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

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)
MOD(s(x), s(y)) -> IFMOD(le(y, x), s(x), s(y))
MOD(s(x), s(y)) -> LE(y, x)
IFMOD(true, s(x), s(y)) -> MOD(minus(x, y), s(y))
IFMOD(true, s(x), s(y)) -> MINUS(x, y)

Furthermore, R contains three SCCs.

R
DPs
→DP Problem 1
Polynomial Ordering
→DP Problem 2
Polo
→DP Problem 3
Polo

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

There are no usable rules for innermost that need to be oriented.

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(LE(x1, x2)) =  x1 POL(s(x1)) =  1 + x1

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 4
Dependency Graph
→DP Problem 2
Polo
→DP Problem 3
Polo

Dependency Pair:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polynomial Ordering
→DP Problem 3
Polo

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

Strategy:

innermost

The following dependency pair can be strictly oriented:

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

Additionally, the following usable rules for innermost can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(IF_MINUS(x1, x2, x3)) =  x2 POL(0) =  0 POL(false) =  0 POL(MINUS(x1, x2)) =  x1 POL(true) =  0 POL(s(x1)) =  1 + x1 POL(le(x1, x2)) =  0

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polo
→DP Problem 5
Dependency Graph
→DP Problem 3
Polo

Dependency Pair:

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

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polo
→DP Problem 3
Polynomial Ordering

Dependency Pairs:

IFMOD(true, s(x), s(y)) -> MOD(minus(x, y), s(y))
MOD(s(x), s(y)) -> IFMOD(le(y, x), s(x), 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))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> ifmod(le(y, x), s(x), s(y))
ifmod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
ifmod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

The following dependency pair can be strictly oriented:

IFMOD(true, s(x), s(y)) -> MOD(minus(x, y), s(y))

Additionally, the following usable rules for innermost can be oriented:

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

Used ordering: Polynomial ordering with Polynomial interpretation:
 POL(0) =  0 POL(false) =  0 POL(MOD(x1, x2)) =  x1 POL(minus(x1, x2)) =  x1 POL(true) =  0 POL(s(x1)) =  1 + x1 POL(IF_MOD(x1, x2, x3)) =  x2 POL(if_minus(x1, x2, x3)) =  x2 POL(le(x1, x2)) =  0

resulting in one new DP problem.

R
DPs
→DP Problem 1
Polo
→DP Problem 2
Polo
→DP Problem 3
Polo
→DP Problem 6
Dependency Graph

Dependency Pair:

MOD(s(x), s(y)) -> IFMOD(le(y, x), s(x), 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))
mod(0, y) -> 0
mod(s(x), 0) -> 0
mod(s(x), s(y)) -> ifmod(le(y, x), s(x), s(y))
ifmod(true, s(x), s(y)) -> mod(minus(x, y), s(y))
ifmod(false, s(x), s(y)) -> s(x)

Strategy:

innermost

Using the Dependency Graph resulted in no new DP problems.

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