Term Rewriting System R:
[Y, X]
le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Innermost Termination of R to be shown.

`   R`
`     ↳Dependency Pair Analysis`

R contains the following Dependency Pairs:

LE(s(X), s(Y)) -> LE(X, Y)
MINUS(s(X), Y) -> IFMINUS(le(s(X), Y), s(X), Y)
MINUS(s(X), Y) -> LE(s(X), Y)
IFMINUS(false, s(X), Y) -> MINUS(X, Y)
QUOT(s(X), s(Y)) -> QUOT(minus(X, Y), s(Y))
QUOT(s(X), s(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:

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

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

Strategy:

innermost

As we are in the innermost case, we can delete all 9 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:

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`
`         ↳Usable Rules (Innermost)`
`       →DP Problem 3`
`         ↳UsableRules`

Dependency Pairs:

IFMINUS(false, s(X), Y) -> MINUS(X, Y)
MINUS(s(X), Y) -> IFMINUS(le(s(X), Y), s(X), Y)

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(minus(X, Y), s(Y)))

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

Dependency Pairs:

IFMINUS(false, s(X), Y) -> MINUS(X, Y)
MINUS(s(X), Y) -> IFMINUS(le(s(X), Y), s(X), Y)

Rules:

le(s(X), 0) -> false
le(0, Y) -> true
le(s(X), s(Y)) -> le(X, Y)

Strategy:

innermost

We number the DPs as follows:
1. IFMINUS(false, s(X), Y) -> MINUS(X, Y)
2. MINUS(s(X), Y) -> IFMINUS(le(s(X), Y), s(X), Y)
and get the following Size-Change Graph(s):
{1} , {1}
2>1
3=2
{2} , {2}
1=2
2=3

which lead(s) to this/these maximal multigraph(s):
{2} , {1}
1>1
2=2
{1} , {2}
2>2
3=3

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 Pair:

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

Rules:

le(0, Y) -> true
le(s(X), 0) -> false
le(s(X), s(Y)) -> le(X, Y)
minus(0, Y) -> 0
minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
quot(0, s(Y)) -> 0
quot(s(X), s(Y)) -> s(quot(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 3`
`         ↳UsableRules`
`           →DP Problem 6`
`             ↳Negative Polynomial Order`

Dependency Pair:

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

Rules:

minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
minus(0, Y) -> 0
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
le(s(X), 0) -> false
le(0, Y) -> true
le(s(X), s(Y)) -> le(X, Y)

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(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
minus(0, Y) -> 0
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
le(s(X), 0) -> false
le(0, Y) -> true
le(s(X), s(Y)) -> le(X, Y)

Used ordering:
Polynomial Order with Interpretation:

POL( QUOT(x1, x2) ) = x1

POL( s(x1) ) = x1 + 1

POL( minus(x1, x2) ) = x1

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

POL( 0 ) = 0

POL( le(x1, x2) ) = 0

POL( false ) = 0

POL( true ) = 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`
`             ↳Neg POLO`
`             ...`
`               →DP Problem 7`
`                 ↳Dependency Graph`

Dependency Pair:

Rules:

minus(s(X), Y) -> ifMinus(le(s(X), Y), s(X), Y)
minus(0, Y) -> 0
ifMinus(true, s(X), Y) -> 0
ifMinus(false, s(X), Y) -> s(minus(X, Y))
le(s(X), 0) -> false
le(0, Y) -> true
le(s(X), s(Y)) -> le(X, Y)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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