Term Rewriting System R:
[x, y]
minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(x, s(y)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

MINUS(s(x), s(y)) -> MINUS(x, y)
LE(s(x), s(y)) -> LE(x, y)
QUOT(x, s(y)) -> IFQUOT(le(s(y), x), x, s(y))
QUOT(x, s(y)) -> LE(s(y), x)
IFQUOT(true, x, y) -> QUOT(minus(x, y), y)
IFQUOT(true, x, y) -> MINUS(x, y)

Furthermore, R contains three SCCs.

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

Dependency Pair:

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

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(x, s(y)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0

Strategy:

innermost

As we are in the innermost case, we can delete all 8 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:

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)`
`       →DP Problem 3`
`         ↳UsableRules`

Dependency Pair:

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

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(x, s(y)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0

Strategy:

innermost

As we are in the innermost case, we can delete all 8 non-usable-rules.

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

Dependency Pair:

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

Rule:

none

Strategy:

innermost

We number the DPs as follows:
1. LE(s(x), s(y)) -> LE(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`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳Usable Rules (Innermost)`

Dependency Pairs:

IFQUOT(true, x, y) -> QUOT(minus(x, y), y)
QUOT(x, s(y)) -> IFQUOT(le(s(y), x), x, s(y))

Rules:

minus(x, 0) -> x
minus(s(x), s(y)) -> minus(x, y)
le(0, y) -> true
le(s(x), 0) -> false
le(s(x), s(y)) -> le(x, y)
quot(x, s(y)) -> ifquot(le(s(y), x), x, s(y))
ifquot(true, x, y) -> s(quot(minus(x, y), y))
ifquot(false, x, y) -> 0

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 3`
`         ↳UsableRules`
`           →DP Problem 6`
`             ↳Narrowing Transformation`

Dependency Pairs:

IFQUOT(true, x, y) -> QUOT(minus(x, y), y)
QUOT(x, s(y)) -> IFQUOT(le(s(y), x), x, s(y))

Rules:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
le(s(x), s(y)) -> le(x, y)
le(0, y) -> true
le(s(x), 0) -> false

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

QUOT(x, s(y)) -> IFQUOT(le(s(y), x), x, s(y))
two new Dependency Pairs are created:

QUOT(s(y''), s(y0)) -> IFQUOT(le(y0, y''), s(y''), s(y0))
QUOT(0, s(y')) -> IFQUOT(false, 0, s(y'))

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 7`
`                 ↳Narrowing Transformation`

Dependency Pairs:

QUOT(s(y''), s(y0)) -> IFQUOT(le(y0, y''), s(y''), s(y0))
IFQUOT(true, x, y) -> QUOT(minus(x, y), y)

Rules:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
le(s(x), s(y)) -> le(x, y)
le(0, y) -> true
le(s(x), 0) -> false

Strategy:

innermost

On this DP problem, a Narrowing SCC transformation can be performed.
As a result of transforming the rule

IFQUOT(true, x, y) -> QUOT(minus(x, y), y)
two new Dependency Pairs are created:

IFQUOT(true, s(x''), s(y'')) -> QUOT(minus(x'', y''), s(y''))
IFQUOT(true, x'', 0) -> QUOT(x'', 0)

The transformation is resulting in one new DP problem:

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 8`
`                 ↳Negative Polynomial Order`

Dependency Pairs:

IFQUOT(true, s(x''), s(y'')) -> QUOT(minus(x'', y''), s(y''))
QUOT(s(y''), s(y0)) -> IFQUOT(le(y0, y''), s(y''), s(y0))

Rules:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
le(s(x), s(y)) -> le(x, y)
le(0, y) -> true
le(s(x), 0) -> false

Strategy:

innermost

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

IFQUOT(true, s(x''), s(y'')) -> QUOT(minus(x'', y''), s(y''))

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

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
le(s(x), s(y)) -> le(x, y)
le(0, y) -> true
le(s(x), 0) -> false

Used ordering:
Polynomial Order with Interpretation:

POL( IFQUOT(x1, ..., x3) ) = x2

POL( s(x1) ) = x1 + 1

POL( QUOT(x1, x2) ) = x1

POL( minus(x1, x2) ) = x1

POL( le(x1, x2) ) = 0

POL( true ) = 0

POL( false ) = 0

This results in one new DP problem.

`   R`
`     ↳DPs`
`       →DP Problem 1`
`         ↳UsableRules`
`       →DP Problem 2`
`         ↳UsableRules`
`       →DP Problem 3`
`         ↳UsableRules`
`           →DP Problem 6`
`             ↳Nar`
`             ...`
`               →DP Problem 9`
`                 ↳Dependency Graph`

Dependency Pair:

QUOT(s(y''), s(y0)) -> IFQUOT(le(y0, y''), s(y''), s(y0))

Rules:

minus(s(x), s(y)) -> minus(x, y)
minus(x, 0) -> x
le(s(x), s(y)) -> le(x, y)
le(0, y) -> true
le(s(x), 0) -> false

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

Innermost Termination of R successfully shown.
Duration:
0:06 minutes