`
`     ↳OC`
`       →TRS2`
`         ↳DPs`
`           →DP Problem 1`
`             ↳Usable Rules (Innermost)`
`           →DP Problem 2`
`             ↳UsableRules`
`           →DP Problem 3`
`             ↳UsableRules`

Dependency Pair:

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

Rules:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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 7 non-usable-rules.

`   R`
`     ↳OC`
`       →TRS2`
`         ↳DPs`
`           →DP Problem 1`
`             ↳UsableRules`
`             ...`
`               →DP Problem 4`
`                 ↳Size-Change Principle`
`           →DP Problem 2`
`             ↳UsableRules`
`           →DP Problem 3`
`             ↳UsableRules`

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:
1. MINUS(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`
`     ↳OC`
`       →TRS2`
`         ↳DPs`
`           →DP Problem 1`
`             ↳UsableRules`
`           →DP Problem 2`
`             ↳Usable Rules (Innermost)`
`           →DP Problem 3`
`             ↳UsableRules`

Dependency Pair:

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

Rules:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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`
`     ↳OC`
`       →TRS2`
`         ↳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(minus(x, y), s(y))

Rules:

minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
pred(s(x)) -> x

Strategy:

innermost

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

QUOT(s(x), s(y)) -> QUOT(minus(x, y), s(y))

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

minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
pred(s(x)) -> x

Used ordering:
Polynomial Order with Interpretation:

POL( QUOT(x1, x2) ) = x1

POL( s(x1) ) = x1 + 1

POL( minus(x1, x2) ) = x1

POL( pred(x1) ) = x1

This results in one new DP problem.

`   R`
`     ↳OC`
`       →TRS2`
`         ↳DPs`
`           →DP Problem 1`
`             ↳UsableRules`
`           →DP Problem 2`
`             ↳UsableRules`
`             ...`
`               →DP Problem 6`
`                 ↳Dependency Graph`
`           →DP Problem 3`
`             ↳UsableRules`

Dependency Pair:

Rules:

minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
pred(s(x)) -> x

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

`   R`
`     ↳OC`
`       →TRS2`
`         ↳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:

pred(s(x)) -> x
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(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`
`     ↳OC`
`       →TRS2`
`         ↳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(minus(x, y), s(y)))
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
pred(s(x)) -> 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(minus(x, y), s(y)))
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
pred(s(x)) -> 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( minus(x1, x2) ) = x1

POL( pred(x1) ) = x1

This results in one new DP problem.

`   R`
`     ↳OC`
`       →TRS2`
`         ↳DPs`
`           →DP Problem 1`
`             ↳UsableRules`
`           →DP Problem 2`
`             ↳UsableRules`
`           →DP Problem 3`
`             ↳UsableRules`
`             ...`
`               →DP Problem 8`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

quot(0, s(y)) -> 0
quot(s(x), s(y)) -> s(quot(minus(x, y), s(y)))
minus(x, 0) -> x
minus(x, s(y)) -> pred(minus(x, y))
pred(s(x)) -> x

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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